JEE Probability Questions

The quadratic is $$f(x)= ax^{2}+2\sqrt{2}\,bx+c$$ with $$a,b,c \in\{1,2,3,4\}$$.

For $$f(x) \gt 0$$ for all real $$x$$, two conditions must hold:
1. Leading coefficient $$a \gt 0$$ (already true because $$a \ge 1$$).
2. Discriminant $$\Delta$$ is negative.

Compute the discriminant:
$$\Delta =(2\sqrt{2}b)^{2}-4ac = 8b^{2}-4ac =4\bigl(2b^{2}-ac\bigr).$$
Hence $$\Delta \lt 0 \Longleftrightarrow 2b^{2}-ac \lt 0 \Longleftrightarrow 2b^{2} \lt ac.$$

Total number of ordered triples $$(a,b,c)$$ is $$4^{3}=64.$$ We now count the favourable triples that satisfy $$2b^{2} \lt ac$$.

Case a = 1:

Condition becomes $$2b^{2} \lt c.$$
• $$b=1$$: $$2\lt c\Rightarrow c=3,4\;(2\text{ choices})$$
• $$b=2,3,4$$: $$2b^{2}\ge 8 \gt 4$$, no solutions.
Favourable triples = 2.

Case a = 2:

Condition becomes $$b^{2} \lt c.$$
• $$b=1$$: $$1\lt c\Rightarrow c=2,3,4\;(3\text{ choices})$$
• $$b=2,3,4$$ give $$b^{2}\ge 4$$, but $$c\le 4$$; no solutions.
Favourable triples = 3.

Case a = 3:

Condition becomes $$2b^{2} \lt 3c.$$ Check each $$c$$:
• $$c=1$$: need $$2b^{2}\lt 3 \Rightarrow b=1\;(1\text{ choice})$$
• $$c=2$$: need $$2b^{2}\lt 6 \Rightarrow b=1\;(1)$$
• $$c=3$$: need $$2b^{2}\lt 9 \Rightarrow b=1,2\;(2)$$
• $$c=4$$: need $$2b^{2}\lt 12 \Rightarrow b=1,2\;(2)$$
Favourable triples = $$1+1+2+2 = 6.$$\

Case a = 4:

Condition becomes $$b^{2} \lt 2c.$$ Check each $$c$$:
• $$c=1$$: $$b^{2}\lt 2 \Rightarrow b=1\;(1)$$
• $$c=2$$: $$b^{2}\lt 4 \Rightarrow b=1\;(1)$$
• $$c=3$$: $$b^{2}\lt 6 \Rightarrow b=1,2\;(2)$$
• $$c=4$$: $$b^{2}\lt 8 \Rightarrow b=1,2\;(2)$$
Favourable triples = $$1+1+2+2 = 6.$$\

Total favourable triples = $$2+3+6+6 = 17.$$

Therefore, required probability $$= \dfrac{17}{64}.$$ Here $$\gcd(17,64)=1$$, so $$m=17,\; n=64.$$ Finally, $$m+n = 17+64 = 81.$$

Answer: 81

Let the individual events be
  $$A$$ : $$S_1$$ solves the problem,   $$B$$ : $$S_2$$ solves the problem,   $$C$$ : $$S_3$$ solves the problem.

Write the probabilities of all eight possible outcome-combinations as

$$\begin{aligned} P(A\,B'\,C') &= x_1, & P(A\,B\,C') &= x_2, & P(A\,B'\,C) &= x_3, & P(A\,B\,C) &= x_4,\\[2pt] P(A'\,B\,C') &= x_5, & P(A'\,B\,C) &= x_6, & P(A'\,B'\,C) &= x_7, & P(A'\,B'\,C') &= x_8. \end{aligned}$$

The eight numbers $$x_1,\dots ,x_8$$ are non-negative and satisfy

$$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8 = 1 \qquad -(1)$$

Step 1 Using $$P(U)=\dfrac12$$
Event $$U$$ is “at least one of $$S_1,S_2,S_3$$ solves”, i.e. $$U = A\cup B\cup C$$, whose complement is $$A'B'C'$$. Hence

$$P(U) = 1-P(A'B'C') = 1-x_8 = \frac12 \;\;\Longrightarrow\;\; x_8 = \frac12. \qquad -(2)$$

Step 2 Using $$P(V)=\dfrac1{10}$$
Event $$V$$ is “$$S_1$$ solves given that neither $$S_2$$ nor $$S_3$$ solves”, that is

$$P(V)=P\!\bigl(A\,|\,B'C'\bigr)=\frac{P(A\,B'\,C')}{P(B'\,C')}=\frac{x_1}{x_1+x_8} =\frac1{10}. $$

Substituting $$x_8=\dfrac12$$ from (2):

$$\frac{x_1}{x_1+\dfrac12}=\frac1{10} \;\;\Longrightarrow\;\;10x_1 = x_1+\frac12 \;\;\Longrightarrow\;\;9x_1 = \frac12 \;\;\Longrightarrow\;\; x_1 = \frac1{18}. \qquad -(3)$$

Step 3 Using $$P(W)=\dfrac1{12}$$
Event $$W$$ is “$$S_2$$ solves and $$S_3$$ does not”, i.e.

$$P(W)=P(B\,C') = P(A\,B\,C') + P(A'\,B\,C') = x_2 + x_5 = \frac1{12}. \qquad -(4)$$

Step 4 Total probability still unaccounted for
From (1), (2) and (3)

$$x_2+x_3+x_4+x_5+x_6+x_7 = 1 - x_8 - x_1 = 1 - \frac12 - \frac1{18} = \frac{4}{9}. \qquad -(5)$$

Step 5 Probability that $$S_3$$ solves
$$S_3$$ succeeds in the four cases $$A\,B'\,C,\;A\,B\,C,\;A'\,B\,C,\;A'\,B'\,C$$ whose probabilities are $$x_3,x_4,x_6,x_7$$ respectively. Adding them, and using (4) and (5):

$$\begin{aligned} P(T) &= x_3+x_4+x_6+x_7 \\[2pt] &= \bigl(x_2+x_3+x_4+x_5+x_6+x_7\bigr) - (x_2+x_5) \\[2pt] &= \frac{4}{9} - \frac{1}{12} \\[2pt] &= \frac{16}{36} - \frac{3}{36} \\[2pt] &= \frac{13}{36}. \end{aligned}$$

Therefore $$P(T)=\dfrac{13}{36}$$.

Option A which is: $$\dfrac{13}{36}$$

There are $$5! = 120$$ different five-letter words that can be formed by using each of the letters A, B, C, D, E exactly once. These words are arranged in ordinary English (lexicographic) order and numbered $$W_1, W_2,\,\dots ,W_{120}$$.

Case 1: Determining the serial number of $$CDBEA$$

Stepwise counting in lexicographic order:

• First letter
 A-words: $$4! = 24$$ words  (serial numbers $$1\!-\!24$$)
 B-words: $$4! = 24$$ words  (serial numbers $$25\!-\!48$$)
 So words beginning with C start at serial $$49$$.

• First two letters $$C\_x$$ (remaining letters A, B, D, E):
 CA-words: $$3! = 6$$ words  (serial $$49\!-\!54$$)
 CB-words: $$3! = 6$$ words  (serial $$55\!-\!60$$)
 CD-words therefore start at serial $$61$$.

• First three letters $$CD\_x$$ (remaining letters A, B, E):
 CDA-words: $$2! = 2$$ words  (serial $$61, 62$$)
 CDB-words therefore start at serial $$63$$.

• First four letters $$CDB\_x$$ (remaining letters A, E):
 CDBA-word: $$1$$ word  (serial $$63$$)
 CDBE-words therefore start at serial $$64$$.

The only CDBE-word is $$CDBEA$$ itself, so
$$n = 64.$$

Case 2: Writing the probability model

Let $$P(W_1)=p.$$ Given $$P(W_n)=2P(W_{n-1})$$ for $$n \gt 1$$, the probabilities form a geometric progression: $$P(W_n)=2^{\,n-1}p.$$

The total probability equals $$1$$: $$\sum_{n=1}^{120}P(W_n)=p\sum_{k=0}^{119}2^k =p\,(2^{120}-1)=1.$$ Hence $$p=\frac{1}{2^{120}-1}.$$

Case 3: Probability of the word $$CDBEA$$

The required probability is $$P(CDBEA)=P(W_{64})=2^{64-1}\,p =\frac{2^{63}}{2^{120}-1}.$$

Comparing with $$P(CDBEA)=\dfrac{2^{\alpha}}{2^{\beta}-1}$$ gives $$\alpha = 63,\quad \beta = 120.$$ Therefore, $$\alpha + \beta = 63 + 120 = 183.$$

Final answer: $$183.$$

Both inequalities restrict $$x, y$$ to small integers.
Ellipse: $$\dfrac{x^{2}}{8} + \dfrac{y^{2}}{20} \lt 1$$ gives $$y^{2} \lt 20$$, so $$y \in \{-4,-3,\dots,4\}$$.
Parabola: $$y^{2} \lt 5x$$ forces $$x \gt 0$$ and, because $$y^{2} \le 16$$, at most $$x = 1, 2$$ are possible (the ellipse allows $$|x| \lt \sqrt{8}\approx 2.83$$).

Checking every admissible $$y$$:

$$\begin{array}{c|c|c} y & \text{Ellipse bound on }|x| & \text{Parabola bound on }x \\ \hline \pm 4 & |x| \lt 1.264 & x \gt 3.2 & \text{No integer }x \\ \pm 3 & |x| \lt 2.098 & x \ge 2 & (2,\pm 3) \\ \pm 2 & |x| \lt 2.529 & x \ge 1 & (1,\pm 2),(2,\pm 2) \\ \pm 1 & |x| \lt 2.758 & x \ge 1 & (1,\pm 1),(2,\pm 1) \\ 0 & |x| \lt 2.828 & x \ge 1 & (1,0),(2,0) \end{array}$$

Thus the set $$X$$ contains exactly 12 lattice points:

$$(2,3),(2,-3);\\ (1,2),(2,2),(1,-2),(2,-2);\\ (1,1),(2,1),(1,-1),(2,-1);\\ (1,0),(2,0).$$

Total ways to choose three distinct points:
$$N_{\text{total}} = {}^{12}C_{3} = 220.$$ This matches the denominators given in the options.

Let us classify the points by the parity of their coordinates
$$(x \bmod 2,\; y \bmod 2):$$

EE (0,0): (2,2),(2,-2),(2,0) ⇒ 3 points
EO (0,1): (2,3),(2,-3),(2,1),(2,-1) ⇒ 4 points
OE (1,0): (1,2),(1,-2),(1,0) ⇒ 3 points
OO (1,1): (1,1),(1,-1) ⇒ 2 points

For any three lattice points the doubled area is
$$\Delta = x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) + x_{3}(y_{1}-y_{2}).$$ Since $$x_i,y_i$$ are integers, $$\Delta$$ is an integer and the (actual) area is $$\dfrac{|\Delta|}{2}$$.

Parity test:
If at least two vertices have identical parity class, the vector joining them has both components even; hence every term in $$\Delta$$ is even and $$\Delta$$ itself is even. If the three vertices lie in three different parity classes, one easily verifies (substituting any representatives) that $$\Delta$$ is odd. Therefore

Area is an integer ⇔ not all three parity classes are different.

Counting triples with three different parity classes:

$$\begin{aligned} \text{EE, EO, OE:}&\; 3 \times 4 \times 3 = 36\\ \text{EE, EO, OO:}&\; 3 \times 4 \times 2 = 24\\ \text{EE, OE, OO:}&\; 3 \times 3 \times 2 = 18\\ \text{EO, OE, OO:}&\; 4 \times 3 \times 2 = 24\\ \hline \text{Total odd } \Delta &= 36+24+18+24 = 102 \end{aligned}$$

Thus triples with even $$\Delta$$ (integer area or zero area): $$N_{\text{even}} = 220 - 102 = 118.$$ We must now discard the degenerate cases (area $$=0$$).

Because all points have $$x = 1$$ or $$x = 2$$, the only straight lines containing three or more of them are the vertical lines $$x = 1$$ and $$x = 2$$:

x = 1 has 5 points ⇒ $${}^{5}C_{3}=10$$ collinear triples.
x = 2 has 7 points ⇒ $${}^{7}C_{3}=35$$ collinear triples.

Total degenerate triples $$N_{\text{col}} = 10 + 35 = 45.$$

Favourable triples (non-collinear with integer area):
$$N_{\text{fav}} = N_{\text{even}} - N_{\text{col}} = 118 - 45 = 73.$$

Probability required:
$$P = \dfrac{N_{\text{fav}}}{N_{\text{total}}} = \dfrac{73}{220}.$$

Option B which is: $$\dfrac{73}{220}$$

The compound statement $$(\sim(P \wedge Q)) \vee ((\sim P) \wedge Q) \Rightarrow ((\sim P) \wedge (\sim Q))$$ is equivalent to what?

By De Morgan's law, $$\sim(P \wedge Q) = \sim P \vee \sim Q$$, so the antecedent becomes $$(\sim P \vee \sim Q) \vee (\sim P \wedge Q).$$

Let $$A = \sim P, B = \sim Q$$. Then the antecedent is $$(A \vee B) \vee (A \wedge \sim B).$$ If $$A = T$$, this expression is true regardless of $$B$$. If $$A = F$$, then $$(F \vee B) \vee (F \wedge \sim B) = B \vee F = B$$. Hence the antecedent simplifies to $$A \vee B = \sim P \vee \sim Q$$.

The full statement thus becomes $$(\sim P \vee \sim Q) \Rightarrow (\sim P \wedge \sim Q)$$. Using $$p \Rightarrow q \equiv \sim p \vee q$$, this is $$\sim(\sim P \vee \sim Q) \vee (\sim P \wedge \sim Q) = (P \wedge Q) \vee (\sim P \wedge \sim Q).$$

This last expression is the biconditional $$P \Leftrightarrow Q = (\sim P \vee Q) \wedge (\sim Q \vee P).$$

The correct answer is Option 1: $$((\sim P) \vee Q) \wedge ((\sim Q) \vee P)$$.

For the product of five rolled numbers to be strictly positive, none of the rolls can be $$0$$ and the number of negative values obtained must be even. The positive faces of the die are $$\{1,2,3\}$$ giving $$3$$ choices, while the negative faces are $$\{-2,-1\}$$ giving $$2$$ choices.

If there are no negative numbers, all five outcomes are positive. The number of such outcomes is

$$\binom{5}{0}(2)^0(3)^5 = 243$$

If there are exactly two negative numbers, the number of outcomes is

$$\binom{5}{2}(2)^2(3)^3 = 1080$$

If there are exactly four negative numbers, the number of outcomes is

$$\binom{5}{4}(2)^4(3)^1 = 240$$

Hence, the total number of favorable outcomes is

$$243+1080+240 = 1563$$

The total number of possible outcomes when the die is rolled five times is

$$6^5 = 7776$$

Therefore, the required probability is

$$\frac{1563}{7776} = \frac{521}{2592}$$

Hence, the required probability is

$$\boxed{\frac{521}{2592}}$$

Given: $$P(X = x) = k(x+1)3^{-x}$$ for $$x = 0, 1, 2, 3, \ldots$$

First, we find k using $$\sum_{x=0}^{\infty} P(X = x) = 1$$.

$$\sum_{x=0}^{\infty} k(x+1)3^{-x} = k \sum_{x=0}^{\infty}(x+1)\left(\frac{1}{3}\right)^x = k \sum_{n=1}^{\infty} n \left(\frac{1}{3}\right)^{n-1}$$

Using the formula $$\sum_{n=1}^{\infty} n r^{n-1} = \frac{1}{(1-r)^2}$$ for $$|r| < 1$$:

$$= k \cdot \frac{1}{(1-1/3)^2} = k \cdot \frac{1}{(2/3)^2} = k \cdot \frac{9}{4}$$

Setting this equal to 1: $$k = \frac{4}{9}$$

Next, we find $$P(X \geq 2) = 1 - P(X=0) - P(X=1)$$.

$$P(X = 0) = \frac{4}{9} \cdot 1 \cdot 1 = \frac{4}{9}$$

$$P(X = 1) = \frac{4}{9} \cdot 2 \cdot \frac{1}{3} = \frac{8}{27}$$

$$P(X \geq 2) = 1 - \frac{4}{9} - \frac{8}{27} = 1 - \frac{12}{27} - \frac{8}{27} = 1 - \frac{20}{27} = \frac{7}{27}$$

The correct answer is Option 1: $$\frac{7}{27}$$.

Let S = smoker, NS = non-smoker, C = lung cancer.

P(S) = 0.25, P(NS) = 0.75

P(C|S) = 27 × P(C|NS). Let P(C|NS) = p, so P(C|S) = 27p.

Using Bayes' theorem:

$$P(S|C) = \frac{P(C|S) \times P(S)}{P(C|S) \times P(S) + P(C|NS) \times P(NS)}$$

$$= \frac{27p \times 0.25}{27p \times 0.25 + p \times 0.75}$$

$$= \frac{6.75p}{6.75p + 0.75p} = \frac{6.75}{7.5} = \frac{27}{30} = \frac{9}{10}$$

So $$P(S|C) = \frac{9}{10} = \frac{k}{10}$$

Therefore $$k = 9$$.

A bag contains 6 balls of different colours.

Now, finding $$p$$ (probability both balls same colour):

Two balls drawn with replacement. The probability that both are the same colour:

$$ p = 6 \times \left(\frac{1}{6}\right)^2 = \frac{6}{36} = \frac{1}{6} $$

Now, finding $$q$$ (probability exactly 3 balls same colour out of 4):

Four balls drawn with replacement. We need exactly 3 of the same colour.

Choose which colour appears 3 times: 6 ways.

Choose which 3 positions out of 4: $$\binom{4}{3} = 4$$ ways.

The 3 chosen positions have probability $$\left(\frac{1}{6}\right)^3$$ each for that colour.

The remaining position must be a different colour: probability $$\frac{5}{6}$$.

$$ q = 6 \times \binom{4}{3} \times \left(\frac{1}{6}\right)^3 \times \frac{5}{6} = 6 \times 4 \times \frac{5}{6^4} = \frac{120}{1296} = \frac{5}{54} $$

Now, finding the ratio $$p : q$$:

$$ \frac{p}{q} = \frac{1/6}{5/54} = \frac{1}{6} \times \frac{54}{5} = \frac{54}{30} = \frac{9}{5} $$

So $$p : q = 9 : 5$$, meaning $$m = 9$$ and $$n = 5$$ (which are co-prime).

$$ m + n = 9 + 5 = 14 $$

A fair $$n$$-faced die ($$n > 1$$) is rolled repeatedly until a number less than $$n$$ appears. We need to find $$n$$ given that the mean number of tosses is $$\frac{n}{9}$$.

Set up the probability model.

On each roll, the probability of getting a number less than $$n$$ (i.e., rolling 1 through $$n-1$$) is:

$$p = \frac{n-1}{n}$$

The probability of rolling $$n$$ (not stopping) is:

$$q = 1 - p = \frac{1}{n}$$

Find the mean (expected value).

The number of tosses until the first success follows a geometric distribution. The mean of a geometric distribution with success probability $$p$$ is:

$$E[X] = \frac{1}{p} = \frac{1}{(n-1)/n} = \frac{n}{n-1}$$

Set up and solve the equation.

$$\frac{n}{n-1} = \frac{n}{9}$$

Since $$n > 1$$, we can divide both sides by $$n$$:

$$\frac{1}{n-1} = \frac{1}{9}$$

$$n - 1 = 9$$

$$n = 10$$

The value of $$n$$ is 10.

Two real numbers $$x$$ and $$y$$ are chosen randomly from $$[0, 60]$$, so the sample space is a square of side 60 whose area is $$60^2 = 3600$$, and the event $$A$$ defined by $$|x - y| \leq a$$ corresponds to the region between the lines $$y = x + a$$ and $$y = x - a$$ within this square.

Since the area outside this region consists of two right triangles each with legs of length $$(60 - a)$$, the probability of $$A$$ is the complement of the ratio of their total area to 3600, namely:

$$ P(A) = 1 - \frac{(60 - a)^2}{60^2} = \frac{3600 - (60 - a)^2}{3600} = \frac{120a - a^2}{3600}. $$

Imposing the condition $$P(A) = \tfrac{11}{36}$$ leads to:

$$ \frac{120a - a^2}{3600} = \frac{11}{36}, \quad 120a - a^2 = 1100, \quad a^2 - 120a + 1100 = 0. $$

By the quadratic formula one finds:

$$ a = \frac{120 \pm \sqrt{14400 - 4400}}{2} = \frac{120 \pm \sqrt{10000}}{2} = \frac{120 \pm 100}{2}, $$

so that $$a = 110$$ or $$a = 10$$, and since $$a$$ must lie within $$[0,60]$$, it follows that $$a = 10$$.

Let $$N$$ denote the number of tosses required to obtain the first head.

Since the probability of obtaining a head is $$\frac{1}{4}$$ and the probability of obtaining a tail is $$\frac{3}{4}$$, $$N$$ follows a geometric distribution with

$$P(N=n)=\left(\frac{3}{4}\right)^{n-1}\left(\frac{1}{4}\right)$$

The quadratic equation is

$$64x^2+5Nx+1=0$$

For the equation to have no real roots, its discriminant must be negative.

$$\Delta=(5N)^2-4(64)(1)<0$$

$$25N^2-256<0$$

$$25N^2<256$$

$$N^2<\frac{256}{25}$$

$$N^2<10.24$$

Since $$N$$ is a positive integer,

$$N\in{1,2,3}$$

Therefore, the required probability is

$$P(N\leq 3)=P(1)+P(2)+P(3)$$

$$P(1)=\left(\frac{3}{4}\right)^0\left(\frac{1}{4}\right)=\frac{1}{4}$$

$$P(2)=\left(\frac{3}{4}\right)^1\left(\frac{1}{4}\right)=\frac{3}{16}$$

$$P(3)=\left(\frac{3}{4}\right)^2\left(\frac{1}{4}\right)=\frac{9}{64}$$

Hence,

$$P(N\leq 3)=\frac{1}{4}+\frac{3}{16}+\frac{9}{64}$$

$$P(N\leq 3)=\frac{16}{64}+\frac{12}{64}+\frac{9}{64}$$

$$P(N\leq 3)=\frac{37}{64}$$

Thus,

$$\frac{p}{q}=\frac{37}{64}$$

giving

$$p=37,\quad q=64$$

Therefore,

$$q-p=64-37=27$$

Hence, the correct answer is

$$27$$

Let $$R$$ be the event of drawing a red ball. The urns $$A, B,$$ and $$C$$ contain:

Urn A: 4 red, 6 black (Total = 10)

Urn B: 5 red, 5 black (Total = 10)

Urn C: $$\lambda$$ red, 4 black (Total = $$\lambda + 4$$)

Since one urn is selected at random, the probability of selecting any urn is $$P(A) = P(B) = P(C) = \frac{1}{3}$$.

The conditional probabilities of drawing a red ball from each urn are:

$$P(R|A) = \frac{4}{10} = 0.4$$

$$P(R|B) = \frac{5}{10} = 0.5$$

$$P(R|C) = \frac{\lambda}{\lambda + 4}$$

Given that the probability of drawing from urn $$C$$ given the ball is red is $$P(C|R) = 0.4$$, we use Bayes' Theorem:

$$P(C|R) = \frac{P(C)P(R|C)}{P(A)P(R|A) + P(B)P(R|B) + P(C)P(R|C)}$$

$$0.4 = \frac{\frac{1}{3} \cdot \frac{\lambda}{\lambda+4}}{\frac{1}{3}(0.4 + 0.5 + \frac{\lambda}{\lambda+4})}$$

$$0.4(0.9 + \frac{\lambda}{\lambda+4}) = \frac{\lambda}{\lambda+4}$$

$$0.36 + \frac{0.4\lambda}{\lambda+4} = \frac{\lambda}{\lambda+4} \implies 0.36 = \frac{0.6\lambda}{\lambda+4}$$

$$0.36\lambda + 1.44 = 0.6\lambda \implies 0.24\lambda = 1.44 \implies \lambda = 6$$

The parabola is $$y^2 = 6x$$. Let an equilateral triangle with one vertex at $$(0,0)$$ have side length $$a$$. The other vertices are $$(x, y)$$ and $$(x, -y)$$. Since the triangle is equilateral, the height is $$\frac{\sqrt{3}}{2}a$$ and the base is $$a$$ (from $$y$$ to $$-y$$, so $$2y = a \implies y = \frac{a}{2}$$).

The vertex $$(x, y)$$ lies on the parabola: $$y^2 = 6x \implies \left(\frac{a}{2}\right)^2 = 6x \implies x = \frac{a^2}{24}$$

The height of the triangle is the $$x$$-coordinate, so $$\frac{\sqrt{3}}{2}a = x$$:

$$\frac{\sqrt{3}}{2}a = \frac{a^2}{24} \implies a = 12\sqrt{3}$$

$$a^2 = (12\sqrt{3})^2 = 144 \times 3 = 432$$

The given square is divided into $$6$$ equal strips in both directions, so the grid contains $$7$$ vertical and $$7$$ horizontal lines.

Hence the number of intersection points is $$7 \times 7 = 49$$. These are the points $$A_1, A_2, \ldots , A_{49}$$.

Two points are called friends when they are directly adjacent in the same row or in the same column, that is, when they share a unit-length edge of the grid.

For a randomly chosen point let the random variable $$X$$ count its number of friends. We have to find $$7E(X)$$.

Instead of evaluating each probability $$p_i$$ separately, we count the total number of “point-friend” incidences in the whole grid and then divide by the number of points.

An unordered adjacent pair contributes $$1$$ friend to each of the two points that form it; thus every such pair contributes $$2$$ to the total $$\sum_{k=1}^{49} X(A_k).$$

Counting adjacent pairs
Horizontal pairs: every one of the $$7$$ rows has $$6$$ adjacent pairs, giving $$7 \times 6 = 42.$$
Vertical pairs: every one of the $$7$$ columns has $$6$$ adjacent pairs, again $$7 \times 6 = 42.$$

Total adjacent pairs $$= 42 + 42 = 84.$$

Therefore the total number of friends counted over all points is
$$\sum_{k=1}^{49} X(A_k) = 2 \times 84 = 168.$$

The expectation is the average per point:
$$E(X) = \frac{168}{49} = \frac{24}{7}.$$

Finally,
$$7E(X) = 7 \times \frac{24}{7} = 24.$$

Thus the required value is $$24$$.

Given curves are

$$y^2=8x+4$$

and

$$x^2+y^2+4\sqrt3\,x-4=0$$

First rewrite them.

Parabola:

$$y^2=8\left(x+\frac12\right)$$

Hence,

$$x=\frac{y^2-4}{8}$$

Circle:

$$x^2+4\sqrt3\,x+y^2-4=0$$

Completing the square,

$$x^2+4\sqrt3\,x+12+y^2=16$$

$$(x+2\sqrt3)^2+y^2=16$$

So the circle has centre

$$(-2\sqrt3,0)$$

and radius

$$4$$

Now find points of intersection.

Substitute

$$x=\frac{y^2-4}{8}$$

in the circle:

$$\left(\frac{y^2-4}{8}\right)^2+y^2+4\sqrt3\left(\frac{y^2-4}{8}\right)-4=0$$

Solving gives

$$y=\pm2$$

and corresponding

$$x=0$$

Hence, intersection points are

$$(0,2)\quad \text{and}\quad (0,-2)$$

Now the smaller enclosed region lies between:

- parabola on the right

- left arc of the circle on the left

For the circle,

$$x=-2\sqrt3+\sqrt{16-y^2}$$

Hence required area is

$$A=\int_{-2}^{2}\left[\frac{y^2-4}{8}-\left(-2\sqrt3+\sqrt{16-y^2}\right)\right]dy$$

Using symmetry,

$$A=2\int_0^2\left(\frac{y^2-4}{8}+2\sqrt3-\sqrt{16-y^2}\right)dy$$

Now,

$$\int_0^2\frac{y^2-4}{8}\,dy =\frac18\left[\frac{y^3}{3}-4y\right]_0^2 =-\frac23$$

Also,

$$\int_0^22\sqrt3\,dy=4\sqrt3$$

Now evaluate

$$\int_0^2\sqrt{16-y^2}\,dy$$

Using standard formula,

$$\int\sqrt{a^2-y^2}\,dy =\frac y2\sqrt{a^2-y^2}+\frac{a^2}{2}\sin^{-1}\frac ya$$

with

$$a=4,$$

$$\int_0^2\sqrt{16-y^2}\,dy =\left[\frac y2\sqrt{16-y^2}+8\sin^{-1}\frac y4\right]_0^2$$

$$=\sqrt{12}+8\cdot\frac{\pi}{6}$$

$$=2\sqrt3+\frac{4\pi}{3}$$

Therefore,

$$A=2\left(-\frac23+4\sqrt3-2\sqrt3-\frac{4\pi}{3}\right)$$

$$=2\left(2\sqrt3-\frac23-\frac{4\pi}{3}\right)$$

$$=4\sqrt3-\frac43-\frac{8\pi}{3}$$

Rearranging,

$$A=\frac{12\sqrt3-4-8\pi}{3}$$

Since area is positive,

$$A=\frac{8\pi+4-12\sqrt3}{3}$$

Hence, the required area is

$$\boxed{\frac13\left(4-12\sqrt3+8\pi\right)}$$.

Bag A has 2 white, 1 black, 3 red balls (6 total). Bag B has 3 black, 2 red, $$n$$ white balls ($$(5+n)$$ total).

Compute the probability of drawing 1 red and 1 black from each bag.

From Bag A: $$P(1R, 1B | A) = \frac{\binom{3}{1}\binom{1}{1}}{\binom{6}{2}} = \frac{3}{15} = \frac{1}{5}$$

From Bag B: $$P(1R, 1B | B) = \frac{\binom{2}{1}\binom{3}{1}}{\binom{5+n}{2}} = \frac{6}{\frac{(5+n)(4+n)}{2}} = \frac{12}{(5+n)(4+n)}$$

Apply Bayes' theorem.

$$P(A | 1R,1B) = \frac{\frac{1}{2} \cdot \frac{1}{5}}{\frac{1}{2} \cdot \frac{1}{5} + \frac{1}{2} \cdot \frac{12}{(5+n)(4+n)}} = \frac{6}{11}$$

Simplify and solve.

$$\frac{\frac{1}{5}}{\frac{1}{5} + \frac{12}{(5+n)(4+n)}} = \frac{6}{11}$$

Cross-multiplying: $$\frac{11}{5} = 6\left(\frac{1}{5} + \frac{12}{(5+n)(4+n)}\right)$$

$$\frac{11}{5} = \frac{6}{5} + \frac{72}{(5+n)(4+n)}$$

$$1 = \frac{72}{(5+n)(4+n)}$$

$$(5+n)(4+n) = 72$$

$$n^2 + 9n + 20 = 72$$

$$n^2 + 9n - 52 = 0$$

$$(n + 13)(n - 4) = 0$$

Since $$n > 0$$, we get $$n = 4$$.

Answer: Option C (n = 4)

A random variable $$X$$ follows $$B(33, p)$$ with $$3P(X=0) = P(X=1)$$.

Find p.

$$P(X=0) = q^{33}$$, $$P(X=1) = 33pq^{32}$$.

$$3q^{33} = 33pq^{32} \implies 3q = 33p \implies q = 11p$$

Since $$p + q = 1$$: $$p + 11p = 1 \implies p = \frac{1}{12},\, q = \frac{11}{12}$$

Compute the ratio $$\frac{P(X=r)}{P(X=r+1)}$$.

$$\frac{P(X=r)}{P(X=r+1)} = \frac{\binom{33}{r}}{\binom{33}{r+1}} \cdot \frac{q}{p} = \frac{r+1}{33-r} \cdot 11$$

Compute $$\frac{P(X=15)}{P(X=18)}$$.

$$\frac{P(X=15)}{P(X=18)} = \frac{P(X=15)}{P(X=16)} \cdot \frac{P(X=16)}{P(X=17)} \cdot \frac{P(X=17)}{P(X=18)}$$

$$= \frac{16 \cdot 11}{18} \cdot \frac{17 \cdot 11}{17} \cdot \frac{18 \cdot 11}{16}$$

$$= \frac{88}{9} \cdot 11 \cdot \frac{99}{8}$$

$$= \frac{88 \times 11 \times 99}{9 \times 8} = \frac{88 \times 11 \times 11}{8} = \frac{10648}{8} = 1331$$

Compute $$\frac{P(X=16)}{P(X=17)}$$.

$$\frac{P(X=16)}{P(X=17)} = \frac{17 \times 11}{17} = 11$$

Final answer.

$$\frac{P(X=15)}{P(X=18)} - \frac{P(X=16)}{P(X=17)} = 1331 - 11 = 1320$$

Answer: Option A (1320)

We are given $$P(B|A) = \frac{2}{5}$$, $$P(A|B) = \frac{1}{7}$$, and $$P(A \cap B) = \frac{1}{9}$$.

From $$P(B|A) = \frac{P(A \cap B)}{P(A)}$$, we get $$P(A) = \frac{P(A \cap B)}{P(B|A)} = \frac{1/9}{2/5} = \frac{5}{18}$$.

From $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$, we get $$P(B) = \frac{P(A \cap B)}{P(A|B)} = \frac{1/9}{1/7} = \frac{7}{9}$$.

Now we check statement $$(S_1)$$: $$P(A' \cup B) = \frac{5}{6}$$.

Using $$P(A' \cup B) = 1 - P(A \cap B') = 1 - [P(A) - P(A \cap B)]$$:

$$P(A' \cup B) = 1 - \left(\frac{5}{18} - \frac{1}{9}\right) = 1 - \left(\frac{5}{18} - \frac{2}{18}\right) = 1 - \frac{3}{18} = 1 - \frac{1}{6} = \frac{5}{6}$$

So $$(S_1)$$ is true.

Now we check statement $$(S_2)$$: $$P(A' \cap B') = \frac{1}{18}$$.

Using De Morgan's law: $$P(A' \cap B') = 1 - P(A \cup B) = 1 - [P(A) + P(B) - P(A \cap B)]$$:

$$P(A' \cap B') = 1 - \left(\frac{5}{18} + \frac{7}{9} - \frac{1}{9}\right) = 1 - \left(\frac{5}{18} + \frac{6}{9}\right) = 1 - \left(\frac{5}{18} + \frac{12}{18}\right) = 1 - \frac{17}{18} = \frac{1}{18}$$

So $$(S_2)$$ is also true.

Hence, both $$(S_1)$$ and $$(S_2)$$ are true, and the correct answer is Option A.

$$X \sim B(7, p)$$ and $$P(X = 3) = 5 \cdot P(X = 4)$$. Find mean + variance.

First, we write the binomial probabilities: $$P(X = 3) = \binom{7}{3} p^3 (1-p)^4 = 35 p^3 (1-p)^4$$ and $$P(X = 4) = \binom{7}{4} p^4 (1-p)^3 = 35 p^4 (1-p)^3$$.

Since $$P(X = 3) = 5 \times P(X = 4)$$, we have $$35 p^3 (1-p)^4 = 5 \times 35 p^4 (1-p)^3$$. Dividing both sides by $$35 p^3 (1-p)^3$$ gives $$1 - p = 5p$$, which leads to $$1 = 6p$$ and hence $$p = \frac{1}{6}$$.

Now we calculate the mean and variance. Using $$np$$ for the mean yields $$7 \times \frac{1}{6} = \frac{7}{6}$$, and using $$np(1-p)$$ for the variance gives $$7 \times \frac{1}{6} \times \frac{5}{6} = \frac{35}{36}$$.

Finally, the sum of the mean and variance is $$\frac{7}{6} + \frac{35}{36} = \frac{42}{36} + \frac{35}{36} = \frac{77}{36}$$.

The answer is Option B: $$\frac{77}{36}$$.

Let the total number of candidates be 100.

Since there are 60 female candidates and 40 male candidates, the total number of qualified candidates is 60% of 100, which equals 60.

Let the number of males qualifying be $$m$$; then the number of females qualifying is $$2m$$. Substituting into the total gives $$2m + m = 60$$, so $$m = 20$$. This gives us females qualifying = 40 and males qualifying = 20.

From the above, the probability that a randomly chosen qualified candidate is female is $$\frac{40}{60} = \frac{2}{3}$$. Therefore, the answer is Option A.

We have a bag with 4 white and 6 black balls. Three balls are drawn at random, and $$X$$ is the number of white balls drawn. We need $$100\sigma^2$$ where $$\sigma^2$$ is the variance of $$X$$.

$$X$$ follows a hypergeometric distribution. The total number of ways to draw 3 balls from 10 is $$\binom{10}{3} = 120$$.

$$P(X = 0) = \frac{\binom{4}{0}\binom{6}{3}}{\binom{10}{3}} = \frac{1 \cdot 20}{120} = \frac{20}{120} = \frac{1}{6}$$

$$P(X = 1) = \frac{\binom{4}{1}\binom{6}{2}}{\binom{10}{3}} = \frac{4 \cdot 15}{120} = \frac{60}{120} = \frac{1}{2}$$

$$P(X = 2) = \frac{\binom{4}{2}\binom{6}{1}}{\binom{10}{3}} = \frac{6 \cdot 6}{120} = \frac{36}{120} = \frac{3}{10}$$

$$P(X = 3) = \frac{\binom{4}{3}\binom{6}{0}}{\binom{10}{3}} = \frac{4 \cdot 1}{120} = \frac{4}{120} = \frac{1}{30}$$

Now we compute $$E(X)$$:

$$E(X) = 0 \cdot \frac{1}{6} + 1 \cdot \frac{1}{2} + 2 \cdot \frac{3}{10} + 3 \cdot \frac{1}{30}$$

$$= 0 + \frac{1}{2} + \frac{3}{5} + \frac{1}{10} = \frac{5 + 6 + 1}{10} = \frac{12}{10} = \frac{6}{5}$$

Now $$E(X^2)$$:

$$E(X^2) = 0 + 1 \cdot \frac{1}{2} + 4 \cdot \frac{3}{10} + 9 \cdot \frac{1}{30}$$

$$= \frac{1}{2} + \frac{12}{10} + \frac{9}{30} = \frac{1}{2} + \frac{6}{5} + \frac{3}{10} = \frac{5 + 12 + 3}{10} = \frac{20}{10} = 2$$

Therefore: $$\sigma^2 = E(X^2) - [E(X)]^2 = 2 - \left(\frac{6}{5}\right)^2 = 2 - \frac{36}{25} = \frac{50 - 36}{25} = \frac{14}{25}$$

So $$100\sigma^2 = 100 \cdot \frac{14}{25} = \frac{1400}{25} = 56$$.

Hence, the correct answer is 56.

We need to find the probability that a randomly chosen 6-digit number formed using only digits 1 and 8 is a multiple of 21, and then compute $$96p$$.

First, each of the 6 positions can be either 1 or 8, so the total number of such 6-digit numbers is $$2^6 = 64$$.

Next, a 6-digit number with digits $$d_1d_2d_3d_4d_5d_6$$ where each $$d_i \in \{1,8\}$$ can be written as

$$N = \sum_{i=1}^{6} d_i \cdot 10^{6-i}\;.$$

Now we set $$d_i = 1 + 7b_i$$ where $$b_i \in \{0,1\}$$ to obtain

$$N = \sum_{i=1}^{6}(1 + 7b_i)\cdot 10^{6-i} = 111111 + 7\sum_{i=1}^{6} b_i \cdot 10^{6-i}\;.$$

Now we check divisibility by 21, noting that $$21 = 3 \times 7$$. For divisibility by 3, the digit sum must be divisible by 3. If $$k$$ of the digits are 8 and the remaining $$(6-k)$$ are 1, then the digit sum is

$$8k + (6-k) = 6 + 7k\;.$$

Since

$$6 + 7k \equiv 0 \pmod{3} \iff 7k \equiv 0 \pmod{3} \iff k \equiv 0 \pmod{3}\;,$$

we conclude that $$k \in \{0,3,6\}$$.

For divisibility by 7, observe that

$$N = 111111 + 7M\quad\text{where}\quad M = \sum_{i=1}^6 b_i\cdot 10^{6-i}\;,$$

and since $$111111 = 7 \times 15873\,$$ it follows that $$N$$ is always divisible by 7.

Therefore, the favorable outcomes occur when $$k \in \{0,3,6\}$$, giving

$$\binom{6}{0} + \binom{6}{3} + \binom{6}{6} = 1 + 20 + 1 = 22\;.$$

Finally, the required probability is

$$p = \frac{22}{64} = \frac{11}{32}\;,$$

and hence

$$96p = 96 \times \frac{11}{32} = 3 \times 11 = 33\;.$$

Hence the answer is $$33$$.

The line $$L$$ is the intersection of the planes

$$lx-y+3(1-l)z-1=0$$ and $$x+2y-z-2=0$$

Any plane passing through $$L$$ is

$$\left(lx-y+3(1-l)z-1\right)+\lambda(x+2y-z-2)=0$$

$$=(l+\lambda)x+(2\lambda-1)y+(3-3l-\lambda)z-(1+2\lambda)=0$$

Given plane is

$$3x-8y+7z-4=0$$

Comparing coefficients,

$$\frac{l+\lambda}{3}=\frac{2\lambda-1}{-8}=\frac{3-3l-\lambda}{7}=\frac{1+2\lambda}{4}$$

Using

$$\frac{2\lambda-1}{-8}=\frac{1+2\lambda}{4},$$

$$4(2\lambda-1)=-8(1+2\lambda)$$

$$8\lambda-4=-8-16\lambda$$

$$24\lambda=-4$$

$$\lambda=-\frac16$$

Now,

$$\frac{l-\frac16}{3}=\frac{1+2\left(-\frac16\right)}{4}$$

$$\frac{l-\frac16}{3}=\frac{2/3}{4}=\frac16$$

$$l-\frac16=\frac12$$

$$l=\frac23$$

Hence, normals of the two planes are

$$\vec n_1=(2,-3,3)$$

and

$$\vec n_2=(1,2,-1)$$

Therefore, direction vector of the line is

$$\vec d=\vec n_1\times\vec n_2$$

$$= \begin{vmatrix} \hat i&\hat j&\hat k\\ 2&-3&3\\ 1&2&-1 \end{vmatrix}$$

$$=\hat i(3-6)-\hat j(-2-3)+\hat k(4+3)$$

$$=-3\hat i+5\hat j+7\hat k$$

Thus,

$$\vec d=(-3,5,7)$$

Let $$\theta$$ be the acute angle between the line and the $$y$$-axis.

Then,

$$\cos\theta=\frac{|5|}{\sqrt{(-3)^2+5^2+7^2}}$$

$$=\frac5{\sqrt{9+25+49}}$$

$$=\frac5{\sqrt{83}}$$

Therefore,

$$\cos^2\theta=\frac{25}{83}$$

Hence,

$$415\cos^2\theta=415\cdot\frac{25}{83}$$

$$=5\cdot25$$

$$=125$$

Therefore, the required value is

$$\boxed{125}$$.

For a binomial distribution with parameters $$n$$ (trials) and $$p$$ (probability of success), the mean is $$\mu = np$$ and the variance is $$\sigma^2 = npq$$ where $$q = 1 - p$$.

We are given that $$\mu + \sigma^2 = 82.5$$ and $$\mu \cdot \sigma^2 = 1350$$. That is, $$np + npq = 82.5$$ and $$np \cdot npq = 1350$$.

From the first equation: $$np(1 + q) = 82.5$$, i.e., $$np(2 - p) = 82.5$$. From the second: $$(np)^2 q = 1350$$, i.e., $$(np)^2(1-p) = 1350$$.

Let $$m = np$$. Then $$m(2-p) = 82.5$$ and $$m^2(1-p) = 1350$$. From the first: $$p = 2 - \frac{82.5}{m}$$, so $$1 - p = \frac{82.5}{m} - 1 = \frac{82.5 - m}{m}$$.

Substituting into the second: $$m^2 \cdot \frac{82.5 - m}{m} = 1350$$, giving $$m(82.5 - m) = 1350$$, so $$82.5m - m^2 = 1350$$, i.e., $$m^2 - 82.5m + 1350 = 0$$.

Multiplying by 2: $$2m^2 - 165m + 2700 = 0$$. Using the quadratic formula: $$m = \frac{165 \pm \sqrt{165^2 - 4 \cdot 2 \cdot 2700}}{4} = \frac{165 \pm \sqrt{27225 - 21600}}{4} = \frac{165 \pm \sqrt{5625}}{4} = \frac{165 \pm 75}{4}$$.

So $$m = \frac{240}{4} = 60$$ or $$m = \frac{90}{4} = 22.5$$.

Case 1: $$m = np = 60$$. Then $$p = 2 - \frac{82.5}{60} = 2 - 1.375 = 0.625 = \frac{5}{8}$$. So $$n = \frac{60}{5/8} = 96$$.

Case 2: $$m = np = 22.5$$. Then $$p = 2 - \frac{82.5}{22.5} = 2 - \frac{11}{3} = -\frac{5}{3}$$, which is negative and invalid.

Hence $$n = 96$$.

Hence, the correct answer is $$\boxed{96}$$.

For the quadratic $$x^2 + 2(a+4)x - 5a + 64 > 0$$ to hold for all $$x \in \mathbb{R}$$, the discriminant must be strictly negative (since the leading coefficient is positive):

$$\Delta = [2(a+4)]^2 - 4 \cdot 1 \cdot (-5a + 64) < 0$$

$$4(a+4)^2 + 4(5a - 64) < 0$$

$$(a+4)^2 + 5a - 64 < 0$$

$$a^2 + 8a + 16 + 5a - 64 < 0$$

$$a^2 + 13a - 48 < 0$$

Factoring: $$(a + 16)(a - 3) < 0$$

This inequality holds for $$-16 < a < 3$$.

Now, $$a$$ is an integer selected from $$[-5, 30]$$. The integers in this range are: $$-5, -4, -3, \ldots, 30$$, giving a total of $$30 - (-5) + 1 = 36$$ integers.

The integers satisfying $$-16 < a < 3$$ within $$[-5, 30]$$ are: $$-5, -4, -3, -2, -1, 0, 1, 2$$, which is 8 integers (from $$-5$$ to $$2$$ inclusive).

The required probability is $$\frac{8}{36} = \frac{2}{9}$$.

This corresponds to Option 2.

Let us denote by $$H$$ the set of patients having a heart ailment and by $$L$$ the set of patients having a lung infection.

We are told that $$P(H)=89\%$$ and $$P(L)=98\%$$, where $$P(\cdot)$$ represents the percentage of the total number of patients.

For two sets, the Inclusion-Exclusion Principle states first:

$$P(H\cup L)=P(H)+P(L)-P(H\cap L).$$

Secondly, because a percentage cannot exceed the whole, we have the bound:

$$P(H\cup L)\le 100\%.$$

Combining these two facts, we write

$$P(H)+P(L)-P(H\cap L)\le 100.$$

Substituting the given values,

$$89+98-K\le 100,$$

where $$K$$ is exactly $$P(H\cap L)$$, the percentage suffering from both ailments.

Now we simplify step by step:

$$187-K\le 100,$$

so

$$-K\le 100-187,$$

hence

$$-K\le -87,$$

and multiplying both sides by $$-1$$ (remembering to reverse the inequality sign) we obtain

$$K\ge 87.$$

Next, we recall another elementary bound: the intersection of two sets can never exceed either of the individual sets. Symbolically,

$$P(H\cap L)\le \min\{P(H),P(L)\}.$$

Because $$P(H)=89\%$$ and $$P(L)=98\%,$$ the smaller of the two is $$89\%,$$ so

$$K\le 89.$$

Putting the two inequalities together, we get the precise interval

$$87\le K\le 89.$$

Thus the only possible integral values of $$K$$ are $$87, 88, 89.$$ Any value outside this closed interval is impossible.

Now we inspect each given option:

Option A: $$\{79,81,83,85\}$$ - none of these values lie in $$[87,89].$$

Option B: $$\{84,87,90,93\}$$ - the element $$87$$ does lie in $$[87,89].$$

Option C: $$\{80,83,86,89\}$$ - the element $$89$$ does lie in $$[87,89].$$

Option D: $$\{84,86,88,90\}$$ - the element $$88$$ does lie in $$[87,89].$$

Therefore, the only set in which $$K$$ cannot possibly fall is Option A.

Hence, the correct answer is Option A.

We are given 400 people divided into three groups: 160 smokers and non-vegetarian (group I), 100 smokers and vegetarian (group II), and 140 non-smokers and vegetarian (group III). Their probabilities of getting a chest disorder are 35%, 20%, and 10% respectively.

Using Bayes' theorem, we need $$P(\text{group I} \mid \text{disorder})$$. The total probability of a randomly chosen person having the disorder is $$P(D) = \frac{160}{400} \times 0.35 + \frac{100}{400} \times 0.20 + \frac{140}{400} \times 0.10$$.

Computing each term: $$\frac{160}{400} \times 0.35 = 0.4 \times 0.35 = 0.14$$, $$\frac{100}{400} \times 0.20 = 0.25 \times 0.20 = 0.05$$, and $$\frac{140}{400} \times 0.10 = 0.35 \times 0.10 = 0.035$$.

So $$P(D) = 0.14 + 0.05 + 0.035 = 0.225$$.

By Bayes' theorem, $$P(\text{group I} \mid D) = \frac{0.14}{0.225} = \frac{140}{225} = \frac{28}{45}$$.

Therefore, the probability that the selected person is a smoker and non-vegetarian is $$\dfrac{28}{45}$$.

The coefficients $$a, b, c$$ are each obtained by throwing a dice, so each takes values from $$\{1, 2, 3, 4, 5, 6\}$$. The total number of outcomes is $$6^3 = 216$$.

For the quadratic $$ax^2 + bx + c = 0$$ to have equal roots, the discriminant must be zero, i.e., $$b^2 - 4ac = 0$$, which gives $$b^2 = 4ac$$.

We now enumerate all valid $$(a, b, c)$$ with $$a, b, c \in \{1, 2, 3, 4, 5, 6\}$$ satisfying $$b^2 = 4ac$$.

If $$b = 2$$, then $$4 = 4ac$$, so $$ac = 1$$. The only possibility is $$(a, c) = (1, 1)$$. That gives 1 case.

If $$b = 4$$, then $$16 = 4ac$$, so $$ac = 4$$. The pairs $$(a, c)$$ with values in $$\{1, 2, 3, 4, 5, 6\}$$ are: $$(1, 4), (2, 2), (4, 1)$$. That gives 3 cases.

If $$b = 6$$, then $$36 = 4ac$$, so $$ac = 9$$. The only valid pair is $$(3, 3)$$, since $$(1, 9)$$ or $$(9, 1)$$ are out of range. That gives 1 case.

For odd values of $$b$$ (i.e., $$b = 1, 3, 5$$), $$b^2$$ is odd, but $$4ac$$ is always even, so no solutions exist.

The total number of favourable outcomes is $$1 + 3 + 1 = 5$$.

Therefore, the required probability is $$\frac{5}{216}$$.

We have a fair die, so the probability of getting a six on any single toss is $$p=\dfrac16$$ and the probability of not getting a six is $$q=\dfrac56$$.

The random variable $$X$$ counts the number of tosses needed until the first six appears. Such a random variable follows the geometric distribution with parameter $$p=\dfrac16$$. For a geometric distribution we know the formula

$$P(X=k)=q^{\,k-1}\,p \quad (k=1,2,3,\ldots).$$

Another useful fact is the “tail” probability. To stay tossing for at least $$n$$ throws, the first $$n-1$$ throws must all be failures (i.e. not six), so

$$P(X\ge n)=q^{\,n-1}.$$

The question asks for the conditional probability

$$P(X\ge5\;|\;X\gt 2).$$

By the definition of conditional probability,

$$P(X\ge5\;|\;X\gt 2)=\dfrac{P(X\ge5\ \text{and}\ X\gt 2)}{P(X\gt 2)}.$$

Notice that the event $$X\ge5$$ automatically implies $$X\gt 2$$, so the intersection “$$X\ge5$$ and $$X\gt 2$$” is simply $$X\ge5$$ itself. Hence

$$P(X\ge5\;|\;X\gt 2)=\dfrac{P(X\ge5)}{P(X\ge3)}.$$

Now we evaluate both probabilities using the tail formula stated above.

First,

$$P(X\ge5)=q^{\,5-1}=q^{\,4}=\left(\dfrac56\right)^4=\dfrac{5^4}{6^4}=\dfrac{625}{1296}.$$

Next,

$$P(X\ge3)=q^{\,3-1}=q^{\,2}=\left(\dfrac56\right)^2=\dfrac{5^2}{6^2}=\dfrac{25}{36}.$$

Substituting these values into the conditional probability expression, we get

$$P(X\ge5\;|\;X\gt 2)=\dfrac{\dfrac{625}{1296}}{\dfrac{25}{36}} =\dfrac{625}{1296}\times\dfrac{36}{25}.$$

We now perform the algebraic simplification step by step. First cancel the common factor $$36$$ between numerator and denominator:

$$\dfrac{625}{1296}\times\dfrac{36}{25} =\dfrac{625}{36\cdot36}\times\dfrac{36}{25} =\dfrac{625}{36}\times\dfrac1{25}.$$

Multiplying the numerators and denominators gives

$$\dfrac{625}{36\times25}=\dfrac{625}{900}.$$

Finally, dividing both numerator and denominator by $$25$$ yields

$$\dfrac{625/25}{900/25}=\dfrac{25}{36}.$$

So the required conditional probability is

$$P(X\ge5\;|\;X\gt 2)=\dfrac{25}{36}.$$

Hence, the correct answer is Option A.

A standard pack has 52 cards with 13 spades. One card is missing. We draw two cards and both are spades. We need to find the probability that the missing card is not a spade.

Let $$S$$ be the event that the missing card is a spade and $$S'$$ that it is not. Let $$E$$ be the event that two drawn cards are both spades.

Using Bayes' theorem: $$P(S' | E) = \frac{P(E | S') \cdot P(S')}{P(E | S') \cdot P(S') + P(E | S) \cdot P(S)}$$.

We have $$P(S) = \frac{13}{52} = \frac{1}{4}$$ and $$P(S') = \frac{39}{52} = \frac{3}{4}$$.

If the missing card is a spade, the remaining deck has 51 cards with 12 spades: $$P(E | S) = \frac{\binom{12}{2}}{\binom{51}{2}} = \frac{66}{1275}$$.

If the missing card is not a spade, the remaining deck has 51 cards with 13 spades: $$P(E | S') = \frac{\binom{13}{2}}{\binom{51}{2}} = \frac{78}{1275}$$.

Substituting: $$P(S' | E) = \frac{\frac{78}{1275} \cdot \frac{3}{4}}{\frac{78}{1275} \cdot \frac{3}{4} + \frac{66}{1275} \cdot \frac{1}{4}} = \frac{78 \times 3}{78 \times 3 + 66 \times 1} = \frac{234}{234 + 66} = \frac{234}{300} = \frac{39}{50}$$.

The required probability is $$\frac{39}{50}$$, which corresponds to option (3).

The digits available are $$3, 3, 4, 4, 4, 5, 5$$. The total number of distinct seven-digit arrangements is $$\frac{7!}{2! \cdot 3! \cdot 2!} = \frac{5040}{2 \cdot 6 \cdot 2} = 210$$.

For the number to be divisible by 2, the last digit must be even. The only even digit available is 4.

Fixing one 4 in the last position, the remaining six digits are $$3, 3, 4, 4, 5, 5$$. The number of distinct arrangements of these six digits is $$\frac{6!}{2! \cdot 2! \cdot 2!} = \frac{720}{8} = 90$$.

Therefore, the probability that the number is divisible by 2 is $$\frac{90}{210} = \frac{3}{7}$$.

We are told that the student answers 8 true-false questions purely by guessing. For every single question the probability of being correct is $$\tfrac12$$ and the probability of being wrong is also $$\tfrac12$$. Therefore, if we let the random variable $$X$$ denote “number of correct answers”, then $$X$$ follows a binomial distribution with parameters $$n=8$$ and $$p=\tfrac12$$, which we write as $$X\sim\text{Binomial}(8,\tfrac12)$$.

The probability mass function of a binomial random variable is given by the formula

$$P(X=k)=\binom{8}{k}\,p^{\,k}(1-p)^{8-k}.$$

Because $$p=\tfrac12$$, the factor $$p^{\,k}(1-p)^{8-k}=(\tfrac12)^8=\tfrac1{256}$$ is the same for every value of $$k$$. Hence

$$P(X=k)=\frac{\binom{8}{k}}{256}.$$

We need the smallest integer $$n$$ such that the probability of getting at least $$n$$ correct answers is less than $$\tfrac12$$. Symbolically we want

$$P(X\ge n)<\frac12,$$

and $$n$$ must be the least integer for which this inequality holds.

We start testing from the middle of the distribution because a binomial with $$p=\tfrac12$$ is symmetric around its mean $$\mu=8\!\times\!\tfrac12=4$$. First record the individual probabilities for $$k=0$$ through $$k=8$$:

$$\begin{aligned} P(X=0)&=\frac{\binom80}{256}=\frac1{256},\\ P(X=1)&=\frac{\binom81}{256}=\frac8{256},\\ P(X=2)&=\frac{\binom82}{256}=\frac{28}{256},\\ P(X=3)&=\frac{\binom83}{256}=\frac{56}{256},\\ P(X=4)&=\frac{\binom84}{256}=\frac{70}{256},\\ P(X=5)&=\frac{\binom85}{256}=\frac{56}{256},\\ P(X=6)&=\frac{\binom86}{256}=\frac{28}{256},\\ P(X=7)&=\frac{\binom87}{256}=\frac8{256},\\ P(X=8)&=\frac{\binom88}{256}=\frac1{256}. \end{aligned}$$

Now accumulate the probabilities from the top down until the total becomes less than $$\tfrac12$$.

Case $$n=4$$:

$$P(X\ge4)=P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)$$

$$=\frac{70+56+28+8+1}{256}=\frac{163}{256}\approx0.6367>\frac12.$$

The condition is not satisfied for $$n=4$$.

Case $$n=5$$:

$$P(X\ge5)=P(X=5)+P(X=6)+P(X=7)+P(X=8)$$

$$=\frac{56+28+8+1}{256}=\frac{93}{256}\approx0.3633<\frac12.$$

The inequality $$P(X\ge5)<\tfrac12$$ is now satisfied, so $$n=5$$ works.

To be sure that 5 is indeed the smallest such integer, we already saw that $$n=4$$ fails; therefore $$n=5$$ is minimal.

Hence, the correct answer is Option A.

We are told that a dice is rolled $$n$$ times. The probability of getting an odd number on a single roll is $$\frac{1}{2}$$ and similarly for an even number.

The probability of getting an odd number exactly $$k$$ times follows a binomial distribution: $$P(X = k) = \binom{n}{k} \left(\frac{1}{2}\right)^n$$.

We are given that the probability of getting an odd number 2 times equals the probability of getting an even number 3 times. So $$\binom{n}{2}\left(\frac{1}{2}\right)^n = \binom{n}{3}\left(\frac{1}{2}\right)^n$$.

This simplifies to $$\binom{n}{2} = \binom{n}{3}$$, that is, $$\frac{n(n-1)}{2} = \frac{n(n-1)(n-2)}{6}$$.

Dividing both sides by $$\frac{n(n-1)}{2}$$, we get $$1 = \frac{n - 2}{3}$$, so $$n = 5$$.

Now we need the probability of getting an odd number an odd number of times, i.e., $$P(X = 1) + P(X = 3) + P(X = 5)$$ with $$n = 5$$.

$$= \left(\frac{1}{2}\right)^5 \left[\binom{5}{1} + \binom{5}{3} + \binom{5}{5}\right] = \frac{1}{32}(5 + 10 + 1) = \frac{16}{32} = \frac{1}{2}$$

Hence, the correct answer is Option D.

We have two persons, $$A$$ and $$B$$. Each of them tosses three fair coins independently. For every single coin, the probability of getting a head is $$\tfrac12$$ and the probability of getting a tail is also $$\tfrac12$$.

Let the random variable $$X$$ denote “number of heads obtained by $$A$$ in the three tosses,” and let $$Y$$ denote “number of heads obtained by $$B$$ in the three tosses.” Both $$X$$ and $$Y$$ follow the same binomial distribution with parameters $$n = 3$$ and $$p = \tfrac12$$.

First we recall the binomial probability formula. For a binomially distributed random variable,

$$\Pr(X = k) \;=\; {^nC_k}\, p^{\,k}\, (1-p)^{\,n-k}.$$

Here $$n = 3$$ and $$p = \tfrac12$$, so for every admissible value $$k = 0,1,2,3$$ we have

$$\Pr(X = k) \;=\; {^3C_k}\,\Bigl(\tfrac12\Bigr)^{k}\,\Bigl(\tfrac12\Bigr)^{3-k} \;=\; {^3C_k}\,\Bigl(\tfrac12\Bigr)^{3} \;=\; {^3C_k}\,\frac{1}{8}.$$

We now list these four probabilities explicitly:

$$$ \begin{aligned} \Pr(X = 0) &= {^3C_0}\,\frac18 \;=\; 1 \times \frac18 \;=\; \frac18,\\[4pt] \Pr(X = 1) &= {^3C_1}\,\frac18 \;=\; 3 \times \frac18 \;=\; \frac38,\\[4pt] \Pr(X = 2) &= {^3C_2}\,\frac18 \;=\; 3 \times \frac18 \;=\; \frac38,\\[4pt] \Pr(X = 3) &= {^3C_3}\,\frac18 \;=\; 1 \times \frac18 \;=\; \frac18. \end{aligned} $$$

The same four numbers are also the probabilities for $$Y$$, because $$B$$ performs the same experiment independently.

The event of interest is “both obtain the same number of heads.” This equality can happen in four mutually exclusive ways:

$$$ (X=0,\,Y=0),\; (X=1,\,Y=1),\; (X=2,\,Y=2),\; (X=3,\,Y=3). $$$

Because the two persons toss their coins independently, the joint probability of any pair $$(X=k,\;Y=k)$$ is the product of the two individual probabilities:

$$$ \Pr(X=k \text{ and } Y=k) \;=\; \Pr(X=k)\,\Pr(Y=k). $$$

We compute these four products one by one:

$$$ \begin{aligned} \Pr(X=0,\,Y=0) &= \frac18 \times \frac18 \;=\; \frac{1}{64},\\[4pt] \Pr(X=1,\,Y=1) &= \frac38 \times \frac38 \;=\; \frac{9}{64},\\[4pt] \Pr(X=2,\,Y=2) &= \frac38 \times \frac38 \;=\; \frac{9}{64},\\[4pt] \Pr(X=3,\,Y=3) &= \frac18 \times \frac18 \;=\; \frac{1}{64}. \end{aligned} $$$

Now we add these four mutually exclusive probabilities to get the total probability that $$A$$ and $$B$$ have the same number of heads:

$$$ \begin{aligned} \Pr(\text{same number of heads}) &= \frac{1}{64} + \frac{9}{64} + \frac{9}{64} + \frac{1}{64}\\[4pt] &= \frac{1 + 9 + 9 + 1}{64}\\[4pt] &= \frac{20}{64}. \end{aligned} $$$

Next we simplify $$\frac{20}{64}$$ by dividing numerator and denominator by their greatest common divisor, $$4$$:

$$$ \frac{20}{64} \;=\; \frac{20 \div 4}{64 \div 4} \;=\; \frac{5}{16}. $$$

So, the probability that both persons get exactly the same number of heads is $$\dfrac{5}{16}$$.

Hence, the correct answer is Option C.

Four dice are thrown and the four numbers are arranged in a $$2 \times 2$$ matrix $$M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$. Since each die shows a value in $$\{1, 2, 3, 4, 5, 6\}$$ and the four positions in the matrix are ordered, the total number of outcomes is $$6^4 = 1296$$.

We need all four entries to be different and $$\det(M) = ad - bc \neq 0$$. The number of ordered arrangements of 4 distinct values chosen from 6 is $$P(6, 4) = 6 \times 5 \times 4 \times 3 = 360$$ (we pick 4 values from 6 in order, since positions $$a, b, c, d$$ are distinguishable).

Now we count how many of these 360 matrices are singular ($$ad = bc$$). We need two disjoint unordered pairs from $$\{1, \ldots, 6\}$$ with the same product. Listing all products of pairs of distinct elements: $$1 \times 2 = 2$$, $$1 \times 3 = 3$$, $$1 \times 4 = 4$$, $$1 \times 5 = 5$$, $$1 \times 6 = 6$$, $$2 \times 3 = 6$$, $$2 \times 4 = 8$$, $$2 \times 5 = 10$$, $$2 \times 6 = 12$$, $$3 \times 4 = 12$$, $$3 \times 5 = 15$$, $$3 \times 6 = 18$$, $$4 \times 5 = 20$$, $$4 \times 6 = 24$$, $$5 \times 6 = 30$$.

The only products that appear for two disjoint pairs are: product $$6$$ from $$\{1, 6\}$$ and $$\{2, 3\}$$, and product $$12$$ from $$\{2, 6\}$$ and $$\{3, 4\}$$. (Other repeated products, if any, would share a common element.)

For each such pair of disjoint pairs, we assign one pair to positions $$(a, d)$$ and the other to $$(b, c)$$: there are 2 ways to decide which pair goes to $$(a, d)$$. Each pair can be placed in its two positions in $$2$$ ways (e.g., $$(a, d) = (1, 6)$$ or $$(6, 1)$$). So each product class gives $$2 \times 2 \times 2 = 8$$ singular matrices.

Total singular matrices with all distinct entries: $$8 + 8 = 16$$. Hence the number of non-singular matrices with all distinct entries is $$360 - 16 = 344$$.

The required probability is $$\frac{344}{1296}$$. Simplify by dividing numerator and denominator by 8: $$\frac{43}{162}$$.

We have 9 distinct balls, and each ball can be dropped in any one of the 4 boxes with equal probability. Thus every ball has 4 independent choices, giving a total of $$4^9$$ equally likely distributions.

We want the event that box $$B_3$$ ends up with exactly 3 balls. To achieve this, we must first select which 3 of the 9 distinct balls enter $$B_3$$. Using the combination formula $$\,^nC_r=\dfrac{n!}{r!(n-r)!}\,$$, the number of ways to make this selection is

$$\binom{9}{3}=84.$$

After fixing those 3 balls in $$B_3$$, the remaining 6 balls are free to go to any of the other three boxes $$B_1, B_2, B_4$$. Each of these 6 balls has 3 choices, so the number of ways to place the remaining balls is $$3^6$$.

Therefore, the total number of favourable distributions is

$$\binom{9}{3}\,3^6.$$

The required probability is hence

$$P=\dfrac{\binom{9}{3}\,3^6}{4^9}.$$

To express this probability in the form $$k\left(\dfrac34\right)^9$$, first note that

$$3^6=\dfrac{3^9}{3^3}.$$

Substituting this inside the expression for $$P$$ we get

$$P=\dfrac{\binom{9}{3}}{3^3}\,\dfrac{3^9}{4^9}= \left(\dfrac{\binom{9}{3}}{3^3}\right)\left(\dfrac34\right)^9.$$

We can now read off

$$k=\dfrac{\binom{9}{3}}{3^3}=\dfrac{84}{27}=\dfrac{28}{9}\approx 3.11.$$

This value of $$k$$ clearly satisfies $$2<k<4,$$ which is exactly the interval described by the set $$\{x\in\mathbb R:\,|x-3|<1\}.$$

Hence, the correct answer is Option A.

We have two independent events A and B with probabilities $$P(A)=p$$ and $$P(B)=2p$$. Because the events are independent, their complements are also independent. Our first task is to express the probability that exactly one of the two events occurs.

By definition, “exactly one of A or B occurs” means the outcome is either “A occurs and B does not” or “B occurs and A does not.” For independent events, we use the multiplication rule $$P(X\cap Y)=P(X)\,P(Y)$$. Hence

$$P(\text{exactly one of }A,B)=P(A\cap B')+P(B\cap A').$$

Substituting the given probabilities and the complements $$P(B')=1-P(B) \quad\text{and}\quad P(A')=1-P(A),$$ we write

$$P(\text{exactly one}) \;=\; P(A)\,P(B') + P(B)\,P(A')$$

$$=\; p\,[1-2p] + (2p)\,[1-p].$$

Expanding each product gives

$$p(1-2p)=p-2p^{2},$$

$$2p(1-p)=2p-2p^{2}.$$

Adding these two expressions we obtain

$$P(\text{exactly one})=(p-2p^{2})+(2p-2p^{2})$$ $$=p-2p^{2}+2p-2p^{2}$$ $$=3p-4p^{2}.$$

The problem states that this probability equals $$\dfrac59$$. Therefore we set

$$3p-4p^{2}=\dfrac59.$$

To clear the fraction, multiply both sides by 9:

$$9(3p-4p^{2})=5,$$ $$27p-36p^{2}=5.$$

Now bring all terms to one side to form a quadratic equation in the standard form $$ax^{2}+bx+c=0$$:

$$-36p^{2}+27p-5=0.$$

Multiplying by -1 (so the leading coefficient is positive) yields

$$36p^{2}-27p+5=0.$$

We next solve this quadratic equation using the quadratic formula. Recall the formula:

For $$ax^{2}+bx+c=0,$$ the roots are $$x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$

Here, $$a=36,\; b=-27,\; c=5$$ (note that b is -27 because the middle term is -27p). Substituting:

Discriminant

$$\Delta=b^{2}-4ac=(-27)^{2}-4(36)(5)=729-720=9.$$

Square root of the discriminant

$$\sqrt{\Delta}=\sqrt{9}=3.$$

Now the two possible roots are

$$p=\dfrac{-(-27)\pm 3}{2\cdot 36}=\dfrac{27\pm 3}{72}.$$

Simplifying each root:

First root: $$p_{1}=\dfrac{27+3}{72}=\dfrac{30}{72}=\dfrac{5}{12}.$$

Second root: $$p_{2}=\dfrac{27-3}{72}=\dfrac{24}{72}=\dfrac{1}{3}.$$

Probabilities must satisfy $$0\le p\le 1$$ and, because $$P(B)=2p,$$ we also need $$2p\le 1,$$ or $$p\le\dfrac12.$$ Both values $$\dfrac{5}{12}\;(=0.416\dots)$$ and $$\dfrac13\;(=0.333\dots)$$ lie in the allowed interval. The question asks for the largest such value, so we choose

$$p=\dfrac{5}{12}.$$

Comparing with the options given, $$\dfrac{5}{12}$$ corresponds to Option C.

Hence, the correct answer is Option C.

Let

$$x_A,x_B,x_C$$ be the probabilities of outcomes belonging only to $$A,B,C$$ respectively.

Let

$$y_{AB},y_{BC},y_{CA}$$ be the probabilities of outcomes belonging to exactly two events, and let $$z=P(A\cap B\cap C)=k^2$$

Then,

$$x_A+x_B+y_{BC}+y_{CA}=1-k\quad\cdots(1)$$

$$x_B+x_C+y_{AB}+y_{CA}=1-2k\quad\cdots(2)$$

$$x_C+x_A+y_{AB}+y_{BC}=1-k\quad\cdots(3)$$

Adding

$$(1),(2),(3),$$

$$2(x_A+x_B+x_C+y_{AB}+y_{BC}+y_{CA})=3-4k$$

So,

$$x_A+x_B+x_C+y_{AB}+y_{BC}+y_{CA}=\frac32-2k$$

Now,

$$P(A\cup B\cup C)=x_A+x_B+x_C+y_{AB}+y_{BC}+y_{CA}+z$$

Substituting

$$z=k^2,$$

$$P(A\cup B\cup C)=\frac32-2k+k^2$$

Rewriting,

$$P(A\cup B\cup C)=k^2-2k+\frac32$$

$$=(k-1)^2+\frac12$$

Since

$$0<k<1,$$

we have

$$0<(k-1)^2<1$$

Therefore,

$$P(A\cup B\cup C)>\frac12$$

Hence, the required probability is

$$\boxed{\frac32-2k+k^2}$$ and it is always greater than $$\boxed{\frac12}$$

We need to find the set $$A$$ of all 4-digit natural numbers with exactly one digit equal to 7. A 4-digit number has digits $$d_1 d_2 d_3 d_4$$ where $$d_1 \in \{1, 2, \ldots, 9\}$$ and $$d_2, d_3, d_4 \in \{0, 1, \ldots, 9\}$$. Exactly one of these digits is 7.

We count $$|A|$$ by considering which position has the digit 7. If $$d_1 = 7$$: the remaining 3 digits are chosen from $$\{0,1,\ldots,9\} \setminus \{7\}$$, giving $$9^3 = 729$$ numbers. If $$d_2 = 7$$ (or $$d_3 = 7$$ or $$d_4 = 7$$): $$d_1$$ is from $$\{1,\ldots,9\}\setminus\{7\} = 8$$ choices, and the other two non-7 positions each have 9 choices. So each gives $$8 \times 9^2 = 648$$ numbers.

Thus $$|A| = 729 + 3 \times 648 = 729 + 1944 = 2673$$.

For the number to leave remainder 2 when divided by 5, the last digit $$d_4$$ must be 2 or 7 (since $$2 \equiv 2 \pmod{5}$$ and $$7 \equiv 2 \pmod{5}$$).

Case 1: $$d_4 = 7$$ (so the 7 is in the units place). Then $$d_1 \in \{1,\ldots,9\}\setminus\{7\}$$ (8 choices), $$d_2, d_3 \in \{0,\ldots,9\}\setminus\{7\}$$ (9 choices each). This gives $$8 \times 9 \times 9 = 648$$ numbers.

Case 2: $$d_4 = 2$$ (so the 7 is in one of the other three positions). Sub-case (a): $$d_1 = 7$$, then $$d_2, d_3 \in \{0,\ldots,9\}\setminus\{7\}$$ giving $$9 \times 9 = 81$$ numbers. Sub-case (b): $$d_2 = 7$$, then $$d_1 \in \{1,\ldots,9\}\setminus\{7\}$$ (8 choices), $$d_3 \in \{0,\ldots,9\}\setminus\{7\}$$ (9 choices), giving $$8 \times 9 = 72$$. Sub-case (c): $$d_3 = 7$$, similarly gives $$8 \times 9 = 72$$.

So Case 2 total = $$81 + 72 + 72 = 225$$.

Total favorable = $$648 + 225 = 873$$.

The probability = $$\frac{873}{2673} = \frac{97}{297}$$ (dividing numerator and denominator by 9).

Therefore, the required probability is $$\dfrac{97}{297}$$.

Given:

- Probability of occurrence of $$0$$ at odd places is

$$\frac13$$

Hence,

$$P(1_{\text{odd}})=\frac23$$

- Probability of occurrence of $$0$$ at even places is

$$\frac12$$

Hence,

$$P(1_{\text{even}})=\frac12$$

We need the probability that the sequence

$$10$$

is followed by

$$01$$

That is, the 4-bit pattern

$$1001$$

Now this sequence can occur in two ways.

Case 1: Sequence starts at an odd position

Then positions are:

Odd, Even, Odd, Even

Hence probability is

$$\frac23\times\frac12\times\frac13\times\frac12$$

$$=\frac1{18}$$

Case 2: Sequence starts at an even position

Then positions are:

Even, Odd, Even, Odd

Hence probability is

$$\frac12\times\frac13\times\frac12\times\frac23$$

$$=\frac1{18}$$

Therefore total probability is

$$\frac1{18}+\frac1{18}$$

$$=\frac19$$

Hence, the required probability is

$$\boxed{\frac19}$$

We need to form a 6-digit integer using digits from $$\{0, 1, 2, 3, 4, 5, 6\}$$ without repetition, and the number must be divisible by 3.

The total sum of all 7 digits is $$0+1+2+3+4+5+6 = 21$$. To form a 6-digit number, we exclude one digit. For the number to be divisible by 3, the sum of the 6 chosen digits must be divisible by 3. Since $$21$$ is divisible by 3, the excluded digit must also be divisible by 3. The digits divisible by 3 from the set are $$\{0, 3, 6\}$$.

The total number of valid 6-digit integers (no repetition, first digit non-zero) from 7 digits: we exclude one digit out of 7. If 0 is excluded, the remaining digits are $$\{1,2,3,4,5,6\}$$, giving $$6! = 720$$ six-digit numbers. If a non-zero digit is excluded, the remaining 6 digits include 0, giving $$6! - 5! = 720 - 120 = 600$$ valid six-digit numbers (subtracting those starting with 0). Total = $$720 + 6 \times 600 = 720 + 3600 = 4320$$.

Now we count favorable cases where we exclude a digit divisible by 3. Case 1: Exclude 0. Remaining digits $$\{1,2,3,4,5,6\}$$, sum = 21, divisible by 3. Number of 6-digit numbers = $$6! = 720$$. Case 2: Exclude 3. Remaining $$\{0,1,2,4,5,6\}$$, sum = 18, divisible by 3. Number of valid 6-digit numbers = $$6! - 5! = 600$$. Case 3: Exclude 6. Remaining $$\{0,1,2,3,4,5\}$$, sum = 15, divisible by 3. Number of valid 6-digit numbers = $$6! - 5! = 600$$.

Total favorable = $$720 + 600 + 600 = 1920$$.

Probability = $$\frac{1920}{4320} = \frac{4}{9}$$.

The answer is $$\frac{4}{9}$$, which corresponds to Option (2).

In a Binomial distribution with $$n = 5$$ trials, let the probability of success be $$p$$ and failure be $$q = 1 - p$$. We are given $$P(X=1) = \binom{5}{1}p\,q^4 = 5p\,q^4 = 0.4096$$ and $$P(X=2) = \binom{5}{2}p^2q^3 = 10p^2q^3 = 0.2048$$.

Dividing the second equation by the first: $$\dfrac{10p^2q^3}{5p\,q^4} = \dfrac{2p}{q} = \dfrac{0.2048}{0.4096} = \dfrac{1}{2}$$. This gives $$q = 4p$$. Substituting into $$p + q = 1$$ yields $$5p = 1$$, so $$p = \dfrac{1}{5}$$ and $$q = \dfrac{4}{5}$$.

We can verify: $$5 \cdot \dfrac{1}{5} \cdot \left(\dfrac{4}{5}\right)^4 = \dfrac{256}{625} = 0.4096$$ and $$10 \cdot \dfrac{1}{25} \cdot \dfrac{64}{125} = \dfrac{640}{3125} = 0.2048$$, both correct.

The probability of exactly 3 successes is $$P(X=3) = \binom{5}{3}p^3q^2 = 10 \cdot \dfrac{1}{125} \cdot \dfrac{16}{25} = \dfrac{160}{3125} = \dfrac{32}{625}$$.

We have the finite set $$S=\{1,2,3,4,5,6\}$$ as both domain and co-domain. An “onto” function from a finite set to itself must be bijective, so every such function is simply a permutation of the six elements. The total number of permutations of six distinct objects is given by the factorial formula $$n!$$; putting $$n=6$$ we obtain the total number of onto functions

$$6!\;=\;720.$$

Now we impose the condition $$g(3)=2g(1).$$ Because the image of $$g(1)$$ must lie in $$S$$, and because doubling that image must also lie in $$S$$, we list the possibilities one by one.

If $$g(1)=1$$, then $$2g(1)=2$$, so we must have $$g(3)=2.$$

If $$g(1)=2$$, then $$2g(1)=4$$, so we must have $$g(3)=4.$$

If $$g(1)=3$$, then $$2g(1)=6$$, so we must have $$g(3)=6.$$

For $$g(1)=4,5,6$$ the number $$2g(1)$$ would be outside $$S$$, so no further choices are admissible. Thus there are exactly three admissible ordered pairs $$\bigl(g(1),g(3)\bigr)$$:

$$\bigl(1,2\bigr),\qquad\bigl(2,4\bigr),\qquad\bigl(3,6\bigr).$$

Fix any one of these three pairs. Two images have now been used up, leaving four distinct images in $$S$$ and four remaining domain elements, namely $$2,4,5,6.$$ To keep the function bijective we must match these four elements with the four unused images in a one-to-one fashion. The number of ways to arrange a bijection between two four-element sets is given by $$4!$$, and we compute

$$4!\;=\;24.$$

Hence, for each of the three admissible choices for $$\bigl(g(1),g(3)\bigr)$$ there are $$24$$ complete bijections. Multiplying, the total number of favorable functions is

$$3\times24\;=\;72.$$

The required probability is therefore the ratio of favorable cases to total cases:

$$\text{Probability} \;=\;\dfrac{72}{720}\;=\;\dfrac{1}{10}.$$

Hence, the correct answer is Option D.

We have a discrete random variable $$X$$ whose probability mass function is given as follows:

$$P(X = 0) = \frac{1}{2}, \qquad P(X = j) = \frac{1}{3^{\,j}} \; \text{for} \; j = 1,2,3,\ldots$$

First we verify that the total probability is $$1$$. For the strictly positive values we must sum the geometric series

$$\sum_{j = 1}^{\infty} \frac{1}{3^{\,j}}.$$

We recall the infinite geometric‐series formula: if $$|r| < 1$$, then

$$\sum_{j = 1}^{\infty} r^{\,j} \;=\; \frac{r}{1 - r}.$$

Here $$r = \dfrac13$$, so

$$\sum_{j = 1}^{\infty} \frac1{3^{\,j}} = \frac{\tfrac13}{1 - \tfrac13} = \frac{\tfrac13}{\tfrac23} = \frac{1}{2}.$$

Adding the probability at $$j = 0$$ gives

$$P(X=0) + \sum_{j=1}^{\infty} P(X=j) = \frac12 + \frac12 = 1,$$

so the distribution is valid.

Now we find the mean (expected value) $$E[X]$$. By definition, for a discrete variable,

$$E[X] \;=\; \sum_{x} x\,P(X=x).$$

Because the term at $$x=0$$ contributes nothing, we have

$$E[X] = \sum_{j = 1}^{\infty} j \,\frac1{3^{\,j}}.$$

Again we use a standard result for an infinite geometric‐like sum. For $$|r| < 1$$,

$$\sum_{j = 1}^{\infty} j\,r^{\,j} \;=\; \frac{r}{(1 - r)^{2}}.$$

Setting $$r = \dfrac13$$ gives

$$E[X] = \frac{\tfrac13}{(1 - \tfrac13)^{2}} = \frac{\tfrac13}{\left(\tfrac23\right)^{2}} = \frac{\tfrac13}{\tfrac49} = \tfrac13 \times \tfrac94 = \frac34.$$

Next we need the probability that $$X$$ is both positive and even. Let us write the positive even values as $$2,4,6,\ldots$$, that is $$X = 2k$$ where $$k = 1,2,3,\ldots$$. Therefore

$$P(X \text{ is positive and even}) = \sum_{k = 1}^{\infty} P(X = 2k) = \sum_{k = 1}^{\infty} \frac1{3^{\,2k}} = \sum_{k = 1}^{\infty} \left(\frac1{3^{2}}\right)^{k} = \sum_{k = 1}^{\infty} \left(\frac1{9}\right)^{k}.$$

This is another geometric series with common ratio $$\dfrac19$$. Using the same formula for the sum, we get

$$\sum_{k = 1}^{\infty} \left(\frac19\right)^{k} = \frac{\tfrac19}{1 - \tfrac19} = \frac{\tfrac19}{\tfrac89} = \frac{1}{8}.$$

We have now obtained both required quantities: the mean $$E[X] = \dfrac34$$ and the probability that $$X$$ is positive and even $$= \dfrac18$$.

Hence, the correct answer is Option B.

First, let us interpret the condition. We want those natural numbers $$n$$ for which the expression $$2^{\,n}-2$$ is a multiple of $$3$$. Writing this in modular arithmetic language, “is a multiple of $$3$$” means “is congruent to $$0$$ modulo $$3$$”. So we require

$$2^{\,n}-2 \equiv 0 \pmod 3.$$

Adding $$2$$ to both sides keeps the equivalence true, therefore

$$2^{\,n}\equiv 2 \pmod 3.$$

Now we study the powers of $$2$$ modulo $$3$$. We begin with the smallest exponents and proceed step by step.

$$\begin{aligned} 2^{1}&=2 &\Longrightarrow&\; 2^{1}\equiv 2 \pmod 3,\\[2pt] 2^{2}&=4 &\Longrightarrow&\; 2^{2}\equiv 1 \pmod 3,\\[2pt] 2^{3}&=8 &\Longrightarrow&\; 2^{3}\equiv 2 \pmod 3,\\[2pt] 2^{4}&=16 &\Longrightarrow&\; 2^{4}\equiv 1 \pmod 3. \end{aligned}$$

We notice a repeating pattern: $$2,\,1,\,2,\,1,\ldots$$ Thus the remainder alternates between $$2$$ and $$1$$ every time we increase the exponent by $$1$$. Consequently,

$$\boxed{2^{\,n}\equiv 2 \pmod 3 \;\Longleftrightarrow\; n \text{ is odd}.}$$

So the given condition is satisfied exactly when $$n$$ is an odd integer.

But the problem restricts us to two-digit numbers. The set of all two-digit natural numbers is $$\{10,11,12,\ldots,99\}$$, and its size is

$$99-10+1 \;=\;90.$$

Among these, we list the odd ones: $$11,13,15,\ldots,99.$$ This is an arithmetic progression with first term $$11$$, common difference $$2$$, and last term $$99$$. Using the standard counting formula for such a progression,

$$\text{Number of odd terms}= \frac{99-11}{2}+1 = \frac{88}{2}+1 =44+1 =45.$$

Therefore, exactly $$45$$ out of the $$90$$ two-digit numbers meet the required criterion.

Probability is defined as $$\text{Probability}= \frac{\text{favourable cases}}{\text{total cases}}.$$ Substituting the values we have just obtained,

$$\text{Probability}= \frac{45}{90}= \frac{1}{2}.$$

Hence, the correct answer is Option C.

For each element of the set $$\{1, 2, 3, 4, 5\}$$, there are four equally likely possibilities when two subsets $$A$$ and $$B$$ are chosen randomly: the element is in both $$A$$ and $$B$$, in $$A$$ only, in $$B$$ only, or in neither. Each possibility has probability $$\frac{1}{4}$$, since each element is independently included in $$A$$ with probability $$\frac{1}{2}$$ and in $$B$$ with probability $$\frac{1}{2}$$.

The total number of ways to choose two subsets is $$2^5 \times 2^5 = 2^{10}$$. We need exactly 2 elements in the intersection $$A \cap B$$. An element belongs to $$A \cap B$$ with probability $$\frac{1}{4}$$, and does not belong to $$A \cap B$$ (i.e., is in at most one of the two subsets) with probability $$\frac{3}{4}$$.

The required probability is $$\binom{5}{2} \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^3 = 10 \cdot \frac{1}{16} \cdot \frac{27}{64} = \frac{270}{1024} = \frac{135}{512} = \frac{135}{2^9}$$.

The correct answer is $$\frac{135}{2^9}$$.

Both dice have faces numbered $$1, 2, 3, 5, 7, 11$$. The total number of outcomes is $$6 \times 6 = 36$$.

We count pairs $$(a, b)$$ with $$a + b \leq 8$$:

When $$a = 1$$: $$b$$ can be $$1, 2, 3, 5, 7$$ (sums: 2, 3, 4, 6, 8 — all $$\leq 8$$; $$1+11=12 > 8$$). That gives 5 outcomes.

When $$a = 2$$: $$b$$ can be $$1, 2, 3, 5$$ (sums: 3, 4, 5, 7; $$2+7=9 > 8$$). That gives 4 outcomes.

When $$a = 3$$: $$b$$ can be $$1, 2, 3, 5$$ (sums: 4, 5, 6, 8; $$3+7=10 > 8$$). That gives 4 outcomes.

When $$a = 5$$: $$b$$ can be $$1, 2, 3$$ (sums: 6, 7, 8; $$5+5=10 > 8$$). That gives 3 outcomes.

When $$a = 7$$: $$b = 1$$ only ($$7+1=8$$; $$7+2=9 > 8$$). That gives 1 outcome.

When $$a = 11$$: no valid $$b$$ ($$11+1=12 > 8$$). That gives 0 outcomes.

Total favourable = $$5 + 4 + 4 + 3 + 1 = 17$$. The probability is $$\frac{17}{36}$$.

The answer is Option B: $$\frac{17}{36}$$.

A chessboard has

$$8\times8=64$$

squares.

Two squares are chosen at random.

Total number of ways:

$$\binom{64}{2}=\frac{64\cdot63}{2}=2016$$

Now count pairs having a common side.

Horizontal adjacent pairs:

Each row has

$$7$$

adjacent pairs.

Since there are

$$8$$

rows,

$$8\times7=56$$

pairs.

Vertical adjacent pairs:

Similarly,

$$8\times7=56$$

pairs.

Hence total favorable pairs:

$$56+56=112$$

Therefore, required probability is

$$\frac{112}{2016}$$

$$=\frac1{18}$$

Hence, the required probability is

$$\boxed{\frac1{18}}$$

Let us denote the six faces of the die by $$1,2,3,4,5,6$$. In an ordinary die the opposite faces are $$1\leftrightarrow 6,\;2\leftrightarrow 5,\;3\leftrightarrow 4$$ so that the numbers on opposite faces always add up to $$7$$. According to the statement, exactly one face is made a little less likely and its opposite face is made equally more likely. Concretely, for some face (let us call it the “particular face”) the probability is $$\dfrac16-x$$, for its opposite face the probability is $$\dfrac16+x$$, while the remaining four faces all keep the usual probability $$\dfrac16$$. We are also told that $$0<x<\dfrac16$$.

Without loss of generality we may assume that the face with reduced probability is $$1$$ and therefore its opposite face $$6$$ has the increased probability. (If we had chosen any other pair, the final calculation of the probability of the total sum $$7$$ would be identical because the set of ordered pairs giving a total of $$7$$ remains the same.) Hence we have

$$\begin{aligned} P(1)&=\frac16-x,\\ P(6)&=\frac16+x,\\ P(2)&=P(3)=P(4)=P(5)=\frac16. \end{aligned}$$

When two such dice are rolled independently, the total equals $$7$$ exactly when the ordered pair of results is one of

$$\bigl(1$$, $$6\bigr)$$, $$\;\bigl(2$$, $$5\bigr)$$, $$\;\bigl(3$$, $$4\bigr)$$, $$\;\bigl(4$$, $$3\bigr)$$, $$\;\bigl(5$$, $$2\bigr)$$, $$\;\bigl(6$$, $$1\bigr).$$

Because the two rolls are independent, the probability of a specific ordered pair is the product of the single-roll probabilities. Therefore the total probability of obtaining a sum of $$7$$ is

$$\begin{aligned} P(\text{sum}=7)&=P(1)P(6)+P(2)P(5)+P(3)P(4)+P(4)P(3)+P(5)P(2)+P(6)P(1)\\ &=2\Bigl[P(1)P(6)+P(2)P(5)+P(3)P(4)\Bigr]. \end{aligned}$$

Now we substitute the individual probabilities:

$$\begin{aligned} P(1)P(6)&=\left(\frac16-x\right)\left(\frac16+x\right) =\left(\frac16\right)^2-x^2 =\frac1{36}-x^2,\\[6pt] P(2)P(5)&=\frac16\cdot\frac16=\frac1{36},\\[6pt] P(3)P(4)&=\frac16\cdot\frac16=\frac1{36}. \end{aligned}$$

Adding these three products we get

$$\begin{aligned} P(1)P(6)+P(2)P(5)+P(3)P(4) =\left(\frac1{36}-x^2\right)+\frac1{36}+\frac1{36} =\frac3{36}-x^2 =\frac1{12}-x^2. \end{aligned}$$

Multiplying by the prefactor $$2$$, the desired probability is

$$P(\text{sum}=7)=2\left(\frac1{12}-x^2\right)=\frac16-2x^2.$$

The problem states that this probability equals $$\dfrac{13}{96}$$. Setting the two expressions equal gives

$$\frac16-2x^2=\frac{13}{96}.$$

To solve for $$x$$ we first write $$\dfrac16$$ with denominator $$96$$:

$$\frac16=\frac{16}{96}.$$

Substituting this, we have

$$\frac{16}{96}-2x^2=\frac{13}{96}.$$

Subtracting $$\dfrac{13}{96}$$ from both sides:

$$\frac{16}{96}-\frac{13}{96}=2x^2.$$

$$\frac{3}{96}=2x^2.$$

Dividing both sides by $$2$$:

$$x^2=\frac{3}{96}\cdot\frac12=\frac{3}{192}=\frac1{64}.$$

Because $$x>0$$, we take the positive square root:

$$x=\sqrt{\frac1{64}}=\frac18.$$

Since $$\dfrac18=\dfrac{1}{8}$$ corresponds to Option C, we conclude:

Hence, the correct answer is Option C.

The probability that a missile is intercepted is $$\frac{1}{3}$$, so the probability that it is not intercepted is $$1 - \frac{1}{3} = \frac{2}{3}$$.

Given that a missile is not intercepted, the probability that it hits the target is $$\frac{3}{4}$$.

Therefore, the probability that a single missile hits the target is $$P(\text{hit}) = P(\text{not intercepted}) \times P(\text{hit} \mid \text{not intercepted}) = \frac{2}{3} \times \frac{3}{4} = \frac{1}{2}$$.

Since three missiles are fired independently, the probability that all three hit the target is $$\left(\frac{1}{2}\right)^3 = \frac{1}{8}$$.

The word EXAMINATION has 11 letters: E, X, A, M, I, N, A, T, I, O, N — with A appearing twice, I appearing twice, and N appearing twice, and the remaining letters M, E, X, T, O each appearing once.

The total number of distinct arrangements of all 11 letters is $$\frac{11!}{2! \cdot 2! \cdot 2!}$$.

To count arrangements where M is fixed at the 4th position, we arrange the remaining 10 letters (E, X, A, I, N, A, T, I, O, N) in the remaining 10 positions. The number of such arrangements is $$\frac{10!}{2! \cdot 2! \cdot 2!}$$.

The required probability is:

$$P = \frac{\dfrac{10!}{2! \cdot 2! \cdot 2!}}{\dfrac{11!}{2! \cdot 2! \cdot 2!}} = \frac{10!}{11!} = \frac{1}{11}$$

We begin with a single throw of a fair coin. In one throw, the probability of getting a head is $$\dfrac12$$ and the probability of getting a tail is also $$\dfrac12$$ because the coin is fair.

When the coin is tossed $$n$$ times, all the throws are independent. The easiest way to find the probability of “at least one head” is to use the complementary event “no head at all,” that is, getting tails in every single throw.

For one throw, the probability of a tail is $$\dfrac12$$. Hence, when the coin is tossed $$n$$ times, the probability of getting a tail on every single throw (that is, zero heads) is $$\left(\dfrac12\right)^n.$$ This uses the multiplication rule for independent events.

Therefore, the probability of getting at least one head in $$n$$ throws is $$ 1-\left(\dfrac12\right)^n. $$

The question tells us that this probability must be at least $$0.9$$, so we set up the inequality: $$ 1-\left(\dfrac12\right)^n \ge 0.9. $$

Rearranging the terms, we isolate the power of $$\dfrac12$$: $$ \left(\dfrac12\right)^n \le 1-0.9 = 0.1. $$

To solve for $$n$$, we take natural logarithms on both sides. (Any logarithm base would work; natural logarithm is convenient.) We state first that for any positive numbers $$a$$ and $$b$$, if $$a<1$$ then $$\ln a<0$$, and dividing by a negative number reverses the inequality sign.

Applying $$\ln$$, we get $$ \ln\!\left(\left(\dfrac12\right)^n\right) \le \ln(0.1). $$ Using the power rule $$\ln(a^k)=k\ln a$$, this becomes $$ n\,\ln\!\left(\dfrac12\right) \le \ln(0.1). $$

Because $$\ln\!\left(\dfrac12\right)<0$$, dividing both sides by this negative number reverses the inequality: $$ n \ge \dfrac{\ln(0.1)}{\ln\!\left(\dfrac12\right)}. $$

We now evaluate the right-hand side numerically: $$ \ln(0.1) \approx -2.302585, \qquad \ln\!\left(\dfrac12\right) \approx -0.693147. $$ Therefore $$ n \ge \dfrac{-2.302585}{-0.693147} \approx 3.321928. $$

Since $$n$$ must be an integer (you cannot toss the coin a fractional number of times), we take the smallest integer not less than $$3.321928$$, which is $$4$$.

We may confirm quickly: for $$n=3$$, the probability of at least one head is $$1-\left(\dfrac12\right)^3 = 1-\dfrac18 = 0.875$$, which is less than $$0.9$$; for $$n=4$$, the probability is $$1-\left(\dfrac12\right)^4 = 1-\dfrac{1}{16} = 0.9375$$, which indeed satisfies the requirement.

So, the answer is $$4$$.

Let us denote the events as follows: $$A$$ is the event that the first unit works and $$B$$ is the event that the second unit works. We are told that the two units behave independently, so their probabilities multiply when we need the intersection of their events.

We have the probabilities of each unit functioning:

$$P(A)=0.9,\qquad P(B)=0.8.$$

If a unit does not work, we denote its failure with a prime. Thus

$$P(A')=1-P(A)=1-0.9=0.1,$$

$$P(B')=1-P(B)=1-0.8=0.2.$$

The instrument fails to operate whenever at least one of the units fails. The failure event for the whole instrument is therefore the union $$A'\cup B'$$. By the Addition Law of Probability, for independent events, we can compute

$$P(A'\cup B')=P(A')+P(B')-P(A'B').$$

But independence gives $$P(A'B')=P(A')P(B')=0.1\times0.2=0.02,$$ so

$$P(A'\cup B')=0.1+0.2-0.02=0.28.$$

There is a simpler equivalent route: the instrument works only when both units work, i.e. on $$AB$$. Hence

$$P(A'\cup B')=1-P(AB)=1-P(A)P(B)=1-0.9\times0.8=1-0.72=0.28,$$

which matches the previous calculation.

We want the conditional probability that only the first unit failed while the second unit still works, given that the instrument has failed. Symbolically, this is

$$p=P(A'B\mid A'\cup B').$$

By definition of conditional probability,

$$P(A'B\mid A'\cup B')=\frac{P(A'B)}{P(A'\cup B')}.$$

Again using independence,

$$P(A'B)=P(A')P(B)=0.1\times0.8=0.08.$$

Substituting the values we have just obtained,

$$p=\frac{0.08}{0.28}=\frac{8}{28}=\frac{2}{7}.$$

The question asks for the value of $$98p$$, so we multiply:

$$98p=98\times\frac{2}{7}=14\times2=28.$$

Hence, the correct answer is Option 28.

Let $$P(B_i) = p_i$$ and $$P(\overline{B_i}) = 1 - p_i = q_i$$ for $$i = 1, 2, 3$$. Since the events are independent, we have:

$$\alpha = p_1 q_2 q_3$$ (only $$B_1$$ occurs), $$\beta = q_1 p_2 q_3$$ (only $$B_2$$ occurs), $$\gamma = q_1 q_2 p_3$$ (only $$B_3$$ occurs), and $$p = q_1 q_2 q_3$$ (none occurs).

We note that $$\frac{\alpha}{p} = \frac{p_1}{q_1}$$, $$\frac{\beta}{p} = \frac{p_2}{q_2}$$, and $$\frac{\gamma}{p} = \frac{p_3}{q_3}$$.

Let $$a = \frac{p_1}{q_1}$$, $$b = \frac{p_2}{q_2}$$, $$c = \frac{p_3}{q_3}$$. Then $$\alpha = ap$$, $$\beta = bp$$, $$\gamma = cp$$.

Substituting into the first equation $$(\alpha - 2\beta)p = \alpha\beta$$, we get $$(ap - 2bp)p = ap \cdot bp$$, which simplifies to $$p^2(a - 2b) = abp^2$$. Since $$p \neq 0$$, we get $$a - 2b = ab$$, so $$a(1 - b) = 2b$$, giving $$a = \frac{2b}{1 - b}$$.

Substituting into the second equation $$(\beta - 3\gamma)p = 2\beta\gamma$$, we get $$(bp - 3cp)p = 2bp \cdot cp$$, which simplifies to $$b - 3c = 2bc$$, so $$b(1 - 2c) = 3c$$, giving $$b = \frac{3c}{1 - 2c}$$.

Now $$1 - b = 1 - \frac{3c}{1 - 2c} = \frac{1 - 5c}{1 - 2c}$$. So $$a = \frac{2 \cdot \frac{3c}{1 - 2c}}{\frac{1 - 5c}{1 - 2c}} = \frac{6c}{1 - 5c}$$.

We need $$\frac{P(B_1)}{P(B_3)} = \frac{p_1}{p_3}$$. Since $$p_i = \frac{a_i}{1 + a_i}$$ (where $$a_1 = a$$, $$a_3 = c$$), we get:

$$\frac{p_1}{p_3} = \frac{a}{1 + a} \cdot \frac{1 + c}{c} = \frac{a(1 + c)}{c(1 + a)}$$

Now $$1 + a = 1 + \frac{6c}{1 - 5c} = \frac{1 + c}{1 - 5c}$$.

So $$\frac{p_1}{p_3} = \frac{\frac{6c}{1 - 5c} \cdot (1 + c)}{c \cdot \frac{1 + c}{1 - 5c}} = \frac{6c(1 + c)(1 - 5c)}{c(1 + c)(1 - 5c)} = 6$$.

Hence, the answer is $$6$$.

Let the probabilities of $$E_1, E_2, E_3$$ be $$p_1, p_2, p_3$$. Since the events are independent:

$$\alpha = p_1(1-p_2)(1-p_3)$$, $$\beta = (1-p_1)p_2(1-p_3)$$, $$\gamma = (1-p_1)(1-p_2)p_3$$, and $$p = (1-p_1)(1-p_2)(1-p_3)$$.

Observe that $$\frac{\alpha}{p} = \frac{p_1}{1-p_1}$$, $$\frac{\beta}{p} = \frac{p_2}{1-p_2}$$, $$\frac{\gamma}{p} = \frac{p_3}{1-p_3}$$.

Let $$x = \frac{p_1}{1-p_1}$$, $$y = \frac{p_2}{1-p_2}$$, $$z = \frac{p_3}{1-p_3}$$.

From $$(\alpha - 2\beta)p = \alpha\beta$$, dividing by $$p^2$$: $$x - 2y = xy$$, so $$x(1-y) = 2y$$, giving $$x = \frac{2y}{1-y}$$ $$-(1)$$.

From $$(\beta - 3\gamma)p = 2\beta\gamma$$, dividing by $$p^2$$: $$y - 3z = 2yz$$, so $$y(1-2z) = 3z$$, giving $$y = \frac{3z}{1-2z}$$ $$-(2)$$.

From $$(2)$$: $$1 - y = \frac{1-2z-3z}{1-2z} = \frac{1-5z}{1-2z}$$. Substituting into $$(1)$$: $$x = \frac{2 \cdot \frac{3z}{1-2z}}{\frac{1-5z}{1-2z}} = \frac{6z}{1-5z}$$.

Now $$1+x = \frac{1-5z+6z}{1-5z} = \frac{1+z}{1-5z}$$ and $$1+z = 1+z$$.

Since $$p_1 = \frac{x}{1+x}$$ and $$p_3 = \frac{z}{1+z}$$: $$\frac{p_1}{p_3} = \frac{x}{1+x} \cdot \frac{1+z}{z} = \frac{x(1+z)}{z(1+x)}$$.

$$= \frac{\frac{6z}{1-5z} \cdot (1+z)}{z \cdot \frac{1+z}{1-5z}} = \frac{6z(1+z)(1-5z)}{z(1+z)(1-5z)} = 6$$.

The answer is 6.

We have a discrete random variable $$X$$ that can take the five values $$1, 2, 3, 4, 5$$ with probabilities $$K, 2K, 2K, 3K, K$$ respectively.

Because these five probabilities must add up to $$1$$, we first write the normalisation condition:

$$K + 2K + 2K + 3K + K = 1.$$

Combining like terms on the left-hand side gives

$$9K = 1.$$

Dividing both sides by $$9$$, we obtain

$$K = \frac{1}{9}.$$

Now we define two events that appear in the required conditional probability.

Event $$A : (1 < X < 4).$$ In words, $$X$$ must be strictly greater than $$1$$ and strictly less than $$4$$, so the possible values are $$X = 2$$ or $$X = 3$$.

Event $$B : (X < 3).$$ This means $$X$$ can be $$1$$ or $$2$$.

The conditional probability we want is

$$p = P\!\bigl(1 < X < 4 \,\bigl|\, X < 3\bigr) = P(A\,|\,B).$$

The definition of conditional probability says

$$P(A\,|\,B) = \frac{P(A \cap B)}{P(B)}.$$

So, we need both $$P(A \cap B)$$ and $$P(B).$$

The intersection $$A \cap B$$ consists of those $$X$$ that satisfy both conditions simultaneously. Event $$A$$ allows $$X = 2, 3,$$ while event $$B$$ allows $$X = 1, 2.$$ The only common value is $$X = 2.$$ Therefore

$$P(A \cap B) = P(X = 2) = 2K.$$

Next, for the denominator, we compute the probability of $$B$$:

$$P(B) = P(X = 1) + P(X = 2) = K + 2K = 3K.$$

Substituting these two results into the conditional-probability formula, we have

$$p = \frac{P(A \cap B)}{P(B)} = \frac{2K}{3K}.$$

The factor $$K$$ cancels, giving

$$p = \frac{2}{3}.$$

Now the problem states that $$5p = \lambda K$$. Substituting $$p = \dfrac{2}{3}$$ on the left and $$K = \dfrac{1}{9}$$ on the right, we get

$$5 \times \frac{2}{3} = \lambda \times \frac{1}{9}.$$

This simplifies to

$$\frac{10}{3} = \frac{\lambda}{9}.$$

Multiplying both sides by $$9$$ to isolate $$\lambda$$, we obtain

$$\lambda = 9 \times \frac{10}{3} = 3 \times 10 = 30.$$

So, the answer is $$30$$.

Let us denote by $$A$$ the set of people who read newspaper A and by $$B$$ the set of people who read newspaper B.

We are told that

$$\text{Percentage in }A = 63\%,$$

$$\text{Percentage in }B = 76\%,$$

and we wish to find a possible value of

$$x\% = \text{Percentage in }A\cap B,$$

that is, the percentage of people who read both newspapers.

We begin with the well-known principle of inclusion-exclusion for two sets. It states

$$|A\cup B| = |A| + |B| - |A\cap B|.$$

Replacing absolute values by percentages, we can write

$$\underbrace{($$ percentage that read at least one of A or B $$)}_{\le 100\%} = 63\% + 76\% - x\%.$$

Because a percentage cannot exceed $$100\%,$$ the left side is at most $$100\%.$$ Hence we obtain the inequality

$$63 + 76 - x \le 100.$$

Simplifying, we have

$$139 - x \le 100,$$

so

$$-x \le 100 - 139 = -39,$$

and multiplying both sides by $$-1$$ (which reverses the inequality sign) gives

$$x \ge 39.$$

Therefore the minimum possible value of $$x$$ is $$39\%.$$ Any smaller value would force the union $$A\cup B$$ above $$100\%,$$ which is impossible.

Next, observe that the intersection cannot be larger than the smaller of the two individual percentages, because one cannot have more people in both sets than are present in a single set. Formally,

$$x \le \min(63,\,76) = 63.$$

Putting the two bounds together, the percentage $$x$$ has to satisfy

$$39 \le x \le 63.$$

Now we test the four answer choices:

A. $$29\%$$    (which is < 39)  →  impossible, reject.

B. $$37\%$$    (which is < 39)  →  impossible, reject.

C. $$65\%$$    (which is > 63)  →  impossible, reject.

D. $$55\%$$    (which lies between 39 and 63)  →  possible, accept.

Hence, the correct answer is Option D.

We have $$10$$ different balls and $$4$$ distinct boxes. When a single ball is dropped, it can go into any one of the four boxes, so for each ball there are $$4$$ choices. Hence, by the Fundamental Principle of Counting, the total number of possible arrangements of the ten balls is

$$4^{10}.$$

Next we count the favourable arrangements in which some two boxes end up with exactly $$2$$ and $$3$$ balls respectively. All other balls can be distributed in any manner among the remaining two boxes.

First, choose which box will finally contain exactly $$2$$ balls. There are $$4$$ possible boxes. After that, choose which different box will finally contain exactly $$3$$ balls. Now only $$3$$ boxes are left, so altogether

$$4 \times 3 = 12$$

ways of selecting the two special boxes.

From the $$10$$ distinct balls we must now pick the two balls that will go into the first chosen box. Stating the combination formula, for selecting $$r$$ objects from $$n$$ distinct objects we use

$$^nC_r \;=\; \frac{n!}{r!(\,n-r\,)!}.$$

Applying it, the number of ways of choosing those two balls is

$$^{10}C_{2}= \frac{10!}{2!\,8!}= \frac{10\times9}{2}=45.$$

After removing these two balls, $$8$$ balls remain. We now select the three balls that will go into the second chosen box:

$$^{8}C_{3}= \frac{8!}{3!\,5!}= \frac{8\times7\times6}{6}=56.$$

Five balls are still left. Each of these five balls can independently be placed in either of the two remaining boxes. Using the rule “each of the $$k$$ objects has $$m$$ choices” we get

$$2^{5}=32$$

ways to arrange the last five balls.

Putting all factors together, the total number of favourable arrangements is

$$ \underbrace{4\times3}_{\text{choice of boxes}} \times \underbrace{^{10}C_{2}}_{\text{choose 2 balls}} \times \underbrace{^{8}C_{3}}_{\text{choose 3 balls}} \times \underbrace{2^{5}}_{\text{arrange last 5 balls}} $$

Substituting the numerical values, we have

$$ 12 \times 45 \times 56 \times 32. $$

Let us multiply step by step:

$$45 \times 56 = 2520,$$

$$12 \times 2520 = 30240,$$

$$30240 \times 32 = 967680.$$

Therefore, the number of favourable cases is $$967\,680.$$

The required probability is the ratio of favourable cases to total cases:

$$ \text{Probability}= \frac{967\,680}{4^{10}}. $$

Now we simplify the denominator. Since $$4 = 2^{2},$$ we have

$$4^{10} = (2^{2})^{10}=2^{20}.$$

So

$$ \text{Probability}= \frac{967\,680}{2^{20}}. $$

To reduce the fraction, notice that $$2^{10}=1024.$$ We divide the numerator and the denominator simultaneously by this factor:

$$ \frac{967\,680}{2^{20}} = \frac{967\,680 \div 1024}{2^{20}\div 1024} = \frac{945}{2^{10}}. $$

No further cancellation is possible, so the simplified probability is

$$\frac{945}{2^{10}}.$$

This matches Option C.

Hence, the correct answer is Option C.

Let us denote by $$A$$ the event “the sum of the two throws is a multiple of 4” and by $$B$$ the event “the score 4 appears at least once”. We wish to find the conditional probability $$P(B\mid A)$$, that is, the probability of $$B$$ given that $$A$$ has already occurred.

By definition of conditional probability we have

$$P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}.$$

First we list every ordered pair $$(x,y)$$, where $$x$$ is the first throw and $$y$$ is the second throw, that belongs to $$A$$. A sum is a multiple of 4 precisely when it equals $$4,8$$ or $$12$$, so we examine these three cases one by one.

For the sum $$4$$:

$$x+y=4\;\Rightarrow\;(x,y)=(1,3),(2,2),(3,1).$$

Hence there are $$3$$ favourable ordered pairs here.

For the sum $$8$$:

$$x+y=8\;\Rightarrow\;(x,y)=(2,6),(3,5),(4,4),(5,3),(6,2).$$

This time we obtain $$5$$ favourable ordered pairs.

For the sum $$12$$:

$$x+y=12\;\Rightarrow\;(x,y)=(6,6).$$

So we get $$1$$ favourable ordered pair in this case.

Summing up all three cases, the total number of elementary outcomes in event $$A$$ is

$$3+5+1=9.$$

Since each ordered pair of two fair die throws is equally likely out of the complete sample space of $$36$$ pairs, we find

$$P(A)=\dfrac{9}{36}=\dfrac{1}{4}.$$

Now we need $$A\cap B$$, the ordered pairs whose sum is a multiple of $$4$$ and that contain at least one $$4$$. Looking back at our list, a score of $$4$$ appears only in the ordered pair $$(4,4)$$ (it belongs to the sum $$8$$ group). No other pair inside $$A$$ contains a $$4$$. Therefore only one outcome satisfies both $$A$$ and $$B$$:

$$(4,4).$$

Consequently, the number of elementary outcomes in $$A\cap B$$ is $$1$$, so

$$P(A\cap B)=\dfrac{1}{36}.$$

Substituting the values of $$P(A\cap B)$$ and $$P(A)$$ into the conditional probability formula, we obtain

$$P(B\mid A)=\dfrac{\dfrac{1}{36}}{\dfrac{1}{4}}=\dfrac{1}{36}\times\dfrac{4}{1}=\dfrac{4}{36}=\dfrac{1}{9}.$$

Hence, the correct answer is Option D.

For a discrete random variable the very first property we use is: “The sum of all the probabilities is equal to 1.” Stated mathematically, if the possible values of $$X$$ are $$x_1,x_2,\dots ,x_n$$ with respective probabilities $$p_1,p_2,\dots ,p_n,$$ then

$$\displaystyle \sum_{i=1}^{n} p_i \;=\;1.$$

Here the values and probabilities are given as

$$\begin{aligned} X &: \;1 \quad 2 \quad 3 \quad 4 \quad 5,\\[2mm] P(X) &: \;k^{2}\quad 2k \quad k \quad 2k \quad 5k^{2}. \end{aligned}$$

Applying the above property, we write

$$k^{2}+2k+k+2k+5k^{2}=1.$$

Now we collect like terms:

$$\bigl(k^{2}+5k^{2}\bigr)+\bigl(2k+k+2k\bigr)=1,$$

$$6k^{2}+5k=1.$$

Next we transpose the 1 to the left side so that every term lies on the same side of the equality:

$$6k^{2}+5k-1=0.$$

This is a quadratic equation in $$k$$ of the standard form $$ax^{2}+bx+c=0.$$ We now invoke the quadratic-formula, which states

$$x=\dfrac{-b\pm\sqrt{\,b^{2}-4ac\,}}{2a},\quad\text{for}\;ax^{2}+bx+c=0.$$

With $$a=6,\;b=5,\;c=-1$$ we substitute:

$$k=\frac{-5\pm\sqrt{\,5^{2}-4\cdot6\cdot(-1)\,}}{2\cdot6}.$$

Inside the square root we simplify step by step:

$$5^{2}=25,\quad -4\cdot6\cdot(-1)=24,$$

so

$$\sqrt{\,25+24\,}=\sqrt{49}=7.$$

Hence

$$k=\frac{-5\pm7}{12}.$$

This yields two numerical possibilities:

$$k=\frac{-5+7}{12}=\frac{2}{12}=\frac16,\qquad k=\frac{-5-7}{12}=\frac{-12}{12}=-1.$$

Because probability values must be non-negative, we discard the negative root. Therefore

$$k=\frac16.$$

We now proceed to the requested probability $$P(X>2).$$ The condition $$X>2$$ corresponds to the outcomes $$X=3,4,5.$$ Using the definition of probability for mutually exclusive events, we add their individual probabilities:

$$P(X>2)=P(X=3)+P(X=4)+P(X=5).$$

Substituting the expressions given in the table, we get

$$P(X>2)=k+2k+5k^{2}.$$

First we combine like terms:

$$k+2k=3k,$$

so

$$P(X>2)=3k+5k^{2}.$$

Now we replace $$k$$ by the value already obtained, $$k=\dfrac16$$:

$$P(X>2)=3\left(\frac16\right)+5\left(\frac16\right)^{2}.$$

We simplify each term separately. For the linear term,

$$3\left(\frac16\right)=\frac36=\frac12.$$

For the quadratic term, first square the denominator: $$\left(\frac16\right)^{2}=\frac1{36}.$$ Then multiply by 5:

$$5\left(\frac1{36}\right)=\frac5{36}.$$

Adding the two fractions with a common denominator of 36, we have

$$\frac12+\frac5{36}=\frac{18}{36}+\frac5{36}=\frac{23}{36}.$$

Thus

$$P(X>2)=\frac{23}{36}.$$

Consulting the options given, the fraction $$\dfrac{23}{36}$$ corresponds to Option D.

Hence, the correct answer is Option D.

We begin by recalling the fundamental fact that tossing an unbiased coin five times produces $$2^{5}=32$$ equally likely sequences.

The random variable $$X$$ takes only four possible values:

$$X=5$$ if the sequence contains five consecutive heads,

$$X=4$$ if it contains four but not five consecutive heads,

$$X=3$$ if it contains exactly three consecutive heads and never four or five in a row,

$$X=-1$$ in every other case.

We now find the probability of each of these four mutually exclusive events.

1. Probability that $$X=5$$

Five consecutive heads can occur in only one way, namely the single sequence $$\text{HHHHH}$$. Thus

$$P(X=5)=\frac{1}{32}.$$

2. Probability that $$X=4$$

We need four heads in a row but not five. The block of four heads can start either at the first toss or at the second toss.

• If it starts at toss 1 we have $$\text{HHHHT}.$$

    • If it starts at toss 2 we have $$\text{THHHH}.$$

There are exactly two such sequences and none of them equals the string of five heads already counted, so

$$P(X=4)=\frac{2}{32}=\frac{1}{16}.$$

3. Probability that $$X=3$$

We require a block of exactly three consecutive heads while forbidding any block of four or five. Let the block HHH start at position k.

    (i) k = 1  ⇒  pattern $$\text{HHH}x_{4}x_{5}.$$
To avoid four heads, $$x_{4}=\text{T}.$$ The fifth toss can be either H or T, giving the two strings $$\text{HHHTH},\; \text{HHHTT}.$$

    (ii) k = 2  ⇒  pattern $$x_{1}\text{HHH}x_{5}.$$
Here $$x_{1}=\text{T}$$ (otherwise we would have four heads) and $$x_{5}=\text{T}$$ for the same reason, yielding the single string $$\text{THHHT}.$$

    (iii) k = 3  ⇒  pattern $$x_{1}x_{2}\text{HHH}.$$
To avoid a block of four heads, $$x_{2}=\text{T}.$$ The first toss $$x_{1}$$ can be H or T, giving $$\text{HTHHH},\; \text{TTHHH}.$$

Counting them all, we have

$$5$$ sequences: \;$$$\text{HHHTH},\, \text{HHHTT},\, \text{THHHT},\, \text{HTHHH}$$$, $$\text{TTHHH}.$$

Thus

$$P(X=3)=\frac{5}{32}.$$

4. Probability that $$X=-1$$

This is simply the complement of the three cases already counted:

$$$P(X=-1)=1-\Bigl[\frac{1}{32}+\frac{2}{32}+\frac{5}{32}\Bigr] =1-\frac{8}{32} =\frac{24}{32} =\frac{3}{4}.$$$

5. Expected value of $$X$$

By definition, the expectation of a discrete random variable is $$E[X]=\sum x_{i}\,P(X=x_{i}).$$ Substituting our four values we get

$$$ \begin{aligned} E[X] &= 5\!\left(\frac{1}{32}\right) +4\!\left(\frac{2}{32}\right) +3\!\left(\frac{5}{32}\right) +(-1)\!\left(\frac{24}{32}\right) \\[6pt] &= \frac{5}{32}+\frac{8}{32}+\frac{15}{32}-\frac{24}{32} \\[6pt] &= \frac{28-24}{32} \\[6pt] &= \frac{4}{32} \\[6pt] &= \frac{1}{8}. \end{aligned} $$$

Hence, the correct answer is Option B.

We are asked to find the probability that the chosen card came from Box 1 after we have been told that the number on the card is a non-prime. This is a question of conditional probability, so we begin by recalling Bayes’ theorem.

Bayes’ theorem for two events $$A$$ and $$B$$ is stated as

$$P(A\mid B)=\dfrac{P(B\mid A)\,P(A)}{P(B\mid A)\,P(A)+P(B\mid A^{\!\!c})\,P(A^{\!\!c})}.$$

In the present situation:

• Event $$A$$ is “the card is drawn from Box 1.”
• Event $$B$$ is “the number on the card is non-prime.”
• Event $$A^{\!\!c}$$ is “the card is drawn from Box 2.”

Thus we want $$P(A\mid B)$$, the probability that the card came from Box 1 given that it is non-prime.

First we note that the choice between the two boxes is made at random, so

$$P(A)=\dfrac12,\qquad P(A^{\!\!c})=\dfrac12.$$

Next we must compute $$P(B\mid A)$$, the probability that the number is non-prime when the card is taken from Box 1. Box 1 contains the integers from $$1$$ to $$30$$.

We list all primes up to $$30$$:

$$2,\,3,\,5,\,7,\,11,\,13,\,17,\,19,\,23,\,29.$$

There are $$10$$ such primes. Since Box 1 has $$30$$ cards altogether, the number of non-primes (including the number $$1$$ and all composites) is

$$30-10=20.$$

Hence

$$P(B\mid A)=\dfrac{20}{30}=\dfrac23.$$

Now we find $$P(B\mid A^{\!\!c})$$, the probability that the number is non-prime when the card comes from Box 2. Box 2 contains the integers from $$31$$ to $$50$$, a total of $$20$$ numbers.

We list the primes between $$31$$ and $$50$$:

$$31,\,37,\,41,\,43,\,47.$$

There are $$5$$ such primes. Therefore the count of non-primes in Box 2 is

$$20-5=15.$$

So

$$P(B\mid A^{\!\!c})=\dfrac{15}{20}=\dfrac34.$$

We have gathered all the ingredients needed for Bayes’ theorem. Substituting into the formula gives

$$P(A\mid B)=\dfrac{\,P(B\mid A)\,P(A)\,}{P(B\mid A)\,P(A)+P(B\mid A^{\!\!c})\,P(A^{\!\!c})}=\dfrac{\left(\dfrac23\right)\!\left(\dfrac12\right)}{\left(\dfrac23\right)\!\left(\dfrac12\right)+\left(\dfrac34\right)\!\left(\dfrac12\right)}.$$

We simplify the numerator first:

$$\left(\dfrac23\right)\!\left(\dfrac12\right)=\dfrac{2}{3}\times\dfrac{1}{2}=\dfrac{2}{6}=\dfrac13.$$

Now we simplify each part of the denominator:

First term in the denominator:

$$\left(\dfrac23\right)\!\left(\dfrac12\right)=\dfrac13$$

Second term in the denominator:

$$\left(\dfrac34\right)\!\left(\dfrac12\right)=\dfrac{3}{4}\times\dfrac{1}{2}=\dfrac{3}{8}.$$

Add the two terms to complete the denominator:

$$\dfrac13+\dfrac38=\dfrac{8}{24}+\dfrac{9}{24}=\dfrac{17}{24}.$$

Thus the conditional probability becomes

$$P(A\mid B)=\dfrac{\dfrac13}{\dfrac{17}{24}}=\dfrac13\times\dfrac{24}{17}=\dfrac{24}{51}=\dfrac{8}{17}.$$

Hence, the correct answer is Option B.

We begin by noting that at every draw the probability of getting an $$A$$ is $$\dfrac{10}{20}= \dfrac12$$ and, similarly, the probability of getting a $$B$$ is also $$\dfrac12$$. Because the card is replaced after each draw, each trial is independent and the two probabilities remain constant throughout the experiment.

We have to keep drawing cards until the second $$A$$ appears. During this process a third $$B$$ may or may not appear. We are asked to find the probability that the second $$A$$ comes before the third $$B$$. In other words, we must calculate the probability that two $$A$$’s are obtained earlier than three $$B$$’s in an infinite sequence of independent trials with $$P(A)=P(B)=\dfrac12$$.

To attack such “race” problems systematically, we introduce a state variable. Let

$$$P(i,j)=\text{Probability that the second }A\text{ beats the third }B\text{ when }i\text{ $$$A$$’s and }j\text{ $$B$$’s have already been drawn.}$$

At the start of the experiment no card has been recorded, so we ultimately want $$P(0,0)$$.

There are obvious boundary states:

• If we ever reach $$i=2$$ (two $$A$$’s) we have already succeeded, so $$P(2,j)=1\quad\text{for }j=0,1,2.$$

• If we ever reach $$j=3$$ (three $$B$$’s) before the second $$A$$, we have failed, so $$P(i,3)=0\quad\text{for }i=0,1.$$

For an interior state $$(i,j)$$ with $$i\lt 2$$ and $$j\lt 3$$, the next draw can be either $$A$$ or $$B$$. Using the Law of Total Probability we write the recursion

$$$P(i,j)=\dfrac12\;P(i+1,\,j)\;+\;\dfrac12\;P(i,\,j+1).$$$

Now we solve these equations step by step, beginning with states closest to the boundaries and working backwards to $$(0,0).$$

1. State $$(1,2)$$
Exactly one $$A$$ and two $$B$$’s have already appeared.
Next draw:

• $$A$$ with probability $$\dfrac12$$ $$\Longrightarrow$$ state $$(2,2)$$ (success).
• $$B$$ with probability $$\dfrac12$$ $$\Longrightarrow$$ state $$(1,3)$$ (failure).

Hence $$P(1,2)=\dfrac12(1)+\dfrac12(0)=\dfrac12.$$

2. State $$(0,2)$$
No $$A$$ yet, two $$B$$’s so far.
Next draw:

• $$A$$ $$\Longrightarrow$$ $$(1,2)$$.
• $$B$$ $$\Longrightarrow$$ $$(0,3)$$ (failure).

Thus $$$P(0,2)=\dfrac12\,P(1,2)+\dfrac12\,(0)=\dfrac12\left(\dfrac12\right)=\dfrac14.$$$

3. State $$(1,1)$$
One $$A$$ and one $$B$$ so far.
Next draw:

• $$A$$ $$\Longrightarrow$$ $$(2,1)$$ (success).
• $$B$$ $$\Longrightarrow$$ $$(1,2)$$.

Hence $$$P(1,1)=\dfrac12(1)+\dfrac12\left(\dfrac12\right)=\dfrac12+\dfrac14=\dfrac34.$$$

4. State $$(0,1)$$
No $$A$$ and one $$B$$ so far.
Next draw:

• $$A$$ $$\Longrightarrow$$ $$(1,1)$$.
• $$B$$ $$\Longrightarrow$$ $$(0,2)$$.

Therefore $$$P(0,1)=\dfrac12\left(\dfrac34\right)+\dfrac12\left(\dfrac14\right)=\dfrac38+\dfrac18=\dfrac12.$$$

5. State $$(1,0)$$
One $$A$$ and no $$B$$ yet.
Next draw:

• $$A$$ $$\Longrightarrow$$ $$(2,0)$$ (success).
• $$B$$ $$\Longrightarrow$$ $$(1,1)$$.

Thus $$$P(1,0)=\dfrac12(1)+\dfrac12\left(\dfrac34\right)=\dfrac12+\dfrac38=\dfrac78.$$$

6. Starting state $$(0,0)$$
No $$A$$ and no $$B$$ have been drawn yet.
Next draw:

• $$A$$ $$\Longrightarrow$$ $$(1,0)$$.
• $$B$$ $$\Longrightarrow$$ $$(0,1)$$.

Therefore $$$P(0,0)=\dfrac12\left(\dfrac78\right)+\dfrac12\left(\dfrac12\right)=\dfrac{7}{16}+\dfrac{4}{16}=\dfrac{11}{16}.$$$

So the probability that the second $$A$$ appears before the third $$B$$ is $$\dfrac{11}{16}$$.

Hence, the correct answer is Option B.

Each throw of a pair of fair dice has $$36$$ equally likely outcomes. We first list the numbers of favourable outcomes for the two critical totals.

$$\displaystyle \text{Sum }6 : (1,5),(2,4),(3,3),(4,2),(5,1)$$  $$\Rightarrow$$ $$5$$ ways $$\displaystyle \text{Sum }7 : (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$$  $$\Rightarrow$$ $$6$$ ways

So we have

$$p=\Pr(\text{sum }6)=\frac{5}{36},\qquad q=\Pr(\text{sum }7)=\frac{6}{36}=\frac16,\qquad r=\Pr(\text{neither }6\text{ nor }7)=1-(p+q)=1-\frac{11}{36}=\frac{25}{36}.$$

A starts the game. The rules are:

• A wins the moment he throws a $$6$$. • B wins the moment he throws a $$7$$. • Any other result leaves the game undecided and the next player throws. • A result of $$7$$ on A’s throw or $$6$$ on B’s throw has no special effect.

On A’s turn the probability that he wins immediately is simply

$$p=\frac{5}{36}.$$

If A fails to throw a $$6$$ (probability $$1-p=\frac{31}{36}$$), B now gets a turn. On B’s turn the probability that he wins immediately is

$$q=\frac{6}{36}=\frac16.$$

If B also fails to win (probability $$1-q=\frac{30}{36}$$), the game returns to exactly the same situation that existed just before A’s first throw: nobody has yet won and it is again A’s turn. We call the pair of throws “A then B” a round.

The probability that a whole round passes without a winner is therefore

$$s=(1-p)(1-q)=\frac{31}{36}\times\frac{30}{36}=\frac{930}{1296}=\frac{155}{216}.$$

The probability that A wins in the first round is $$p$$. If the first round is neutral (probability $$s$$) and the game enters a second round, A can again win on his next throw with probability $$p$$, and so on. Thus the overall probability that A finally wins is the infinite geometric sum

$$\Pr(A\text{ wins})=p\bigl(1+s+s^{2}+s^{3}+\dots\bigr).$$

For an infinite geometric series with first term $$1$$ and common ratio $$s$$ ($$|s|\lt 1$$), the sum is $$\dfrac{1}{1-s}$$. Stating the formula:

$$1+s+s^{2}+s^{3}+\dots=\frac{1}{1-s}.$$ Substituting $$p=\dfrac{5}{36}$$ and $$s=\dfrac{155}{216}$$:

$$\Pr(A\text{ wins})=\frac{5}{36}\times\frac{1}{1-\frac{155}{216}} =\frac{5}{36}\times\frac{1}{\frac{61}{216}} =\frac{5}{36}\times\frac{216}{61} =\frac{5\times216}{36\times61} =\frac{5\times6}{61} =\frac{30}{61}.$$

Hence, the correct answer is Option D.

We are told that each of the five machines fails independently and that for any single machine the probability of being out of service on a particular day is $$\frac14$$. Therefore, for one machine the probability of being in service is $$1-\frac14=\frac34$$.

If we let the random variable $$X$$ denote the number of machines that are out of service on a given day, then $$X$$ follows a binomial distribution with parameters $$n=5$$ and $$p=\frac14$$. The general binomial probability formula is

$$P(X=r)=\binom{n}{r}\,p^{\,r}\,(1-p)^{\,n-r}.$$

Here the phrase “at most two machines will be out of service’’ means the event $$X\le 2$$, i.e. $$X=0,1,2$$. Thus we need

$$P(X\le 2)=P(X=0)+P(X=1)+P(X=2).$$

We now evaluate each of these three terms one by one.

For $$r=0$$ machines out of service we have

$$P(X=0)=\binom{5}{0}\Bigl(\tfrac14\Bigr)^{0}\Bigl(\tfrac34\Bigr)^{5} =1\cdot1\cdot\Bigl(\tfrac34\Bigr)^{5} =\Bigl(\tfrac34\Bigr)^{5}.$$

For $$r=1$$ machine out of service we have

$$P(X=1)=\binom{5}{1}\Bigl(\tfrac14\Bigr)^{1}\Bigl(\tfrac34\Bigr)^{4} =5\cdot\Bigl(\tfrac14\Bigr)\cdot\Bigl(\tfrac34\Bigr)^{4}.$$

For $$r=2$$ machines out of service we have

$$P(X=2)=\binom{5}{2}\Bigl(\tfrac14\Bigr)^{2}\Bigl(\tfrac34\Bigr)^{3} =10\cdot\Bigl(\tfrac14\Bigr)^{2}\cdot\Bigl(\tfrac34\Bigr)^{3}.$$

Adding these three results gives the total probability:

$$ \begin{aligned} P(X\le 2) &=\Bigl(\tfrac34\Bigr)^{5} +5\Bigl(\tfrac14\Bigr)\Bigl(\tfrac34\Bigr)^{4} +10\Bigl(\tfrac14\Bigr)^{2}\Bigl(\tfrac34\Bigr)^{3}.\\ \end{aligned} $$

Because the final answer must be expressed in the form $$\bigl(\tfrac34\bigr)^{3}k$$, we factor $$\bigl(\tfrac34\bigr)^{3}$$ from every term:

$$ \begin{aligned} P(X\le 2) &=\Bigl(\tfrac34\Bigr)^{3} \left[ \Bigl(\tfrac34\Bigr)^{2} +5\Bigl(\tfrac14\Bigr)\Bigl(\tfrac34\Bigr) +10\Bigl(\tfrac14\Bigr)^{2} \right]. \end{aligned} $$

We now simplify each bracketed expression separately.

First term inside brackets:

$$\Bigl(\tfrac34\Bigr)^{2}=\frac{9}{16}.$$

Second term inside brackets:

$$5\Bigl(\tfrac14\Bigr)\Bigl(\tfrac34\Bigr) =5\cdot\frac14\cdot\frac34 =5\cdot\frac{3}{16} =\frac{15}{16}.$$

Third term inside brackets:

$$10\Bigl(\tfrac14\Bigr)^{2} =10\cdot\frac{1}{16} =\frac{10}{16} =\frac{5}{8}.$$

Now we add these three fractions, being careful to use a common denominator of 16:

$$ \begin{aligned} \frac{9}{16} + \frac{15}{16} + \frac{5}{8} &=\frac{9}{16} + \frac{15}{16} + \frac{10}{16}\\ &=\frac{9+15+10}{16}\\ &=\frac{34}{16}\\ &=\frac{17}{8}. \end{aligned} $$

Therefore

$$P(X\le 2)=\Bigl(\tfrac34\Bigr)^{3}\cdot\frac{17}{8}.$$

Comparing with the required form $$\Bigl(\frac34\Bigr)^{3}k$$ we see directly that

$$k=\frac{17}{8}.$$

Hence, the correct answer is Option A.

Let us denote the probability of an event by the symbol $$\text P(\,\cdot\,)$$.

We have two events $$A$$ and $$B$$. The statement “exactly one of them occurs” refers to the event in which either $$A$$ happens alone or $$B$$ happens alone, but not both together. In probability notation, this “exactly one” event is $$A \triangle B=(A\setminus B)\cup(B\setminus A)$$, and its probability is known to be $$\frac{2}{5}$$, that is

$$\text P(A\triangle B)=\frac{2}{5}.$$

Next, the phrase “$$A$$ or $$B$$ occurs” means the union $$A\cup B$$. Its probability is given as $$\frac12$$, so we know

$$\text P(A\cup B)=\frac12.$$

We now translate the verbal conditions into algebraic equations involving the unknown probability $$\text P(A\cap B)$$, which is what we want to find.

First, recall the relation that expresses the probability of the union of two events:

$$\text P(A\cup B)=\text P(A)+\text P(B)-\text P(A\cap B).$$

Second, recall the formula for the probability that exactly one of the two events occurs. Exactly one occurs when we add the individual probabilities and then subtract twice the overlap (because the overlap contributes to both $$A$$ and $$B$$ but must be excluded entirely in the “exactly one” count). Thus,

$$\text P(A\triangle B)=\text P(A)+\text P(B)-2\,\text P(A\cap B).$$

Let us set

$$x=\text P(A)+\text P(B)\quad\text{and}\quad y=\text P(A\cap B).$$

The two factual statements now become a pair of linear equations:

1. From the “exactly one” condition $$x-2y=\frac25.$$

2. From the “union” condition $$x-y=\frac12.$$

We have a very small system of two equations in the two unknowns $$x$$ and $$y$$:

$$\begin{aligned} x-2y &=\frac25,\\ x-y &=\frac12. \end{aligned}$$

To isolate $$y$$, subtract the first equation from the second. Performing that subtraction step by step, we have

$$\bigl(x-y\bigr)-\bigl(x-2y\bigr)=\frac12-\frac25.$$

Expanding the left‐hand side, the $$x$$ terms cancel and we get

$$-y+2y\;=\;y.$$

On the right‐hand side, put both fractions over the common denominator $$10$$:

$$\frac12=\frac5{10},\quad\frac25=\frac4{10},\quad\text{so}\quad\frac12-\frac25=\frac5{10}-\frac4{10}=\frac1{10}.$$

Therefore, the entire subtraction gives

$$y=\frac1{10}.$$

But $$y=\text P(A\cap B)$$, so we have found

$$\text P(A\cap B)=\frac1{10}=0.10.$$

Hence, the correct answer is Option D.

We are given two independent events A and B with probabilities

$$P(A)=\frac13, \qquad P(B)=\frac16.$$

Because A and B are independent, the fundamental fact we shall use again and again is

$$P(A\cap B)=P(A)\,P(B).$$

Independence also carries over to complements; that is, A is independent of B′ and A′ is independent of both B and B′. Now we test each option one by one.

Option A asks for the conditional probability $$P\!\left(A\;|\;B\right).$$ The definition of conditional probability is

$$P\!\left(A\;|\;B\right)=\frac{P(A\cap B)}{P(B)}.$$

Substituting the values, we have

$$P(A\cap B)=P(A)\,P(B)=\frac13\cdot\frac16=\frac1{18},$$

so

$$P\!\left(A\;|\;B\right)=\frac{\dfrac1{18}}{\dfrac16}=\frac1{18}\times\frac61=\frac6{18}=\frac13.$$

The option claims $$\frac23,$$ therefore Option A is false.

Option B requires the conditional probability $$P\!\left(A\;|\;B'\right),$$ where B′ is the complement of B. First, evaluate $$P(B')$$:

$$P(B')=1-P(B)=1-\frac16=\frac56.$$

Because A is independent of B′, we write

$$P(A\cap B')=P(A)\,P(B')=\frac13\cdot\frac56=\frac5{18}.$$

Applying the conditional-probability formula,

$$P\!\left(A\;|\;B'\right)=\frac{P(A\cap B')}{P(B')}=\frac{\dfrac5{18}}{\dfrac56}=\frac5{18}\times\frac65=\frac6{18}=\frac13.$$

This exactly matches the value stated in Option B, so Option B is true.

Option C asks for $$P\!\left(A'\;|\;B'\right).$$ First compute $$P(A')$$:

$$P(A')=1-P(A)=1-\frac13=\frac23.$$

Since A′ and B′ are independent,

$$P(A'\cap B')=P(A')\,P(B')=\frac23\cdot\frac56=\frac{10}{18}=\frac59.$$

Then

$$P\!\left(A'\;|\;B'\right)=\frac{P(A'\cap B')}{P(B')}=\frac{\dfrac59}{\dfrac56}=\frac59\times\frac65=\frac{30}{45}=\frac23.$$

Option C claims $$\frac13,$$ so it is incorrect.

Option D involves $$P\!\left(A\;|\;A\cup B\right).$$ First, notice that

$$A\cap(A\cup B)=A,$$

so the numerator of the conditional probability is simply $$P(A)=\frac13.$$ Next we need $$P(A\cup B).$$ Using the addition rule,

$$P(A\cup B)=P(A)+P(B)-P(A\cap B).$$

We already have $$P(A\cap B)=\frac1{18},$$ hence

$$P(A\cup B)=\frac13+\frac16-\frac1{18}=\frac6{18}+\frac3{18}-\frac1{18}=\frac8{18}=\frac49.$$

Now compute the conditional probability:

$$P\!\left(A\;|\;A\cup B\right)=\frac{P(A)}{P(A\cup B)}=\frac{\dfrac13}{\dfrac49}=\frac13\times\frac94=\frac{9}{12}=\frac34.$$

The option lists $$\frac14,$$ so Option D is false.

Only Option B agrees with the actual calculation.

Hence, the correct answer is Option 2.

We have to determine the conditional probability $$P\!\left((E_2^{\,C}\cap E_3^{\,C})\,\big/\,E_1\right)\!.$$

First recall the definition of conditional probability:

$$P(A/B)=\dfrac{P(A\cap B)}{P(B)},\qquad\text{provided }P(B)>0.$$

Taking $$A=(E_2^{\,C}\cap E_3^{\,C})\quad\text{and}\quad B=E_1,$$ we obtain

$$P\!\left((E_2^{\,C}\cap E_3^{\,C})\big/ E_1\right)=\dfrac{P\!\left(E_1\cap E_2^{\,C}\cap E_3^{\,C}\right)}{P(E_1)}.$$

Hence our immediate task is to evaluate $$P\!\left(E_1\cap E_2^{\,C}\cap E_3^{\,C}\right).$$ Observe that

$$E_1\cap E_2^{\,C}\cap E_3^{\,C}=E_1\setminus(E_2\cup E_3),$$

so by the subtraction rule of probability,

$$P\!\left(E_1\cap E_2^{\,C}\cap E_3^{\,C}\right)=P(E_1)-P\!\left(E_1\cap(E_2\cup E_3)\right).$$

Using the formula for the probability of a union,

$$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y),$$

with $$X=E_1\cap E_2,\;Y=E_1\cap E_3,$$ we have

$$P\!\left(E_1\cap(E_2\cup E_3)\right)=P(E_1\cap E_2)+P(E_1\cap E_3)-P(E_1\cap E_2\cap E_3).$$

The problem states that $$P(E_1\cap E_2\cap E_3)=0,$$ so the last term vanishes. Therefore

$$P\!\left(E_1\cap(E_2\cup E_3)\right)=P(E_1\cap E_2)+P(E_1\cap E_3).$$

Because the events are pairwise independent, we can employ the independence property $$P(E_i\cap E_j)=P(E_i)\,P(E_j)$$ for every pair. Thus

$$P(E_1\cap E_2)=P(E_1)P(E_2),\qquad P(E_1\cap E_3)=P(E_1)P(E_3).$$

Substituting into the previous expression,

$$P\!\left(E_1\cap(E_2\cup E_3)\right)=P(E_1)P(E_2)+P(E_1)P(E_3).$$

Now place this back into the earlier subtraction expression:

$$P\!\left(E_1\cap E_2^{\,C}\cap E_3^{\,C}\right)=P(E_1)-\big[P(E_1)P(E_2)+P(E_1)P(E_3)\big].$$

Factoring out the common term $$P(E_1)$$ yields

$$P\!\left(E_1\cap E_2^{\,C}\cap E_3^{\,C}\right)=P(E_1)\Bigl[1-P(E_2)-P(E_3)\Bigr].$$

Returning to the conditional probability and dividing by $$P(E_1)\;(\text{which is}>0),$$ we get

$$P\!\left((E_2^{\,C}\cap E_3^{\,C})/E_1\right)=\dfrac{P(E_1)\bigl[1-P(E_2)-P(E_3)\bigr]}{P(E_1)}=1-P(E_2)-P(E_3).$$

Noting that the complement of $$E_3$$ has probability $$P(E_3^{\,C})=1-P(E_3),$$ we can rewrite the expression as

$$1-P(E_2)-P(E_3)=\bigl[1-P(E_3)\bigr]-P(E_2)=P(E_3^{\,C})-P(E_2).$$

This matches option D in the list provided.

Hence, the correct answer is Option D.

Let us denote the 11 consecutive natural numbers by $$1,2,3,\dots ,11$$. Because only the relative positions matter, any other block of 11 consecutive numbers would give exactly the same count, so this labelling is allowed.

The total number of unordered selections of three numbers from these 11 is given by the combination formula

$$^nC_r=\frac{n!}{r!\,(n-r)!},$$

where $$n=11$$ and $$r=3$$. Substituting, we obtain

$$^{11}C_3=\frac{11!}{3!\,8!}=\frac{11\times10\times9}{3\times2\times1}=165.$$

Now we must count how many of those 165 triplets form an increasing arithmetic progression (A.P.) with positive common difference. Suppose the first term of such a progression is $$x$$ and the common difference is $$d$$, where $$d\ge 1$$. Then the three terms are

$$x,\;x+d,\;x+2d.$$

All three terms must lie in the set $$\{1,2,\dots ,11\}$$, so the largest term must satisfy

$$x+2d\le 11.$$

For a given positive integer $$d$$, the smallest permissible value of $$x$$ is $$1$$, and the largest is $$11-2d$$. Hence, for that particular $$d$$, the number of admissible values of $$x$$ equals

$$11-2d.$$

This quantity is positive only as long as $$11-2d\ge 1\; \Longrightarrow\; 2d\le 10\; \Longrightarrow\; d\le 5.$$ Therefore $$d$$ can take the integer values $$1,2,3,4,5$$. We now list the counts for each $$d$$ and add them.

For $$d=1$$: admissible $$x=1\text{ to }9$$, giving $$11-2(1)=9$$ progressions.
For $$d=2$$: admissible $$x=1\text{ to }7$$, giving $$11-2(2)=7$$ progressions.
For $$d=3$$: admissible $$x=1\text{ to }5$$, giving $$11-2(3)=5$$ progressions.
For $$d=4$$: admissible $$x=1\text{ to }3$$, giving $$11-2(4)=3$$ progressions.
For $$d=5$$: admissible $$x=1\text{ to }1$$, giving $$11-2(5)=1$$ progression.

Adding these gives the total number of favourable triplets:

$$9+7+5+3+1 = 25.$$

The required probability is therefore

$$\text{Probability}=\frac{\text{favourable cases}}{\text{total cases}}=\frac{25}{165}.$$

Simplifying the fraction, we divide numerator and denominator by their highest common factor, which is $$5$$:

$$\frac{25}{165}=\frac{25\div5}{165\div5}=\frac{5}{33}.$$

Hence, the correct answer is Option C.

We have the individual probabilities

$$P(A)=0.6,\qquad P(B)=0.4,\qquad P(C)=0.5.$$

First, from the union of A and B we can determine their intersection. The addition (union) rule of two events is

$$P(A\cup B)=P(A)+P(B)-P(A\cap B).$$

Substituting the given numbers,

$$0.8 = 0.6 + 0.4 - P(A\cap B).$$

So

$$P(A\cap B)=0.6+0.4-0.8 = 0.2.$$

The triple intersection is given directly:

$$P(A\cap B\cap C)=0.2.$$

The intersection of A and C is also supplied:

$$P(A\cap C)=0.3.$$

Let us denote

$$\beta = P(B\cap C),\qquad \alpha = P(A\cup B\cup C).$$

To involve $$\beta$$ and $$\alpha$$ simultaneously, we invoke the general addition (union) rule for three events:

$$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C).$$

Substituting every known value step by step,

$$\alpha = 0.6 + 0.4 + 0.5 \;-\; 0.2 \;-\; 0.3 \;-\; \beta \;+\; 0.2.$$

Simplifying the numeric part carefully,

$$\alpha = 1.5 - 0.2 - 0.3 - \beta + 0.2 = 1.2 - \beta.$$

The problem states that $$\alpha$$ must satisfy

$$0.85 \le \alpha \le 0.95.$$

Replacing $$\alpha$$ by $$1.2-\beta$$, we obtain the compound inequality

$$0.85 \le 1.2 - \beta \le 0.95.$$

We solve each side separately.

Left-hand side:

$$0.85 \le 1.2 - \beta \;\;\Longrightarrow\;\; -\beta \le 1.2 - 0.85 = 0.35 \;\;\Longrightarrow\;\; \beta \le 0.35.$$

Right-hand side:

$$1.2 - \beta \le 0.95 \;\;\Longrightarrow\;\; -\beta \le 0.95 - 1.2 = -0.25 \;\;\Longrightarrow\;\; \beta \ge 0.25.$$

Combining the two inequalities we get

$$0.25 \le \beta \le 0.35.$$

Finally, we should check that this range is also permissible with elementary constraints such as

$$\beta \le P(B)=0.4,\qquad \beta \le P(C)=0.5,\qquad \beta \ge P(A\cap B\cap C)=0.2,$$

all of which are clearly satisfied by every value in the interval $$[0.25,0.35]$$.

Therefore $$\beta$$ must lie in the interval $$[0.25,0.35].$$

Among the given choices, this corresponds precisely to Option B.

Hence, the correct answer is Option B.

First we note that a 5-digit number runs from 10000 to 99999, so its left-most digit can be 1,2,…,9. Hence the total number of 5-digit numbers is

$$9\times10^4=90000.$$

Our task is to count how many of these 90000 numbers use exactly two distinct digits. Both chosen digits must appear at least once inside the 5 positions.

We begin by choosing the two digits. Choosing two different digits out of the ten possible digits 0-9 is done by the combination formula

$$\,^nC_r=\frac{n!}{r!(n-r)!},$$

so

$$\binom{10}{2}=45$$

unordered pairs of digits are possible. Now we split these 45 pairs into two convenient classes, depending on whether 0 is one of the chosen digits.

Class 1: Neither digit is 0. There are 9 non-zero digits, so the number of such unordered pairs is

$$\binom{9}{2}=36.$$

Fix any one of these 36 pairs, say {a,b} with a≠b and both a,b∈{1,…,9}. We form 5-length strings from {a,b} such that both a and b occur and the first digit is non-zero (already guaranteed here). Without any restriction each of the 5 positions could be chosen in 2 ways, giving

$$2^5=32$$

strings. From these 32 we must discard the two monochromatic strings aaaaa and bbbbb in which both digits do not appear. Hence

$$32-2=30$$

valid numbers arise from any fixed pair {a,b}. Multiplying by the 36 choices of the pair gives

$$36\times30=1080$$

valid 5-digit numbers in Class 1.

Class 2: Exactly one digit is 0. Here the pair is {0,b} with b∈{1,…,9}. There are

$$9$$

such pairs. For each pair the first digit of the number cannot be 0 (otherwise we would not have a 5-digit number), so the first digit is forced to be b. The remaining four positions can be filled with either 0 or b, but 0 must appear at least once among them so that both digits occur. For the last four places we have

$$2^4=16$$

total strings, of which exactly one, namely bbbb, contains no 0. Therefore the admissible strings for the last four places are

$$16-1=15.$$\

Thus each pair {0,b} contributes 15 numbers, and altogether Class 2 supplies

$$9\times15=135$$

valid 5-digit numbers.

Adding the two classes together, the number of favourable 5-digit numbers is

$$1080+135=1215.$$

The required probability is therefore

$$\frac{\text{favourable}}{\text{total}}=\frac{1215}{90000}.$$

We now simplify this fraction. Both numerator and denominator are divisible by 45, because

$$1215=45\times27,\qquad 90000=45\times2000.$$

So we get

$$\frac{1215}{90000}=\frac{27}{2000}.$$

To match the form given in the options we multiply numerator and denominator by 5:

$$\frac{27}{2000}=\frac{27\times5}{2000\times5}=\frac{135}{10000}=\frac{135}{10^4}.$$

This expression appears as Option A.

Hence, the correct answer is Option A.

We throw four fair dice in one go and we are interested in the event that “at least two dice show a three or a five”. Let us first analyse a single throw of the four dice and then extend the result to the 27 independent repetitions.

For any single die, a face can be $$1,2,3,4,5,6$$, all equally likely. The favourable faces for our event are $$3$$ or $$5$$, so for one die we have

$$P(\text{face is }3\text{ or }5)=\frac{2}{6}=\frac13.$$

Correspondingly,

$$P(\text{face is neither }3\text{ nor }5)=1-\frac13=\frac23.$$

Let the random variable $$X$$ denote “the number of dice (out of four) that show a 3 or a 5” in one single throw of the four dice. Because each die acts independently, $$X$$ follows a binomial distribution with parameters $$n=4$$ and $$p=\frac13$$. Symbolically,

$$X\sim\text{Binomial}(4,\tfrac13).$$

We need the probability that at least two dice among the four are favourable, i.e.

$$P(\,X\ge 2\,)=1-P(\,X=0\,)-P(\,X=1\,).$$

Using the binomial probability formula $$P(X=k)=\binom{n}{k}p^{\,k}(1-p)^{\,n-k},$$ we compute each term step-by-step:

First, $$P(X=0)=\binom{4}{0}\Bigl(\tfrac13\Bigr)^{0}\Bigl(\tfrac23\Bigr)^{4}=1\cdot1\cdot\Bigl(\tfrac23\Bigr)^{4}=\left(\frac23\right)^{4}=\frac{16}{81}.$$

Next, $$P(X=1)=\binom{4}{1}\Bigl(\tfrac13\Bigr)^{1}\Bigl(\tfrac23\Bigr)^{3}=4\cdot\frac13\cdot\left(\frac23\right)^{3}=4\cdot\frac13\cdot\frac{8}{27}=4\cdot\frac{8}{81}=\frac{32}{81}.$$

Now we combine these values:

$$P(X\ge 2)=1-\frac{16}{81}-\frac{32}{81}=1-\frac{48}{81}=\frac{33}{81}=\frac{11}{27}.$$

This probability represents the chance of a “success” (at least two favourable faces) in one four-dice throw. Let us denote this success probability by $$p_s=\frac{11}{27}.$$

We are told that the four dice are thrown independently a total of 27 times. Let the random variable $$Y$$ count “the number of throws (out of 27) in which at least two dice show a three or a five.” Each throw is independent and has success probability $$p_s$$, so

$$Y\sim\text{Binomial}(27, p_s).$$

For a binomial random variable, the expected value is given by the well-known formula

$$E(Y)=n\,p,$$

where $$n$$ is the number of trials and $$p$$ the success probability per trial. Substituting $$n=27$$ and $$p=p_s=\frac{11}{27},$$ we obtain

$$E(Y)=27\left(\frac{11}{27}\right)=11.$$

So, on average, we expect the desired event to occur eleven times in the 27 repetitions.

So, the answer is $$11$$.

We are told that every bomb hits the target with probability $$p = 0.5$$, and each bomb acts independently of the others. Let us drop $$n$$ bombs and define a random variable $$X$$ that counts the number of hits among these $$n$$ bombs. Because each bomb can be thought of as a Bernoulli trial (Hit = success, Miss = failure), $$X$$ follows the binomial distribution with parameters $$n$$ and $$p$$. The probability mass function for a binomially distributed variable is stated by the formula

$$\Pr(X = k) = \binom{n}{k}\,p^{\,k}\,(1-p)^{\,n-k}, \qquad k = 0,1,2,\dots ,n.$$

The target is completely destroyed if we obtain at least two hits, i.e. if $$X \ge 2$$. We therefore want

$$\Pr(X \ge 2) \ge 0.99.$$

Instead of summing many terms to find $$\Pr(X \ge 2)$$ directly, it is easier to use the complement rule. We have the basic identity

$$\Pr(X \ge 2) = 1 - \Pr(X \le 1) = 1 - \bigl[\Pr(X = 0) + \Pr(X = 1)\bigr].$$

So our requirement translates into

$$1 - \bigl[\Pr(X = 0) + \Pr(X = 1)\bigr] \;\ge\; 0.99.$$

Rearranging, this is equivalent to

$$\Pr(X = 0) + \Pr(X = 1) \;\le\; 0.01.$$

Now we calculate the two individual probabilities using the binomial formula.

First, for zero hits:

$$\Pr(X = 0) = \binom{n}{0}\,p^{\,0}\,(1-p)^{\,n} = 1 \times 1 \times (1-0.5)^{\,n} = 0.5^{\,n}.$$

Next, for exactly one hit:

$$\Pr(X = 1) = \binom{n}{1}\,p^{\,1}\,(1-p)^{\,n-1} = n \times 0.5 \times 0.5^{\,n-1} = n \times 0.5^{\,n}.$$

Adding these two probabilities gives

$$\Pr(X = 0) + \Pr(X = 1) = 0.5^{\,n} + n \times 0.5^{\,n} = (1+n)\,0.5^{\,n}.$$

Our inequality $$\Pr(X = 0) + \Pr(X = 1) \le 0.01$$ therefore becomes

$$(1+n)\,0.5^{\,n} \;\le\; 0.01.$$

Because $$0.5^{\,n} = \dfrac{1}{2^{\,n}}$$, we can rewrite the left side as

$$(1+n)\,\dfrac{1}{2^{\,n}} = \dfrac{1+n}{2^{\,n}},$$

so the inequality we need to satisfy is

$$\dfrac{1+n}{2^{\,n}} \;\le\; 0.01.$$

To find the minimum integer $$n$$ meeting this condition, we test successive values of $$n$$ starting from small integers and stop as soon as the inequality holds.

• For $$n = 10$$, $$\dfrac{1+10}{2^{\,10}} = \dfrac{11}{1024} \approx 0.010742 \;>\; 0.01.$$ This is slightly larger than $$0.01$$, so ten bombs are not enough.

• For $$n = 11$$, $$\dfrac{1+11}{2^{\,11}} = \dfrac{12}{2048} \approx 0.005859 \;<\; 0.01.$$ Now the inequality is satisfied and we have reached the required confidence level.

Because $$n = 10$$ fails while $$n = 11$$ succeeds, the smallest possible number of bombs that achieves the desired $$99\%$$ chance of completely destroying the target is $$n = 11$$.

So, the answer is $$11$$.

First, we note that the probability of the man hitting the target in one shot is given as $$\dfrac{1}{10}$$. Therefore, the probability of his missing the target in a single shot is

$$1-\dfrac{1}{10}=\dfrac{9}{10}\,.$$

Now, assume he fires $$n$$ independent shots. Because the shots are independent, the probability that he misses every one of the $$n$$ shots is the product of the individual miss-probabilities. Hence, the required probability of missing all $$n$$ shots is

$$\left(\dfrac{9}{10}\right)^n\,.$$

We are interested in the complementary event, namely, the event that he hits the target at least once in those $$n$$ shots. For any two complementary events $$A$$ and $$\overline{A}$$, we have the basic rule

$$P(A)=1-P(\overline{A}).$$

Applying this rule with $$A$$ = “at least one hit” and $$\overline{A}$$ = “no hit at all,” we obtain

$$P(\text{at least one hit}) = 1-\left(\dfrac{9}{10}\right)^n.$$

The question demands that this probability be greater than $$\dfrac{1}{4}$$. Therefore, we set up the inequality

$$1-\left(\dfrac{9}{10}\right)^n > \dfrac{1}{4}.$$

Subtracting $$1$$ from both sides (and remembering to change the sign), we get

$$-\left(\dfrac{9}{10}\right)^n > -\dfrac{3}{4}.$$

Multiplying both sides by $$-1$$ reverses the inequality sign, giving

$$\left(\dfrac{9}{10}\right)^n < \dfrac{3}{4}.$$

To isolate $$n$$ we take natural logarithms (or common logarithms; any positive base other than 1 works). Writing with natural logs, we have

$$\ln\!\left(\dfrac{9}{10}\right)^n < \ln\!\left(\dfrac{3}{4}\right).$$

Using the power rule for logarithms, $$\ln(a^b)=b\ln a$$, we get

$$n\;\ln\!\left(\dfrac{9}{10}\right) < \ln\!\left(\dfrac{3}{4}\right).$$

Observe that $$\dfrac{9}{10}=0.9$$ and $$\dfrac{3}{4}=0.75,$$ both of which lie between 0 and 1. Hence their natural logarithms are negative. When dividing by a negative number, the direction of the inequality reverses. Therefore, dividing both sides by $$\ln(0.9)$$ gives

$$n > \dfrac{\ln\!\left(\dfrac{3}{4}\right)}{\ln\!\left(\dfrac{9}{10}\right)}.$$

We now compute the numerical value:

$$\ln\!\left(\dfrac{3}{4}\right)=\ln(0.75)\approx -0.287682072,$$

$$\ln\!\left(\dfrac{9}{10}\right)=\ln(0.9)\approx -0.105360516.$$

Substituting these into the inequality, we get

$$n > \dfrac{-0.287682072}{-0.105360516}\approx 2.731.$$ Since $$n$$ must be a whole number (the man cannot fire a fractional shot), we choose the least integer greater than $$2.731$$, which is $$3$$.

So, the answer is $$3$$.

First, we list the three simple statements exactly as given and decide whether each one is true or false.

$$P : \; 5 \text{ is a prime number}$$ Because a prime number has exactly two distinct positive divisors, namely 1 and itself, and $$5$$ satisfies this definition, we have $$P = \text{True}.$$

$$Q : \; 7 \text{ is a factor of } 192$$ A number $$a$$ is a factor of $$b$$ if and only if $$b \div a$$ is an integer. Computing $$\dfrac{192}{7}=27.428571\ldots$$ which is not an integer, so $$7$$ is not a factor of $$192$$. Hence $$Q = \text{False}.$$

$$R : \; \text{LCM of } 5 \text{ and } 7 \text{ is } 35$$ Since $$5$$ and $$7$$ are coprime, their least common multiple equals their product, i.e. $$\text{LCM}(5,7)=5 \times 7 = 35.$$ Therefore $$R = \text{True}.$$

Next, we evaluate each compound statement in the options by substituting these truth values and applying the standard logical rules.

Option A: $$P \vee (\sim Q \wedge R)$$ First find each component: $$\sim Q = \text{not False} = \text{True},$$ so $$\sim Q \wedge R = \text{True} \wedge \text{True} = \text{True}.$$ Now use the value of $$P$$ together with the disjunction rule: $$P \vee (\sim Q \wedge R) = \text{True} \vee \text{True} = \text{True}.$$

Option B: $$(P \wedge Q) \vee (\sim R)$$ Compute the conjunction $$P \wedge Q$$: $$P \wedge Q = \text{True} \wedge \text{False} = \text{False}.$$ Next negate $$R$$: $$\sim R = \text{not True} = \text{False}.$$ Finally take their disjunction: $$(P \wedge Q) \vee (\sim R) = \text{False} \vee \text{False} = \text{False}.$$

Option C: $$(\sim P) \vee (Q \wedge R)$$ Negate $$P$$: $$\sim P = \text{not True} = \text{False}.$$ Find $$Q \wedge R$$: $$Q \wedge R = \text{False} \wedge \text{True} = \text{False}.$$ Now the disjunction: $$(\sim P) \vee (Q \wedge R) = \text{False} \vee \text{False} = \text{False}.$$

Option D: $$(\sim P) \wedge (\sim Q \wedge R)$$ From above, $$\sim P = \text{False}$$ and $$\sim Q \wedge R = \text{True}$$. Therefore the conjunction gives $$(\sim P) \wedge (\sim Q \wedge R) = \text{False} \wedge \text{True} = \text{False}.$$

Summarizing the evaluations:

$$\text{Option A} = \text{True}, \quad \text{Option B} = \text{False}, \quad \text{Option C} = \text{False}, \quad \text{Option D} = \text{False}.$$

Only Option A yields the truth value “True”.

Hence, the correct answer is Option A.

We have the statement $$P(n): n^{2}-n+41$$ is prime, and we wish to test it for the two particular natural numbers $$n = 3$$ and $$n = 5$$ that appear in the options.

First we take $$n = 3$$. Substituting $$3$$ into the expression we get

$$\begin{aligned} P(3) &= 3^{2}-3+41 \\ &= 9-3+41 \\ &= 6+41 \\ &= 47. \end{aligned}$$

Now we check whether $$47$$ is prime. A prime number is a positive integer greater than $$1$$ that has no positive divisors other than $$1$$ and itself. The possible prime divisors less than $$\sqrt{47}$$ are $$2,3,5$$. Clearly, $$47$$ is not divisible by $$2$$ because it is odd, not divisible by $$3$$ because the sum of its digits $$4+7=11$$ is not a multiple of $$3$$, and not divisible by $$5$$ because it does not end in $$0$$ or $$5$$. Hence $$47$$ has no divisors other than $$1$$ and $$47$$, so $$47$$ is prime. Therefore $$P(3)$$ is true.

Next we take $$n = 5$$. Substituting $$5$$ into the expression gives

$$\begin{aligned} P(5) &= 5^{2}-5+41 \\ &= 25-5+41 \\ &= 20+41 \\ &= 61. \end{aligned}$$

We again test for primality. The square root of $$61$$ is a little less than $$8$$, so we only need to test divisibility by the primes $$2,3,5,7$$. The number $$61$$ is odd, so it is not divisible by $$2$$. The sum of its digits is $$6+1=7$$, which is not a multiple of $$3$$, so it is not divisible by $$3$$. It does not end in $$0$$ or $$5$$, so it is not divisible by $$5$$. Finally, dividing by $$7$$ gives $$61/7 = 8.714\ldots$$, which is not an integer, so it is not divisible by $$7$$. Thus $$61$$ has no divisors other than $$1$$ and $$61$$, so $$61$$ is prime. Therefore $$P(5)$$ is also true.

We have shown that $$P(3)$$ is true and $$P(5)$$ is true. Comparing with the given options, the only option that matches this outcome is:

Option C: Both $$P(3)$$ and $$P(5)$$ are true.

Hence, the correct answer is Option C.

We begin with the symbolic sentence

$$p \;\oplus\; q \wedge \sim p \;\odot\; q$$

The tilde $$\sim$$ (negation) always has the highest precedence. Among the binary connectives we remember the usual rule “AND binds more strongly than OR”. Therefore the fixed connective $$\wedge$$ (which is written explicitly in the statement) is evaluated before whichever symbol will finally be placed in $$\oplus$$. In the same way, the symbol $$\odot$$ will first be replaced by either $$\wedge$$ or $$\vee$$ and then be evaluated inside the parentheses that we shall introduce for clarity.

So, whatever the final choice of $$\oplus$$ and $$\odot$$ is, the expression can be read as

$$p \;\oplus\; \bigl(q \wedge (\sim p \;\odot\; q)\bigr).$$

We are told that, after the two replacements are made, the above statement must be identically equal to the simple conjunction $$p \wedge q$$. There are only four possible ordered pairs

$$(\oplus,\;\odot)\;=\; (\vee,\wedge),\; (\wedge,\wedge),\; (\vee,\vee),\; (\wedge,\vee).$$

We shall substitute each pair one by one, simplify by ordinary Boolean algebra, and see which pair really gives $$p\wedge q$$.

1. $$(\oplus,\odot)=(\vee,\wedge)$$

$$\begin{aligned} p\vee\bigl(q\wedge(\sim p\wedge q)\bigr) &=p\vee\bigl(q\wedge\sim p\wedge q\bigr)\\[2mm] &=p\vee\bigl(q\wedge q\wedge\sim p\bigr)\\[2mm] &=p\vee(q\wedge\sim p)\\[2mm] &\;=\;(p\vee q)\wedge(p\vee\sim p)\qquad\text{[distributive law]}\\[2mm] &=p\vee q. \end{aligned}$$

This is $$p\vee q$$, not $$p\wedge q$$, so the first pair fails.

2. $$(\oplus,\odot)=(\wedge,\wedge)$$

$$\begin{aligned} p\wedge\bigl(q\wedge(\sim p\wedge q)\bigr) &=p\wedge q\wedge\sim p\wedge q\\[2mm] &=p\wedge\sim p\wedge q\\[2mm] &=\text{False}. \end{aligned}$$

This is the zero (contradiction) function, not $$p\wedge q$$, so the second pair is also rejected.

3. $$(\oplus,\odot)=(\vee,\vee)$$

$$\begin{aligned} p\vee\bigl(q\wedge(\sim p\vee q)\bigr) &=p\vee\bigl((q\wedge\sim p)\vee(q\wedge q)\bigr)\\[2mm] &=p\vee\bigl((q\wedge\sim p)\vee q\bigr)\\[2mm] &=p\vee q. \end{aligned}$$

Again the result is $$p\vee q\neq p\wedge q$$, so the third pair does not work either.

4. $$(\oplus,\odot)=(\wedge,\vee)$$

$$\begin{aligned} p\wedge\bigl(q\wedge(\sim p\vee q)\bigr) &=p\wedge\Bigl(q\wedge\bigl[(\sim p\vee q)\bigr]\Bigr)\\[2mm] &\text{Use the distributive law inside the square brackets:}\\[2mm] &\qquad q\wedge(\sim p\vee q) =(q\wedge\sim p)\vee(q\wedge q) =(q\wedge\sim p)\vee q\\[2mm] &\qquad\qquad =q\vee(q\wedge\sim p)=q.\\[4mm] \therefore\; p\wedge\bigl(q\wedge(\sim p\vee q)\bigr) &=p\wedge q. \end{aligned}$$

This time the simplified form is exactly $$p\wedge q$$, the target expression. Hence only the ordered pair $$\bigl(\wedge,\;\vee\bigr)$$ satisfies the requirement.

Therefore the Boolean sentence is equivalent to $$p\wedge q$$ only when

$$ (\oplus,\;\odot)=\;(\wedge,\;\vee). $$

Hence, the correct answer is Option 4.

We have to simplify the statement $$[\sim(\sim p \vee q) \vee (p \wedge r)] \wedge (\sim q \wedge r).$$

First recall De Morgan’s law, which states $$\sim(A \vee B)\equiv (\sim A)\wedge(\sim B).$$ Applying this with $$A=\sim p$$ and $$B=q,$$ we obtain

$$\sim(\sim p \vee q)\equiv (\sim(\sim p))\wedge(\sim q)\equiv p \wedge \sim q.$$

Substituting this result back, the expression becomes

$$[(p \wedge \sim q) \vee (p \wedge r)] \wedge (\sim q \wedge r).$$

Now we focus on the disjunction inside the first bracket. Taking out the common factor $$p,$$ we use the distributive law $$X\wedge Y \,\vee\, X\wedge Z \equiv X\wedge(Y\vee Z).$$ Thus,

$$(p \wedge \sim q) \vee (p \wedge r)\equiv p \wedge (\sim q \vee r).$$

Hence the whole statement is now

$$\bigl[p \wedge (\sim q \vee r)\bigr] \wedge (\sim q \wedge r).$$

The connective $$\wedge$$ is associative and commutative, so we can rearrange terms without changing truth-value:

$$p \wedge (\sim q \vee r) \wedge \sim q \wedge r.$$

Observe that the factor $$\sim q \wedge r$$ already guarantees both $$\sim q$$ and $$r$$ are true, and whenever these two are true, the disjunction $$\sim q \vee r$$ is automatically true. Consequently the extra factor $$\sim q \vee r$$ is redundant inside the conjunction. Formally,

$$\bigl(\sim q \wedge r\bigr)\ \wedge\ (\sim q \vee r)\ \equiv\ \sim q \wedge r.$$

So the expression simplifies further to

$$p \wedge \sim q \wedge r.$$

Finally we may write this in the same order as option B:

$$(p \wedge r) \wedge \sim q.$$

This is exactly the statement given in option B. Hence, the correct answer is Option B.

We have two fair dice, each die showing one of the numbers 1 to 6 with equal probability. Because the dice are independent, the total number of equally likely ordered outcomes is $$6 \times 6 = 36.$$

First we identify the three mutually exclusive cases mentioned in the question and count how many outcomes belong to each case.

Case 1 - Doublet (both dice show the same number)
The possible doublets are $$(1,1),(2,2),(3,3),(4,4),(5,5),(6,6).$$
So, the number of favourable outcomes is $$6.$$

Case 2 - Sum equals 9
We list all ordered pairs giving a total of 9:
$$(3,6),\;(4,5),\;(5,4),\;(6,3).$$
Hence, the number of favourable outcomes is $$4.$$
(Notice that none of these are doublets, so there is no overlap with Case 1.)

Case 3 - Any other outcome
The remaining outcomes are $$36-6-4 = 26.$$

Next we assign the monetary result to each case:

• Doublet → gain of Rs. 15
• Sum 9 → gain of Rs. 12
• Any other outcome → loss of Rs. 6

The expectation (expected gain) is calculated using the formula for the expected value of a discrete random variable: $$E = \sum (\text{gain or loss}) \times (\text{probability of that outcome}).$$

We now compute the probabilities and the corresponding contributions one by one.

Probability of a doublet $$P(\text{doublet}) = \dfrac{6}{36} = \dfrac16.$$ Contribution to expectation $$\dfrac16 \times 15 = \dfrac{15}{6} = 2.5.$$

Probability of sum 9 $$P(\text{sum }9) = \dfrac{4}{36} = \dfrac19.$$ Contribution to expectation $$\dfrac19 \times 12 = \dfrac{12}{9} = \dfrac43 \approx 1.333\ldots$$

Probability of any other outcome $$P(\text{other}) = \dfrac{26}{36} = \dfrac{13}{18}.$$ Contribution to expectation $$\dfrac{13}{18} \times (-6) = -\dfrac{78}{18} = -\dfrac{13}{3} \approx -4.333\ldots$$

Now we add all three contributions to obtain the overall expected gain:

$$E = 2.5 + \dfrac43 - \dfrac{13}{3}.$$ We first convert every term to a common denominator of 6: $$2.5 = \dfrac{15}{6},\qquad \dfrac43 = \dfrac{8}{6},\qquad -\dfrac{13}{3} = -\dfrac{26}{6}.$$ So, $$E = \dfrac{15}{6} + \dfrac{8}{6} - \dfrac{26}{6} = \dfrac{15 + 8 - 26}{6} = \dfrac{-3}{6} = -\dfrac12.$$

The expectation is $$-\dfrac12,$$ i.e., an average loss of Rs. 0.50 per throw.

Hence, the correct answer is Option A.

Let the universal set be the whole class of 60 students. We denote by $$n(\text{NCC})$$ the number of students who opted for NCC, by $$n(\text{NSS})$$ the number who opted for NSS, and by $$n(\text{NCC} \cap \text{NSS})$$ the number who opted for both NCC and NSS.

According to the question, we have $$n(\text{NCC}) = 40,$$ $$n(\text{NSS}) = 30,$$ $$n(\text{NCC} \cap \text{NSS}) = 20.$$

First, we find the total number of students who opted for at least one of the two activities. For two sets, the principle of inclusion-exclusion states

$$n(\text{NCC} \cup \text{NSS}) = n(\text{NCC}) + n(\text{NSS}) - n(\text{NCC} \cap \text{NSS}).$$

Substituting the given numbers, we get $$n(\text{NCC} \cup \text{NSS}) = 40 + 30 - 20 = 50.$$

Now, the students who opted for neither NCC nor NSS are outside this union. So their count is obtained by subtracting the size of the union from the total class strength:

$$n(\text{neither}) = 60 - n(\text{NCC} \cup \text{NSS}) = 60 - 50 = 10.$$

The experiment is to choose one student at random from the class. The probability that the selected student belongs to the “neither” category equals the ratio of favourable outcomes to total outcomes, namely

$$\text{Probability} = \frac{n(\text{neither})}{\text{total students}} = \frac{10}{60} = \frac{1}{6}.$$

Hence, the correct answer is Option A.

Let us place the ground set in front of us: $$S=\{1,2,3,\ldots ,20\}.$$

The task is to pick a subset $$B\subseteq S$$ whose elements add up to $$203.$$ Because every element of $$S$$ is either in $$B$$ or in its complement $$B^c,$$ the following identity of sums holds:

$$\displaystyle\sum_{x\in S}x=\sum_{x\in B}x+\sum_{x\in B^c}x.$$

The left-hand sum is the total of the first twenty natural numbers. Using the familiar formula for an arithmetic progression,

$$1+2+\cdots +20=\frac{20\cdot 21}{2}=210.$$

We are demanding $$\displaystyle\sum_{x\in B}x=203,$$ so substitution gives

$$210=203+\sum_{x\in B^c}x,$$

which rearranges step by step to

$$\sum_{x\in B^c}x=210-203=7.$$

Thus a subset $$B$$ is “nice’’ exactly when its complement has total $$7.$$ The mapping $$B\longleftrightarrow B^c$$ is a perfect one-to-one correspondence between subsets of sum $$203$$ and subsets of sum $$7,$$ so counting either family gives the same result. Because $$7$$ is a very small number, it is far easier to enumerate the complements.

We now list every subset of $$S$$ whose elements are distinct and add to $$7.$$ Begin with singletons:

$$\{7\}\quad\text{sum}=7.$$

Move on to pairs of distinct elements. We require $$a+b=7,$$ with $$1\le a

$$1+6=7\;\Longrightarrow\;\{1,6\},$$ $$2+5=7\;\Longrightarrow\;\{2,5\},$$ $$3+4=7\;\Longrightarrow\;\{3,4\}.$$

Next, examine triples. We need three different positive integers whose sum is $$7.$$ The only possibility is

$$1+2+4=7\;\Longrightarrow\;\{1,2,4\}.$$

Trying to use four or more distinct numbers fails immediately, because $$1+2+3+4=10>7,$$ and any larger collection would have an even bigger sum.

Therefore the entire catalogue of subsets of $$S$$ whose sum is $$7$$ consists of

$$\bigl\{\; \{7\},\; \{1,6\},\; \{2,5\},\; \{3,4\},\; \{1,2,4\}\;\bigr\}.$$

Exactly five subsets appear in that list, so there are $$5$$ “nice’’ complements and consequently $$5$$ “nice’’ subsets whose sum is $$203.$$

The sample space in the problem is the set of all subsets of $$S,$$ and a set with $$20$$ elements has exactly $$2^{20}$$ subsets. The required probability is therefore

$$\text{Probability}=\frac{\text{number of nice subsets}}{\text{total number of subsets}}=\frac{5}{2^{20}}.$$

Hence, the correct answer is Option B.

Let us begin with the very first random event. An unbiased coin is tossed. For any unbiased coin we have the fundamental fact $$P(\text{Head})=\frac12 \quad\text{and}\quad P(\text{Tail})=\frac12.$$

Now, we split the entire experiment into two separate branches because what happens after the coin depends on whether we see a head or a tail.

Branch 1 (Head): A pair of unbiased dice is rolled. Each die has six faces, so the ordered pair of results has $$6\times 6 = 36$$ equally likely possibilities. We are asked to find the probability that the sum of the two dice is either $$7$$ or $$8$$.

First we list all ordered pairs giving these sums.

Sum $$7$$: $$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\,$$  ⇒ $$6$$ ways.

Sum $$8$$: $$(2,6),(3,5),(4,4),(5,3),(6,2)\,$$  ⇒ $$5$$ ways.

So the count of favourable outcomes on the dice is $$6+5=11.$$ Therefore, using $$P=\dfrac{\text{favourable outcomes}}{\text{total outcomes}},$$ we get

$$P(\text{sum is }7\text{ or }8\;|\;\text{Head})=\frac{11}{36}.$$

Multiplying with the earlier coin probability, the overall contribution from this branch is

$$P(\text{noted number is }7\text{ or }8\;\&\;\text{Head}) \;=\; P(\text{Head})\times P(\text{sum }7\text{ or }8|\text{Head}) \;=\; \frac12 \times \frac{11}{36} \;=\; \frac{11}{72}.$$

Branch 2 (Tail): A single card is drawn at random from nine cards numbered $$1,2,3,\dots,9.$$ Because the deck is well-shuffled, every card is equally likely, so

$$P(\text{any specified card})=\frac19.$$

There are exactly two favourable cards—those showing $$7$$ or $$8$$—so by adding the probabilities (or simply counting favourable numbers) we have

$$P(\text{card is }7\text{ or }8\;|\;\text{Tail})=\frac{2}{9}.$$

Multiplying by the coin probability for a tail, the total contribution from this branch equals

$$P(\text{noted number is }7\text{ or }8\;\&\;\text{Tail}) \;=\; P(\text{Tail})\times P(\text{card }7\text{ or }8|\text{Tail}) \;=\; \frac12 \times \frac{2}{9} \;=\; \frac19.$$

For a common denominator, rewrite $$\frac19=\frac{8}{72}.$$

Finally, because the two branches are mutually exclusive and cover all possibilities, we add their probabilities:

$$\begin{aligned} P(\text{noted number is }7\text{ or }8) &= \frac{11}{72} + \frac{8}{72} \\ &= \frac{19}{72}. \end{aligned}$$

Hence, the correct answer is Option 2.

First we note the contents of the urn at the beginning. There are $$5$$ red balls and $$2$$ green balls, so in all $$7$$ balls.

We name the events so that the later algebra will be clear:

$$R_1 : \text{``the first ball drawn is red''}$$

$$G_1 : \text{``the first ball drawn is green''}$$

$$R_2 : \text{``the second ball drawn is red''}$$

We are required to find the probability $$P(R_2)$$, that is, the probability that the second ball is red.

According to the Law of Total Probability (which states $$P(B)=P(A_1)P(B|A_1)+P(A_2)P(B|A_2)+\dots$$ for a complete set of events $$A_i$$), we can write

$$ P(R_2)=P(R_1)\,P(R_2|R_1)+P(G_1)\,P(R_2|G_1). $$

So we shall calculate each factor step by step.

First-draw probabilities:

$$ P(R_1)=\frac{5}{7}, \qquad P(G_1)=\frac{2}{7}. $$

Now we handle the composition changes that happen after the first draw.

Case 1: The first ball is red ($$R_1$$). A red ball is removed, and a green ball is added. So the urn now contains:

Red balls: $$5-1=4$$

Green balls: $$2+1=3$$

Total balls remain $$7$$ (because one was taken out and one was put in).

Hence, given $$R_1$$, the conditional probability that the second ball is red is

$$ P(R_2|R_1)=\frac{4}{7}. $$

Case 2: The first ball is green ($$G_1$$). A green ball is removed, and a red ball is added. So the urn now contains:

Red balls: $$5+1=6$$

Green balls: $$2-1=1$$

Total balls again $$7$$.

Hence, given $$G_1$$, the conditional probability that the second ball is red is

$$ P(R_2|G_1)=\frac{6}{7}. $$

We now substitute all these numbers into the total-probability expression:

$$ P(R_2)=\frac{5}{7}\times\frac{4}{7}+\frac{2}{7}\times\frac{6}{7}. $$

We multiply term by term:

$$ \frac{5}{7}\times\frac{4}{7}=\frac{20}{49},\qquad \frac{2}{7}\times\frac{6}{7}=\frac{12}{49}. $$

Adding the two fractions (same denominator $$49$$) gives

$$ P(R_2)=\frac{20}{49}+\frac{12}{49}=\frac{32}{49}. $$

Therefore the probability that the second ball is red equals $$\dfrac{32}{49}$$.

Hence, the correct answer is Option C.

Let us list every assumption clearly. Each child is equally likely to be a boy (B) or a girl (G), so the probability of either outcome for a single birth is $$\tfrac12$$. There are two families and each family has two children, giving altogether $$2+2=4$$ children whose genders are independent.

First we establish the complete sample space. Because each of the four children can be B or G, the total number of possible gender strings is $$2^4 = 16.$$ We may write a convenient enumeration:

$$\begin{array}{cccc} 1:&\; GGGG &\quad& 9:&\; BGGG\\ 2:&\; GGGB && 10:&\; GBGG\\ 3:&\; GGBG && 11:&\; GGBB\\ 4:&\; GBGG && 12:&\; BGBG\\ 5:&\; GBGB && 13:&\; BBBG\\ 6:&\; GBBG && 14:&\; BBGB\\ 7:&\; BGGB && 15:&\; BGBB\\ 8:&\; BG BG && 16:&\; BBBB \end{array}$$

Now we introduce two relevant events:

Event $$A$$: “all four children are girls”.
Event $$B$$: “at least two of the four children are girls”.

We want $$P(A\mid B)$$, the conditional probability of $$A$$ given $$B$$. By the definition of conditional probability,

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}.$$

Observe that if all four children are girls, the condition “at least two are girls” is automatically satisfied, so $$A\subseteq B$$ and therefore $$A\cap B=A$$. Thus the numerator simplifies to $$P(A)$$.

We now compute the cardinalities (counts) of the events inside the 16-point sample space.

Counting the outcomes of event A:
There is exactly one outcome in which every child is a girl, namely $$GGGG$$. Hence

$$|A|=1.$$

Counting the outcomes of event B:
“At least two girls” means “two, three, or four girls.” It is usually easier to count its complement, “fewer than two girls,” i.e. either zero or one girl, and subtract from the total.

• Zero girls: all boys, $$BBBB$$ - that is 1 outcome.
• Exactly one girl: choose which of the four positions is G; there are $$\binom{4}{1}=4$$ such arrangements.

Hence the number of outcomes with fewer than two girls is $$1+4=5.$$

Therefore the number with at least two girls is

$$|B| = 16 - 5 = 11.$$

Converting counts to probabilities:
Because every outcome among the 16 is equally likely, the probability of any specific set of outcomes is simply (number of favorable outcomes)/16. Thus

$$P(A)=\frac{|A|}{16} = \frac{1}{16}, \qquad P(B)=\frac{|B|}{16} = \frac{11}{16}.$$

Applying the conditional-probability formula:

$$ P(A\mid B)=\frac{P(A\cap B)}{P(B)} =\frac{P(A)}{P(B)} =\frac{\tfrac{1}{16}}{\tfrac{11}{16}} =\frac{1}{11}. $$

Hence, the correct answer is Option C.

We have an admission-test screening in which a candidate attempts $$n=50$$ independent problems. For every single problem, the probability that the candidate solves it is given as $$\tfrac45$$. Consequently, the probability that the candidate does not solve (fails to solve) a problem is

$$q=1-\tfrac45=\tfrac15.$$

Let the random variable $$X$$ denote the number of problems the candidate is unable to solve out of the fifty. Because each problem is independent and has the same probability of failure $$q=\tfrac15$$, the distribution of $$X$$ is binomial with parameters $$n=50$$ and $$p=q=\tfrac15$$. The probability-mass function of a binomial random variable is stated first:

$$P(X=k)=\binom{n}{k}p^{\,k}(1-p)^{\,n-k}.$$

Here we require the probability that the candidate is unable to solve less than two problems, i.e.

$$P(X<2)=P(X=0)+P(X=1).$$

First term: $$P(X=0)$$. Substituting $$k=0$$ in the formula, we get

$$P(X=0)=\binom{50}{0}\Bigl(\tfrac15\Bigr)^{0}\Bigl(1-\tfrac15\Bigr)^{50} =1\cdot1\cdot\Bigl(\tfrac45\Bigr)^{50} =\Bigl(\tfrac45\Bigr)^{50}.$$

Second term: $$P(X=1)$$. Substituting $$k=1$$ gives

$$P(X=1)=\binom{50}{1}\Bigl(\tfrac15\Bigr)^{1}\Bigl(1-\tfrac15\Bigr)^{49} =50\cdot\tfrac15\cdot\Bigl(\tfrac45\Bigr)^{49} =10\Bigl(\tfrac45\Bigr)^{49}.$$

Adding the two probabilities. Factorising the common power $$\Bigl(\tfrac45\Bigr)^{49}$$ we have

$$P(X<2)=\Bigl(\tfrac45\Bigr)^{50}+10\Bigl(\tfrac45\Bigr)^{49} =\Bigl(\tfrac45\Bigr)^{49}\!\left[\tfrac45+10\right].$$

Convert the bracket to a single fraction:

$$\tfrac45+10=\tfrac45+\tfrac{50}{5}=\tfrac{4+50}{5}=\tfrac{54}{5}.$$

So,

$$P(X<2)=\frac{54}{5}\Bigl(\tfrac45\Bigr)^{49}.$$

This expression matches Option D in the given choices.

Hence, the correct answer is Option D.

Let us denote the probabilities of the four persons hitting the target by $$p_1=\dfrac12,\;p_2=\dfrac13,\;p_3=\dfrac14,\;p_4=\dfrac18.$$

We want the probability that the target is hit by at least one of them. A convenient way to calculate this is to first find the probability that none of them hits the target, and then subtract that value from 1.

The probability that a person misses the target is obtained by subtracting the probability of a hit from 1. Thus we have

$$q_1 = 1 - p_1 = 1 - \dfrac12 = \dfrac12,$$

$$q_2 = 1 - p_2 = 1 - \dfrac13 = \dfrac23,$$

$$q_3 = 1 - p_3 = 1 - \dfrac14 = \dfrac34,$$

$$q_4 = 1 - p_4 = 1 - \dfrac18 = \dfrac78.$$

Because the shots are independent, the probability that all four persons miss the target is the product of their individual miss-probabilities:

$$P(\text{all miss}) \;=\; q_1 \times q_2 \times q_3 \times q_4 \;=\; \dfrac12 \times \dfrac23 \times \dfrac34 \times \dfrac78.$$

Step-by-step multiplication gives

$$\dfrac12 \times \dfrac23 = \dfrac{1}{3},$$

$$\dfrac{1}{3} \times \dfrac34 = \dfrac14,$$

$$\dfrac14 \times \dfrac78 = \dfrac{7}{32}.$$

So we have

$$P(\text{all miss}) = \dfrac{7}{32}.$$

Now we apply the complementary probability formula

$$P(\text{at least one hit}) = 1 - P(\text{all miss}).$$

Substituting the value just obtained,

$$P(\text{at least one hit}) = 1 - \dfrac{7}{32} = \dfrac{32}{32} - \dfrac{7}{32} = \dfrac{25}{32}.$$

Hence, the correct answer is Option D.

We are told that the probability of the shooter hitting the target in a single, independent shot is $$p=\frac{1}{3}$$.

First, we recall the complementary rule of probability: if an event has probability $$p$$ of occurring in one trial, then the probability of it not occurring in that trial is $$1-p$$. Hence, the probability of missing the target in one shot is

$$q \;=\; 1-p \;=\; 1-\frac{1}{3} \;=\; \frac{2}{3}.$$

Because the shots are independent, the probability of missing the target in every one of $$n$$ shots is the product of the individual miss-probabilities, that is

$$q^n \;=\;\left(\frac{2}{3}\right)^n.$$

The probability of hitting the target at least once in those $$n$$ shots is obtained by subtracting the “miss-all” probability from $$1$$. Stated formally,

$$P(\text{at least one hit}) \;=\; 1 - P(\text{no hit in }n\text{ shots}) \;=\; 1 - \left(\frac{2}{3}\right)^n.$$

The problem demands that this probability be greater than $$\frac{5}{6}$$. Therefore we must solve the inequality

$$1 - \left(\frac{2}{3}\right)^n \;>\; \frac{5}{6}.$$

Now we move the term $$\left(\frac{2}{3}\right)^n$$ to the right:

$$-\left(\frac{2}{3}\right)^n \;>\; \frac{5}{6} - 1.$$

Since $$\frac{5}{6} - 1 = -\frac{1}{6}$$, we can multiply both sides by $$-1$$, remembering to reverse the inequality sign:

$$\left(\frac{2}{3}\right)^n \;<\; \frac{1}{6}.$$

To discover the smallest integer $$n$$ that satisfies this strict inequality, we now examine successive powers of $$\frac{2}{3}$$ and compare each to $$\frac{1}{6}$$.

For $$n=1$$:

$$\left(\frac{2}{3}\right)^1 = \frac{2}{3} \approx 0.667 \;>\; \frac{1}{6}\approx0.167.$$ So the inequality is not satisfied.

For $$n=2$$:

$$\left(\frac{2}{3}\right)^2 = \frac{4}{9} \approx 0.444 \;>\; 0.167.$$ Still not satisfied.

For $$n=3$$:

$$\left(\frac{2}{3}\right)^3 = \frac{8}{27} \approx 0.296 \;>\; 0.167.$$ Still too large.

For $$n=4$$:

$$\left(\frac{2}{3}\right)^4 = \frac{16}{81} \approx 0.198 \;>\; 0.167.$$ Again the inequality fails.

For $$n=5$$:

$$\left(\frac{2}{3}\right)^5 = \frac{32}{243} \approx 0.132 \;<\; 0.167.$$ Here, the inequality is finally satisfied.

Thus, the smallest integer $$n$$ that makes $$\left(\frac{2}{3}\right)^n < \frac{1}{6}$$ true is $$n=5$$.

Therefore, the shooter must fire at least $$5$$ independent shots for the probability of hitting the target at least once to exceed $$\frac{5}{6}$$.

Hence, the correct answer is Option B.

For a single throw of a fair die, the probability of getting either 5 or 6 is $$\dfrac{2}{6}=\dfrac13$$ and the probability of getting any other number (1, 2, 3, 4) is $$\dfrac{4}{6}=\dfrac23.$$

The man follows this rule: he stops at the first appearance of 5 or 6, but in any case throws at most three times. We compute every possible stopping scenario, its probability, and the total money won or lost in that scenario.

Let $$p=\dfrac13 \;(\text{success}), \qquad q=\dfrac23 \;(\text{failure}).$$

Case 1: Success on the first throw.
Probability $$=p=\dfrac13,$$
Money outcome $$=+100.$$

Case 2: Failure on the first throw and success on the second.
Probability $$=q\,p=\dfrac23\cdot\dfrac13=\dfrac29,$$
Money outcome $$=-50 \;(\text{first throw}) + 100 \;(\text{second throw}) = +50.$$

Case 3: Failures on the first two throws and success on the third.
Probability $$=q^{2}p=\left(\dfrac23\right)^{2}\!\!\cdot\dfrac13=\dfrac49\cdot\dfrac13=\dfrac{4}{27},$$
Money outcome $$=-50-50+100=0.$$

Case 4: Failures on all three throws.
Probability $$=q^{3}=\left(\dfrac23\right)^{3}=\dfrac{8}{27},$$
Money outcome $$=-50-50-50=-150.$$

Now we apply the definition of expected value: $$E=\sum (\text{probability})\times(\text{money outcome}).$$

So,

$$\begin{aligned} E &= \left(\dfrac13\right)(100) \;+\; \left(\dfrac29\right)(50) \;+\; \left(\dfrac{4}{27}\right)(0) \;+\; \left(\dfrac{8}{27}\right)(-150) \\[6pt] &= \dfrac{100}{3} \;+\; \dfrac{100}{9} \;+\; 0 \;-\; \dfrac{1200}{27}. \end{aligned}$$

To add these, we convert each term to a common denominator of 27:

$$ \dfrac{100}{3}=\dfrac{900}{27},\qquad \dfrac{100}{9}=\dfrac{300}{27},\qquad -\dfrac{1200}{27} \text{ is already over } 27. $$

Adding, we obtain

$$E=\dfrac{900}{27}+\dfrac{300}{27}-\dfrac{1200}{27}=0.$$

The expected value is zero rupees; the game is fair in the long run.

Hence, the correct answer is Option D.

First recall the definition of “the experiment ends on the fifth throw.” It means that

$$\text{4th throw}=4,\qquad \text{5th throw}=4$$

and that no pair of consecutive throws equal to $$4,4$$ has appeared before this pair. In other words, the pair made by the 4th and 5th throws must be the first occurrence of two successive fours.

Denote the five outcomes by $$x_1,x_2,x_3,x_4,x_5.$$ We have the two fixed conditions

$$x_4=4,\qquad x_5=4.$$

To prevent an earlier “double four,” the following extra conditions are necessary:

1. The pair $$(x_1,x_2)$$ must not be $$(4,4).$$ 2. Because $$x_4=4$$, the pair $$(x_3,x_4)$$ would be $$(4,4)$$ if $$x_3=4$$. Therefore we must force

$$x_3\ne 4.$$

No other restriction is needed, because once $$x_3\ne4$$ the pair $$(x_2,x_3)$$ automatically cannot be $$(4,4).$$

We now count the number of possible ordered triples $$(x_1,x_2,x_3)$$ satisfying these two conditions.

A. Start with the unrestricted count: each $$x_i$$ (for $$i=1,2$$) can be any of the six faces, while $$x_3$$ can be any face except 4. Hence

$$6 \times 6 \times 5 = 180$$

total triples before exclusion.

B. Next exclude the forbidden triples where $$(x_1,x_2)=(4,4).$$ If $$x_1=4$$ and $$x_2=4$$, then $$x_3$$ still has the same five possibilities (1,2,3,5,6). Thus the number of disallowed triples is

$$1 \times 1 \times 5 = 5.$$

C. Subtracting, the number of valid triples is

$$180-5=175.$$

For every one of these 175 admissible triples, the fourth and fifth throws are already fixed as $$4,4$$, giving exactly one full sequence of five throws. Therefore the total number of favourable five-throw sequences equals

$$175.$$

Because the die is fair, all $$6^5$$ sequences of five throws are equally likely, and the desired probability is

$$\dfrac{175}{6^5}.$$

Hence, the correct answer is Option B.

We know that if a random variable $$X$$ follows a binomial distribution, we write $$X \sim \text{Binomial}(n,p)$$, where

$$\text{Mean} = np \quad\text{and}\quad \text{Variance} = npq,$$

with $$q = 1-p$$.

According to the statement, the mean equals $$8$$ and the variance equals $$4$$. So we set up the two standard relations:

$$np = 8 \qquad\text{(1)}$$

$$npq = 4 \qquad\text{(2)}$$

From equation (1) we already have $$np = 8$$. Now divide equation (2) by equation (1):

$$\frac{npq}{np} \;=\; \frac{4}{8} \;\Longrightarrow\; q = \frac{1}{2}.$$

Because $$p + q = 1$$, we immediately get

$$p = 1 - q = 1 - \frac12 = \frac12.$$

Substituting $$p = \dfrac12$$ into equation (1), we find $$n$$:

$$n \Bigl(\frac12\Bigr) = 8 \;\Longrightarrow\; n = 16.$$

Thus $$X \sim \text{Binomial}(16,\tfrac12).$$ For such a distribution each term has probability

$$P(X = r) = \binom{16}{r}\Bigl(\frac12\Bigr)^r\Bigl(\frac12\Bigr)^{16-r} = \frac{\binom{16}{r}}{2^{16}}.$$

We are asked for $$P(X \leq 2)$$, i.e. the sum of the probabilities for $$r = 0,1,2$$:

$$P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$$

$$\;\;\;= \frac{\binom{16}{0}}{2^{16}} + \frac{\binom{16}{1}}{2^{16}} + \frac{\binom{16}{2}}{2^{16}}.$$

Now we compute each binomial coefficient explicitly:

$$\binom{16}{0} = 1,$$

$$\binom{16}{1} = 16,$$

$$\binom{16}{2} = \frac{16 \times 15}{2} = 120.$$

Adding them gives

$$1 + 16 + 120 = 137.$$

Hence

$$P(X \le 2) = \frac{137}{2^{16}}.$$

Comparing with the required form $$\dfrac{k}{2^{16}},$$ we identify

$$k = 137.$$

Hence, the correct answer is Option D.

Let us denote by $$n$$ the number of independent tosses of a fair coin we plan to make. Because the coin is fair, the probability of getting a head in a single toss is $$\dfrac12$$, and the probability of getting a tail is also $$\dfrac12$$.

We wish to find the smallest $$n$$ for which the probability of obtaining at least one head exceeds $$99\%$$, that is, is greater than $$0.99$$.

Instead of counting heads directly, it is easier to count the complementary event: “no head at all”, which means all tosses show tails. The rule of complementary probability states

$$P(\text{at least one head}) \;=\; 1 \;-\; P(\text{no head}).$$

For $$n$$ independent tosses, getting a tail every time has probability

$$P(\text{no head}) \;=\; \Bigl(\dfrac12\Bigr)^n,$$

because we multiply $$\dfrac12$$ with itself $$n$$ times.

Hence, using the complementary rule,

$$P(\text{at least one head}) \;=\; 1 \;-\; \Bigl(\dfrac12\Bigr)^n.$$

The requirement that this probability be more than $$99\%$$ translates to the inequality

$$1 \;-\; \Bigl(\dfrac12\Bigr)^n \;>\; 0.99.$$ Subtracting $$1$$ from both sides and changing the sign gives

$$-\Bigl(\dfrac12\Bigr)^n \;>\; -0.01,$$

which simplifies to

$$\Bigl(\dfrac12\Bigr)^n \;<\; 0.01.$$

To solve for $$n$$, we take logarithms. Stating the logarithm rule first: for any positive number $$a$$ and any base, $$\log(a^n)=n\log a.$$ Applying this rule, we write

$$\log\!\Bigl(\Bigl(\dfrac12\Bigr)^n\Bigr) \;<\; \log(0.01).$$

Using the logarithm rule mentioned above, we obtain

$$n\,\log\!\Bigl(\dfrac12\Bigr) \;<\; \log(0.01).$$

Now, $$\log\!\Bigl(\dfrac12\Bigr)$$ is negative (since $$\dfrac12<1$$). Dividing by a negative number reverses the inequality sign. Hence,

$$n \;>\; \dfrac{\log(0.01)}{\log\!\bigl(\dfrac12\bigr)}.$$

We evaluate the logs in base 10 for convenience:

$$\log_{10}(0.01) \;=\; -2,$$

$$\log_{10}\!\Bigl(\dfrac12\Bigr) \;=\; \log_{10}(1) - \log_{10}(2) \;=\; 0 - 0.3010 \;=\; -0.3010.$$

Substituting these numerical values, we have

$$n \;>\; \dfrac{-2}{-0.3010} \;=\; \dfrac{2}{0.3010} \;\approx\; 6.644.$$

The inequality demands $$n$$ be strictly greater than $$6.644$$, so the smallest integer satisfying it is

$$n = 7.$$

Therefore, we must toss the fair coin at least seven times to make the probability of getting at least one head exceed $$99\%$$.

Hence, the correct answer is Option D.

We want the probability of getting at least one head to be at least 90 %. A fair coin has probability $$\tfrac12$$ of showing a head and the same probability $$\tfrac12$$ of showing a tail on every toss.

It is often easier to work with the complementary event. The complementary event to “at least one head in $$n$$ tosses” is “no head at all in $$n$$ tosses”, i.e. all the $$n$$ tosses show tails.

For a single toss, the probability of getting a tail is $$\tfrac12$$. Because successive tosses are independent, the probability that all $$n$$ tosses are tails is

$$\left(\tfrac12\right)^n.$$

The required probability of getting at least one head is therefore

$$1-\left(\tfrac12\right)^n.$$

We are told this probability must be at least 90 %, that is $$0.90$$. Translating the verbal condition into an inequality, we write

$$1-\left(\tfrac12\right)^n \;\ge\; 0.9.$$

Rearranging, we subtract 1 from both sides and then multiply by −1; remembering that multiplying an inequality by a negative reverses the inequality sign, we get

$$\left(\tfrac12\right)^n \;\le\; 0.1.$$

To solve for $$n$$ we take natural logarithms (any logarithm base could be used, but using one base consistently is essential). First we state the logarithm rule we are about to use:

If $$a>0$$ and $$0<b<1$$, then $$b^a=c$$ implies $$a=\dfrac{\ln c}{\ln b}.$$

Applying this rule to our inequality

$$\left(\tfrac12\right)^n \le 0.1,$$

we obtain

$$n \,\ln\!\left(\tfrac12\right) \;\le\; \ln(0.1).$$

Because $$\ln\!\left(\tfrac12\right)$$ is negative, dividing by it reverses the inequality once again:

$$n \;\ge\; \dfrac{\ln(0.1)}{\ln\!\left(\tfrac12\right)}.$$

Now we evaluate the logarithms numerically:

$$\ln(0.1) = -2.302585\ldots,$$

$$\ln\!\left(\tfrac12\right) = -0.693147\ldots.$$

Substituting these values, $$ n \ge \dfrac{-2.302585\ldots}{-0.693147\ldots} = 3.321928\ldots $$

The number of tosses $$n$$ must be an integer, and it must be at least the value just obtained, so we round up to the next whole number:

$$n = 4.$$

We can verify quickly: with 4 tosses the probability of at least one head is

$$1-\left(\tfrac12\right)^4 =1-\dfrac1{16} =\dfrac{15}{16} =0.9375,$$ which indeed exceeds 0.90. For 3 tosses the probability is only $$1-\tfrac18=0.875,$$ which is insufficient.

Hence, the correct answer is Option B.

First we remind ourselves of the basic facts about a standard deck. A full deck has 52 cards in total, out of which exactly 4 are aces. When we draw a card and then replace it before the next draw, the two draws are statistically independent and each draw is always made from a fresh 52-card deck. Hence the single-draw probabilities never change.

Let us denote by $$p$$ the probability of drawing an ace in one single draw. Because there are 4 aces out of 52 cards, we have

$$ p \;=\; \frac{\text{number of favourable cards}}{\text{total number of cards}} \;=\; \frac{4}{52} \;=\; \frac{1}{13}. $$

Correspondingly, the probability of not drawing an ace in a single draw is

$$ 1-p \;=\; 1-\frac{1}{13} \;=\; \frac{12}{13}. $$

The random variable $$X$$ counts how many aces appear in the two independent draws. It can take the values 0, 1, or 2. The exercise asks us to find

$$ P(X=1) + P(X=2). $$

We shall compute the two probabilities separately and then add them.

Computation of $$P(X=2)$$: Both draws must show an ace. Because the draws are independent, we multiply the single-draw probabilities:

$$ P(X=2) \;=\; p \times p \;=\; \left(\frac{1}{13}\right)\!\left(\frac{1}{13}\right) \;=\; \frac{1}{169}. $$

Computation of $$P(X=1)$$: Exactly one draw must show an ace and the other draw must not. There are two distinct orders in which this can happen—

(i) ace on the first draw and non-ace on the second draw, or

(ii) non-ace on the first draw and ace on the second draw.

We add the probabilities of these two mutually exclusive cases.

For case (i):

$$ \text{Prob(case (i))} \;=\; p \times (1-p) \;=\; \frac{1}{13}\times\frac{12}{13} \;=\; \frac{12}{169}. $$

For case (ii):

$$ \text{Prob(case (ii))} \;=\; (1-p) \times p \;=\; \frac{12}{13}\times\frac{1}{13} \;=\; \frac{12}{169}. $$

Adding these two gives

$$ P(X=1) \;=\; \frac{12}{169} + \frac{12}{169} \;=\; \frac{24}{169}. $$

Now we find the required sum:

$$ P(X=1) + P(X=2) \;=\; \frac{24}{169} + \frac{1}{169} \;=\; \frac{25}{169}. $$

Hence, the correct answer is Option D.

We have the finite set $$\{1,2,3,\ldots ,11\}$$ from which two distinct integers are chosen at random, each unordered pair being equally likely.

The total of the two selected numbers is already known to be even. An even sum can arise in exactly two ways: either both chosen numbers are even or both are odd. Our task is to find the probability that the first of these possibilities occurs, under the condition that the sum is even.

Let us list the parity composition of the set. The even numbers are $$2,4,6,8,10,$$ so there are $$5$$ evens. The odd numbers are $$1,3,5,7,9,11,$$ so there are $$6$$ odds.

Number of unordered pairs where both numbers are even:

First, we state the combination formula: choosing $$r$$ objects from $$n$$ distinct objects is counted by $$\displaystyle \binom{n}{r}=\frac{n!}{r!(n-r)!}.$$ Using this, the count of pairs of evens is

$$\binom{5}{2}=\frac{5!}{2!\,3!}=10.$$

Number of unordered pairs where both numbers are odd:

$$\binom{6}{2}=\frac{6!}{2!\,4!}=15.$$

Because an even sum must be obtained from either two evens or two odds, every pair with even sum has now been enumerated. Hence the total number of unordered pairs whose sum is even equals

$$10+15=25.$$

Now we invoke the definition of conditional probability. If event $$E$$ denotes “sum is even” and event $$A$$ denotes “both numbers are even,” then

$$P(A\mid E)=\frac{\text{Number of pairs in }A}{\text{Number of pairs in }E} =\frac{10}{25} =\frac{2}{5}.$$

So the conditional probability that both selected numbers are even, given that their sum is even, equals $$\dfrac{2}{5}.$$

Hence, the correct answer is Option C.

Let us begin by taking the total population of the city as $$100$$ units so that every percentage mentioned in the question can be treated directly as that many “people out of 100”.

We are told that $$25\%$$ of the population reads newspaper A and $$20\%$$ reads newspaper B, while $$8\%$$ reads both A and B. Using these figures we first separate the population into mutually exclusive groups.

For the readers of only A, we subtract the overlap from the total A-readers:

We have $$\text{Only A readers}=25-8=17.$$ Thus $$17\%$$ of the population reads A but not B.

Similarly, for the readers of only B we subtract the overlap from the total B-readers:

We have $$\text{Only B readers}=20-8=12.$$ Thus $$12\%$$ of the population reads B but not A.

The readers of both papers have already been given as $$8\%.$$ All the above groups account for $$17+12+8=37$$ out of the $$100$$ people. The remaining $$100-37=63$$ people read neither A nor B, but they will not concern us further because the advertisement-looking percentages are given only for the three reading groups.

Now we incorporate the information about advertisements. The question specifies three different percentages:

• $$30\%$$ of those who read A but not B look into advertisements.
• $$40\%$$ of those who read B but not A look into advertisements.
• $$50\%$$ of those who read both A and B look into advertisements.

We multiply each percentage with the corresponding group size to find the actual proportion of the whole population that looks into advertisements from each group.

For the “only A” group:

$$\text{Ads from only A}=30\%\times17=\frac{30}{100}\times17=5.1.$$

For the “only B” group:

$$\text{Ads from only B}=40\%\times12=\frac{40}{100}\times12=4.8.$$

For the “both A and B” group:

$$\text{Ads from both}=50\%\times8=\frac{50}{100}\times8=4.$$

We now add these three mutually exclusive contributions to obtain the overall percentage of the city’s population that looks into advertisements:

$$\text{Total ad-lookers}=5.1+4.8+4=13.9.$$

So, $$13.9\%$$ of the entire population looks into advertisements.

Hence, the correct answer is Option C.

We are given the compound statement $$\bigl((p \wedge \sim q) \wedge (p \wedge r)\bigr)\;\rightarrow\;(\sim p \,\vee\, q)$$ and we are told that this entire implication is false.

First, recall the fundamental truth-table rule for an implication:

$$A \rightarrow B$$ is false exactly when $$A$$ is true and $$B$$ is false.

Applying this rule, let us denote

$$A = (p \wedge \sim q) \wedge (p \wedge r), \qquad B = \sim p \,\vee\, q.$$

Because the whole implication is false, we must have

$$A \text{ is true} \quad\text{and}\quad B \text{ is false.}$$

We analyse these two conditions one after another.

Condition 1: antecedent $$A$$ is true.

Write $$A$$ more clearly:

$$A = (p \wedge \sim q) \wedge (p \wedge r).$$

For a conjunction to be true, every part must be true. So we need

$$p \wedge \sim q \text{ is true} \quad\text{and}\quad p \wedge r \text{ is true.}$$

Each of these is itself a conjunction, so we split further:

• From $$p \wedge \sim q$$ being true we get

$$p \text{ is true} \quad\text{and}\quad \sim q \text{ is true.}$$

• From $$p \wedge r$$ being true we get

$$p \text{ is true} \quad\text{and}\quad r \text{ is true.}$$

Combining all these sub-conditions we have so far

$$p = \text{T}, \quad \sim q = \text{T}, \quad r = \text{T}.$$

The statement $$\sim q = \text{T}$$ immediately gives

$$q = \text{F}.$$

Condition 2: consequent $$B$$ is false.

Now look at

$$B = \sim p \,\vee\, q.$$

For a disjunction $$X \vee Y$$ to be false, both $$X$$ and $$Y$$ must be false. Therefore

$$\sim p = \text{F} \quad\text{and}\quad q = \text{F}.$$

We already have $$q = \text{F}$$ from the first condition, so that part is consistent. The requirement $$\sim p = \text{F}$$ means

$$p = \text{T},$$

which also matches what we obtained earlier. Thus both conditions are satisfied simultaneously by exactly one set of truth values:

$$p = \text{T}, \quad q = \text{F}, \quad r = \text{T}.$$

These correspond to the option <T, F, T>.

Hence, the correct answer is Option B.

We are given the Boolean expression $$\sim(p \vee q) \vee (\sim p \wedge q)$$ and we must find an equivalent, simpler form.

First recall De Morgan’s law, which states that $$\sim (A \vee B)=\sim A \wedge \sim B.$$

Applying this law to the first part, we have:

$$\sim(p \vee q)=\sim p \wedge \sim q.$$

Substituting this result into the original expression, we get

$$\bigl(\sim p \wedge \sim q\bigr)\;\; \vee \;\;(\sim p \wedge q).$$

Now observe that $$\sim p$$ is common in both terms of the disjunction. Using the distributive law $$A\wedge B \;\; \vee \;\; A\wedge C = A\wedge (B\vee C),$$ we factor out $$\sim p$$:

$$\bigl(\sim p \wedge \sim q\bigr)\;\; \vee \;\;(\sim p \wedge q) = \sim p \wedge \bigl(\,\sim q \vee q\bigr).$$

Inside the parentheses we have $$\sim q \vee q,$$ which is always true (a tautology). Denote the Boolean constant “true” by $$T$$. Hence

$$\sim q \vee q = T.$$

Substituting this back, we obtain

$$\sim p \wedge T.$$

In Boolean algebra, any proposition ANDed with $$T$$ remains unchanged, so

$$\sim p \wedge T = \sim p.$$

Thus, the original expression simplifies completely to $$\sim p.$$

Looking at the given options, $$\sim p$$ corresponds to Option B.

Hence, the correct answer is Option B.

Let the number of children in each family be $$n$$. Hence, the total number of children belonging to the two families together is $$2n$$.

Because no child can receive more than one ticket, distributing the three tickets is equivalent to choosing exactly three distinct children out of the total pool. Thus,

$$\text{Total ways} \;=\;\binom{2n}{3}.$$

The required event is that all the tickets go only to the children of family B. Since family B has $$n$$ children, the number of favourable selections equals the number of ways of choosing any three children out of these $$n$$, namely

$$\text{Favourable ways} \;=\;\binom{n}{3}.$$

By definition of probability,

$$P(\text{all three to B}) \;=\;\frac{\binom{n}{3}}{\binom{2n}{3}}.$$

We are told that this probability equals $$\dfrac{1}{12}$$. Therefore,

$$\frac{\binom{n}{3}}{\binom{2n}{3}} \;=\;\frac{1}{12}.$$

First we write the combinations in factorial form. Recalling the formula $$\binom{r}{k}=\dfrac{r!}{k!(r-k)!},$$ we get

$$\frac{\dfrac{n(n-1)(n-2)}{6}}{\dfrac{2n(2n-1)(2n-2)}{6}}=\frac{1}{12}.$$

The common factor $$6$$ cancels out, giving

$$\frac{n(n-1)(n-2)}{2n(2n-1)(2n-2)}=\frac{1}{12}.$$

We now cancel one factor of $$n$$ present in both numerator and denominator:

$$\frac{(n-1)(n-2)}{2(2n-1)(2n-2)}=\frac{1}{12}.$$

Cross-multiplying,

$$12(n-1)(n-2)=2(2n-1)(2n-2).$$

Dividing both sides by $$2$$ to simplify,

$$6(n-1)(n-2)=(2n-1)(2n-2).$$

Now we expand each side. First,

$$(n-1)(n-2)=n^{2}-3n+2,$$

so

$$6(n-1)(n-2)=6(n^{2}-3n+2)=6n^{2}-18n+12.$$

Next,

$$(2n-1)(2n-2)=4n^{2}-6n+2.$$

Equating the two expressions, we have

$$6n^{2}-18n+12=4n^{2}-6n+2.$$

Moving every term to the left side yields

$$6n^{2}-18n+12-4n^{2}+6n-2=0,$$

which simplifies to

$$2n^{2}-12n+10=0.$$

Dividing by $$2$$ throughout,

$$n^{2}-6n+5=0.$$

We solve the quadratic equation $$n^{2}-6n+5=0$$ using the quadratic-formula result $$n=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}.$$ Here $$a=1,\,b=-6,\,c=5$$, so

$$n=\frac{6\pm\sqrt{(-6)^{2}-4\cdot1\cdot5}}{2}=\frac{6\pm\sqrt{36-20}}{2}=\frac{6\pm4}{2}.$$

This gives two possible roots:

$$n=\frac{6+4}{2}=5 \quad\text{or}\quad n=\frac{6-4}{2}=1.$$

However, $$n=1$$ is impossible because we must be able to hand all three tickets to children of family B, so family B must have at least three children. Consequently, the only viable solution is $$n=5$$.

Therefore, each family has $$5$$ children.

Hence, the correct answer is Option B.

First, we note the contents of the bag before any drawing takes place. There are $$4$$ red balls and $$6$$ black balls, giving a total of $$10$$ balls.

We draw one ball at random and observe its colour. Let us compute the probability of each possible colour for this first draw. Since all balls are equally likely to be chosen, we have

$$P(\text{first ball is red})=\frac{4}{10}=\frac{2}{5},\qquad P(\text{first ball is black})=\frac{6}{10}=\frac{3}{5}.$$

After seeing the colour, we return that same ball and also place two extra balls of the same colour back into the bag. We now follow both cases separately in complete detail.

Case 1 - The first ball is red.

We removed one red, so the reds momentarily drop to $$3$$, but we immediately return the original red along with two more red balls. Thus the count of red balls becomes

$$3+3=6.$$

The black balls remain unchanged at $$6.$$ Hence the total number of balls in the bag is now

$$6+6=12.$$

The probability that the second (new) draw gives a red ball under this condition is therefore

$$P(\text{second red}\mid\text{first red})=\frac{6}{12}=\frac{1}{2}.$$

Case 2 - The first ball is black.

We removed one black, leaving $$5$$ black balls. Putting the original black back along with two more blacks adds $$3$$ blacks, so the black count increases to

$$5+3=8.$$

The number of red balls stays at $$4.$$ Thus the total number of balls is again

$$4+8=12.$$

The probability that the second draw gives a red ball in this situation is

$$P(\text{second red}\mid\text{first black})=\frac{4}{12}=\frac{1}{3}.$$

Now we combine these two conditional probabilities using the Law of Total Probability, which states

$$P(A)=P(B_1)\,P(A\mid B_1)+P(B_2)\,P(A\mid B_2)$$

when $$\{B_1,B_2\}$$ form a partition of the sample space. Here, $$A$$ is “second ball is red”, $$B_1$$ is “first ball is red”, and $$B_2$$ is “first ball is black”. Substituting the numbers we have already calculated,

$$\begin{aligned} P(\text{second red}) &=P(\text{first red})\,P(\text{second red}\mid\text{first red}) +P(\text{first black})\,P(\text{second red}\mid\text{first black})\\[6pt] &=\left(\frac{2}{5}\right)\left(\frac{1}{2}\right) +\left(\frac{3}{5}\right)\left(\frac{1}{3}\right)\\[6pt] &=\frac{2}{5}\cdot\frac{1}{2}+\frac{3}{5}\cdot\frac{1}{3}\\[6pt] &=\frac{1}{5}+\frac{1}{5}\\[6pt] &=\frac{2}{5}. \end{aligned}$$

Thus the probability that the ball drawn in the second draw is red equals $$\frac{2}{5}.$$

Hence, the correct answer is Option C.

We have two boxes and the first action is to pick one of them at random. Because no preference is mentioned, the probability of choosing either box is $$\frac12$$.

Let the event $$E$$ denote “exactly one white ball and one red ball are obtained in the two draws.” Let the event $$B$$ denote “the chosen box is box B.” Our aim is to find the conditional probability $$P(B\,|\,E)$$.

According to the definition of conditional probability,

$$P(B\,|\,E)=\dfrac{P(B\cap E)}{P(E)}.$$

We now compute the two probabilities in the fraction, first for “one white and one red” from each box, and then for the overall event.

Step 1: Probability of drawing one white and one red from box A.

Box A contains 2 white, 3 red and 2 black balls, a total of $$2+3+2 = 7$$ balls.

The number of ways of choosing any 2 balls from 7 is given by the combination formula $$^nC_r=\dfrac{n!}{r!\,(n-r)!},$$ so $$\binom{7}{2}.$$ Evaluating, $$\binom{7}{2}=\dfrac{7\cdot6}{2\cdot1}=21.$$

The favourable ways for exactly one white and one red are obtained by choosing 1 white out of 2 and 1 red out of 3, so $$\binom{2}{1}\binom{3}{1}=2\cdot3=6.$$

Thus,

$$P(E\text{ from }A)=\dfrac{6}{21}=\dfrac{2}{7}.$$

Step 2: Probability of drawing one white and one red from box B.

Box B contains 4 white, 2 red and 3 black balls, a total of $$4+2+3 = 9$$ balls.

The total number of pairs that can be chosen is $$\binom{9}{2}=\dfrac{9\cdot8}{2\cdot1}=36.$$

The favourable selections (1 white, 1 red) are counted by $$\binom{4}{1}\binom{2}{1}=4\cdot2=8.$$

Therefore,

$$P(E\text{ from }B)=\dfrac{8}{36}=\dfrac{2}{9}.$$

Step 3: Probability of the event $$E$$ overall.

Because each box is chosen with probability $$\frac12$$ and draws are made independently thereafter, the law of total probability gives

$$P(E)=P(A)\,P(E\text{ from }A)+P(B)\,P(E\text{ from }B).$$

Substituting $$P(A)=P(B)=\frac12$$ along with the results from Steps 1 and 2, we get

$$P(E)=\frac12\cdot\frac{2}{7}+\frac12\cdot\frac{2}{9}.$$

Combine the two fractions inside the parentheses first:

$$\frac{2}{7}+\frac{2}{9}=\frac{2\cdot9}{7\cdot9}+\frac{2\cdot7}{9\cdot7}=\frac{18+14}{63}=\frac{32}{63}.$$

Now multiply by $$\frac12$$:

$$P(E)=\frac12\cdot\frac{32}{63}=\frac{16}{63}.$$

Step 4: Joint probability $$P(B\cap E).$$

This is simply the product of choosing box B and then getting the desired colour combination:

$$P(B\cap E)=P(B)\,P(E\text{ from }B)=\frac12\cdot\frac{2}{9}=\frac{1}{9}.$$

Step 5: Required conditional probability.

Using the formula from the beginning,

$$P(B\,|\,E)=\dfrac{P(B\cap E)}{P(E)}=\dfrac{\frac{1}{9}}{\frac{16}{63}}.$$

To simplify, multiply numerator and denominator:

$$P(B\,|\,E)=\frac{1}{9}\times\frac{63}{16}=\frac{63}{144}.$$

We now reduce the fraction by dividing numerator and denominator by their greatest common divisor, which is 9:

$$\frac{63\div9}{144\div9}=\frac{7}{16}.$$

Hence, the correct answer is Option A.

Let us denote by $$P_X$$ the probability that player X finally wins the game and by $$P_Y$$ the probability that player Y finally wins the game. Because somebody must eventually show a head, we obviously have

$$P_X+P_Y=1.$$

According to the statement, the chances of the two players are the same, that is

$$P_X=P_Y.$$

Substituting $$P_Y=1-P_X$$ into the above equality we get

$$P_X=1-P_X \;\;\Longrightarrow\;\; 2P_X=1 \;\;\Longrightarrow\;\; P_X=\dfrac12.$$

So our task is to find the value of $$p$$ (the probability of head for X’s biased coin) that makes $$P_X=\dfrac12.$$

Now we analyse the game step by step. Player X starts the first toss.

Formula we use: When a process can return to its initial state, we can write a recurrence $$P_X=\text{(probability X wins immediately)}+\text{(probability game resets)}\times P_X.$$

We have:

1. X wins right away if he tosses a head on his first attempt. The probability of this event is simply $$p.$$

2. If X fails (i.e. gets a tail) and Y also fails (i.e. gets a tail), the situation is exactly the same as at the very beginning, with X to play next.    • Probability that X fails = $$1-p.$$    • Probability that Y fails = $$1-\dfrac12=\dfrac12.$$    • Hence, probability that both fail consecutively = $$(1-p)\times\dfrac12.$$    • After this double failure, the game is in its initial state, so the chance that X eventually wins from there is again $$P_X.$$

Putting these cases together, the recurrence relation becomes

$$P_X \;=\; p + \Bigl[(1-p)\times\dfrac12\Bigr]\;P_X.$$

We now solve this equation for $$P_X$$ in terms of $$p.$$

First distribute the bracket:

$$P_X = p + \dfrac{1-p}{2}\,P_X.$$

Bring the term containing $$P_X$$ on the right to the left-hand side:

$$P_X - \dfrac{1-p}{2}\,P_X = p.$$

Factor out $$P_X$$ on the left:

$$P_X\Bigl[\,1-\dfrac{1-p}{2}\Bigr]=p.$$

Simplify the expression inside the brackets:

$$1-\dfrac{1-p}{2}=\dfrac{2}{2}-\dfrac{1-p}{2}= \dfrac{2-(1-p)}{2}= \dfrac{1+p}{2}.$$

So the equation now reads

$$P_X \times \dfrac{1+p}{2}=p.$$

Isolate $$P_X$$:

$$P_X=\dfrac{2p}{1+p}.$$

But earlier we proved that $$P_X=\dfrac12.$$ Setting the two expressions equal gives

$$\dfrac{2p}{1+p}=\dfrac12.$$

Cross-multiplying, we get

$$4p = 1 + p.$$

Subtract $$p$$ from both sides:

$$4p - p = 1 \;\;\Longrightarrow\;\; 3p = 1.$$

Finally, dividing by 3 gives

$$p = \dfrac13.$$

This value exactly matches Option A.

Hence, the correct answer is Option A.

We have three events $$A,\;B,\;C$$ that are pair-wise independent, so by definition we know

$$P(A\cap B)=P(A)\,P(B),\qquad P(A\cap C)=P(A)\,P(C),\qquad P(B\cap C)=P(B)\,P(C).$$

It is given that the simultaneous occurrence of all three events is impossible, that is

$$P(A\cap B\cap C)=0,$$

and also that $$P(C)>0.$$

We are asked to find the conditional probability

$$P\!\left[(\bar A\cap\bar B)\mid C\right].$$

By the definition of conditional probability,

$$P\!\left[(\bar A\cap\bar B)\mid C\right]= \frac{P\!\big((\bar A\cap\bar B)\cap C\big)}{P(C)}.$$

We now compute the numerator $$P((\bar A\cap\bar B)\cap C).$$ Observe that

$$(\bar A\cap\bar B)\cap C = C\setminus\big((A\cup B)\cap C\big) = C-\big((A\cap C)\cup(B\cap C)\big).$$

Using the addition rule for probabilities, we have

$$P\!\big((A\cap C)\cup(B\cap C)\big) = P(A\cap C)+P(B\cap C)-P(A\cap B\cap C).$$

Substituting the given information and the results from pair-wise independence:

$$P(A\cap C)=P(A)P(C),\qquad P(B\cap C)=P(B)P(C),\qquad P(A\cap B\cap C)=0.$$

So

$$P\!\big((A\cap C)\cup(B\cap C)\big) = P(A)P(C)+P(B)P(C)-0 = P(C)\big(P(A)+P(B)\big).$$

Hence

$$P\big((\bar A\cap\bar B)\cap C\big) = P(C)-P\!\big((A\cap C)\cup(B\cap C)\big) = P(C)-P(C)\big(P(A)+P(B)\big).$$

Taking $$P(C)$$ common, we get

$$P\big((\bar A\cap\bar B)\cap C\big) = P(C)\big[1-P(A)-P(B)\big] = P(C)\big[(1-P(A))-P(B)\big].$$

Recalling that $$1-P(A)=P(\bar A),$$ we have

$$P\big((\bar A\cap\bar B)\cap C\big) = P(C)\big[P(\bar A)-P(B)\big].$$

Now divide by $$P(C)$$ to obtain the desired conditional probability:

$$P\!\left[(\bar A\cap\bar B)\mid C\right] = \frac{P(C)\big[P(\bar A)-P(B)\big]}{P(C)} = P(\bar A)-P(B).$$

Thus, the required value equals $$P(\bar A)-P(B),$$ which matches Option A.

Hence, the correct answer is Option A.

We are asked to find the probability that, when an unbiased coin is tossed eight times, we get at least one Head and at least one Tail.

First, recall the basic probability formula: if all outcomes are equally likely, then

Probability $$=\frac{$$ Number of favourable outcomes $$}{$$ Total number of possible outcomes $$}.$$

However, counting directly the “favourable outcomes” (all sequences that contain both Heads and Tails) is lengthy. A common technique is to use the Complement Rule, which states:

$$P($$ Event $$) = 1 - P($$ Complement of the event $$).$$

Here the event we want is “at least one Head and at least one Tail.” Its complement is “no Head or no Tail,” which translates to “all Heads” or “all Tails.”

Now, let us determine the total number of possible outcomes when a coin is tossed eight times. Each toss has 2 possible results (H or T). Hence, by the rule of product, the total number of sequences of length 8 is

$$2^8 = 256.$$

Because the coin is unbiased, every sequence is equally likely, so the probability of any single sequence is

$$\frac{1}{256}.$$

Next, we count the outcomes in the complement:

• All eight tosses are Heads: there is exactly one such sequence, namely HHHHHHHH.
• All eight tosses are Tails: again, there is exactly one sequence, namely TTTTTTTT.

Thus, the number of complementary outcomes is

$$1 + 1 = 2.$$

Therefore, the probability of the complement event (“all Heads or all Tails”) is

$$P(\text{all Heads or all Tails}) = \frac{2}{256}.$$

We can simplify this fraction step by step:

First divide numerator and denominator by 2:

$$\frac{2}{256} = \frac{1}{128}.$$

Now we apply the Complement Rule:

$$\begin{aligned} P(\text{at least one Head and at least one Tail}) &= 1 - P(\text{all Heads or all Tails}) \\[4pt] &= 1 - \frac{1}{128}. \end{aligned}$$

We convert the whole number 1 to a fraction with denominator 128 to combine easily:

$$1 = \frac{128}{128}.$$

Subtracting term-by-term, we obtain

$$\frac{128}{128} - \frac{1}{128} = \frac{127}{128}.$$

So the required probability is

$$\frac{127}{128}.$$

Hence, the correct answer is Option A.

Let us denote the single-event probabilities as $$P(A),\;P(B),\;P(C)$$ and the pairwise intersection probabilities as $$P(A\cap B),\;P(B\cap C),\;P(C\cap A)$$. For ease of writing we put

$$x=P(A)+P(B)+P(C),\qquad y=P(A\cap B)+P(B\cap C)+P(C\cap A).$$

First recall the standard result for two events:

Probability that exactly one of the two events $$A$$ or $$B$$ occurs is given by

$$P(\text{exactly one of }A\text{ or }B)=P(A)+P(B)-2P(A\cap B).$$

We now apply this formula to the three statements given in the question.

For $$A$$ and $$B$$ we have

$$P(A)+P(B)-2P(A\cap B)=\frac14.$$ For $$B$$ and $$C$$ we have

$$P(B)+P(C)-2P(B\cap C)=\frac14.$$ For $$C$$ and $$A$$ we have

$$P(C)+P(A)-2P(C\cap A)=\frac14.$$

Adding these three equations term-by-term, we obtain

$$\bigl[P(A)+P(B)+P(C)\bigr]+\bigl[P(A)+P(B)+P(C)\bigr]\; -\;2\bigl[P(A\cap B)+P(B\cap C)+P(C\cap A)\bigr] =\frac14+\frac14+\frac14.$$

That is

$$2\bigl[x-y\bigr]=\frac34\quad\Longrightarrow\quad x-y=\frac38.$$ Hence

$$x-y=\frac38. \quad -(1)$$

Now use the inclusion-exclusion formula for three events:

$$P(A\cup B\cup C)=x-y+P(A\cap B\cap C).$$

We are given $$P(A\cap B\cap C)=\dfrac1{16},$$ so substituting this and the value from (1) gives

$$P(A\cup B\cup C)=\frac38+\frac1{16} =\frac{6}{16}+\frac{1}{16} =\frac{7}{16}.$$

Thus the probability that at least one of the three events occurs equals $$\dfrac{7}{16}.$$

Hence, the correct answer is Option B.

The group contains $$10$$ men and $$5$$ women, so in total $$15$$ people. Four-member committees are to be formed with the condition that each committee must contain at least one woman. We wish to find the probability that such a committee has more women than men.

When a committee of four has more women than men, the number of women $$W$$ is greater than the number of men $$M$$. Because $$W+M=4$$, the only possible distributions are:

$$\begin{aligned} W=3,\;M=1 \quad\text{or}\quad W=4,\;M=0 \end{aligned}$$

First, the size of the sample space is the total number of four-member committees that contain at least one woman. Start with the number of unrestricted four-member committees chosen from all $$15$$ people:

$$\binom{15}{4}=\frac{15\times14\times13\times12}{4\times3\times2\times1}=1365\qquad -(1)$$

The only committees that violate the “at least one woman” condition are the all-male committees. The number of such all-male committees is:

$$\binom{10}{4}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210\qquad -(2)$$

Subtracting equation $$-(2)$$ from equation $$-(1)$$ gives the size of the sample space:

$$1365-210=1155\qquad -(3)$$

Next, count the favourable committees.

Case 1: 3 women and 1 man.
Choose $$3$$ women out of $$5$$ and $$1$$ man out of $$10$$:

$$\binom{5}{3}\times\binom{10}{1}=10\times10=100\qquad -(4)$$

Case 2: 4 women and 0 men.
Choose $$4$$ women out of $$5$$:

$$\binom{5}{4}=5\qquad -(5)$$

Adding equations $$-(4)$$ and $$-(5)$$ gives the total number of favourable committees:

$$100+5=105\qquad -(6)$$

The required probability is the quotient of equations $$-(6)$$ and $$-(3)$$:

$$\frac{105}{1155}\qquad -(7)$$

Simplify the fraction by dividing numerator and denominator by $$105$$:

$$\frac{105\div105}{1155\div105}=\frac{1}{11}\qquad -(8)$$

Therefore, the probability that a four-member committee satisfying the “at least one woman” condition has more women than men is $$\dfrac{1}{11}$$.

Hence, the correct answer is Option C.

We have to choose two different numbers from the finite set $$\{0,1,2,3,\ldots ,10\}$$ and we want both their sum and their absolute difference to be multiples of $$4$$.

First, let us count the total number of unordered pairs of distinct numbers that can be chosen from the eleven-element set. The formula for combinations gives

$$^ {11}C_{2} \;=\;\frac{11\times10}{2}=55.$$

So there are $$55$$ possible pairs in all.

Now we must find how many of these pairs satisfy both conditions simultaneously:

$$a+b\equiv0\;(\text{mod }4)\quad\text{and}\quad|a-b|\equiv0\;(\text{mod }4).$$

Because the absolute difference is a multiple of $$4$$, the two numbers must give the same remainder when divided by $$4$$; in other words, they lie in the same congruence class modulo $$4$$. Let us therefore write the common remainder as $$r$$, with

$$r\in\{0,1,2,3\}\quad\text{and}\quad a\equiv r\;(\text{mod }4),\; b\equiv r\;(\text{mod }4).$$

Next, insist that their sum is also a multiple of $$4$$:

$$a+b\equiv r+r=2r\equiv0\;(\text{mod }4).$$

So we need $$2r\equiv0\;(\text{mod }4).$$ Let us examine the four possible values of $$r$$:

$$\begin{aligned} r=0&\;:\;2r=0\equiv0\pmod4\quad\text{(allowed)},\\ r=1&\;:\;2r=2\not\equiv0\pmod4\quad\text{(rejected)},\\ r=2&\;:\;2r=4\equiv0\pmod4\quad\text{(allowed)},\\ r=3&\;:\;2r=6\equiv2\pmod4\quad\text{(rejected)}. \end{aligned}$$

Hence only the two residue classes $$r=0$$ and $$r=2$$ satisfy both conditions. We must therefore list the members of the original set that fall into these two classes:

Residue $$0$$ modulo $$4$$: $$0,4,8$$

Residue $$2$$ modulo $$4$$: $$2,6,10$$

All pairs chosen from within the same class already have a difference divisible by $$4$$, and since the class itself is admissible, their sum is automatically a multiple of $$4$$ as well. Thus every distinct pair drawn from each of these two groups is favourable.

Number of favourable pairs from the residue-$$0$$ group:

$$^3C_2=\frac{3\times2}{2}=3.$$

Number of favourable pairs from the residue-$$2$$ group:

$$^3C_2=3.$$

Total favourable pairs:

$$3+3=6.$$

Finally, the required probability equals the ratio of favourable pairs to total pairs:

$$\text{Probability}=\frac{6}{55}.$$

Hence, the correct answer is Option A.

Let the probabilities of the two independent events be denoted by

$$P(E)=p,\;P(F)=q\qquad -(1)$$

Because the events are independent, the probability of their simultaneous occurrence is simply the product of their individual probabilities. The question tells us that this probability equals $$\frac{1}{12}$$, so

$$pq=\frac{1}{12}\qquad -(2)$$

It is also given that neither event happens with probability $$\frac{1}{2}$$. For independent events, the probability that both fail is the product of their individual failure probabilities:

$$P(E'\cap F')=(1-p)(1-q)=\frac{1}{2}\qquad -(3)$$

Now expand the left‐hand side of $$(3)$$:

$$(1-p)(1-q)=1-p-q+pq\qquad -(4)$$

Substituting $$pq$$ from $$(2)$$ into $$(4)$$ gives

$$1-p-q+\frac{1}{12}=\frac{1}{2}\qquad -(5)$$

Move all constant terms to one side and the variables to the other:

$$1+\frac{1}{12}-\frac{1}{2}=p+q\qquad -(6)$$

Convert everything to a common denominator $$12$$ inside $$(6)$$:

$$\frac{12}{12}+\frac{1}{12}-\frac{6}{12}=p+q\qquad -(7)$$

Simplifying $$(7)$$ gives

$$\frac{7}{12}=p+q\qquad -(8)$$

So we now have two simultaneous relations, $$(2)$$ and $$(8)$$:

$$pq=\frac{1}{12},\qquad p+q=\frac{7}{12}\qquad -(9)$$

Solve for one variable in terms of the other using $$(8)$$; let us write

$$q=\frac{7}{12}-p\qquad -(10)$$

Substitute $$(10)$$ into $$(2)$$:

$$p\left(\frac{7}{12}-p\right)=\frac{1}{12}\qquad -(11)$$

Multiply both sides of $$(11)$$ by $$12$$ to remove denominators:

$$7p-12p^{2}=1\qquad -(12)$$

Rearrange $$(12)$$ into the standard quadratic form:

$$12p^{2}-7p+1=0\qquad -(13)$$

Compute the discriminant $$D$$ of $$(13)$$:

$$D=(-7)^{2}-4(12)(1)=49-48=1\qquad -(14)$$

The square root of the discriminant is $$\sqrt{D}=1$$, so the quadratic formula yields

$$p=\frac{7\pm1}{2\cdot12}=\frac{7\pm1}{24}\qquad -(15)$$

This gives two possible values:

$$p=\frac{8}{24}=\frac{1}{3}\quad\text{or}\quad p=\frac{6}{24}=\frac{1}{4}\qquad -(16)$$

Using $$(10)$$ to find the corresponding $$q$$ values:

If $$p=\frac{1}{3}$$, then $$q=\frac{7}{12}-\frac{1}{3}=\frac{7}{12}-\frac{4}{12}=\frac{3}{12}=\frac{1}{4}\qquad -(17)$$

If $$p=\frac{1}{4}$$, then $$q=\frac{7}{12}-\frac{1}{4}=\frac{7}{12}-\frac{3}{12}=\frac{4}{12}=\frac{1}{3}\qquad -(18)$$

Thus the unordered pair $$\{p,q\}$$ is $$\left\{\frac{1}{3},\frac{1}{4}\right\}$$, and either event could take either value. The required ratio is

$$\frac{P(E)}{P(F)}=\frac{p}{q}\qquad -(19)$$

When $$p=\frac{1}{3},\;q=\frac{1}{4}$$, equation $$(19)$$ gives

$$\frac{1/3}{1/4}=\frac{1}{3}\times4=\frac{4}{3}\qquad -(20)$$

If the roles are reversed, $$(19)$$ would give $$\frac{1}{4}/\frac{1}{3}=\frac{3}{4}$$, which is not among the answer choices. Hence the admissible value that matches an option is

$$\frac{P(E)}{P(F)}=\frac{4}{3}\qquad -(21)$$

Hence, the correct answer is Option A.

The probability that person P hits the target is given to be $$\dfrac{3}{4}$$.

Similarly, the probability that person Q hits the target is $$\dfrac{1}{2}$$, and the probability that person R hits the target is $$\dfrac{5}{8}$$.

All three persons act independently, so the probabilities of their individual successes or failures can be multiplied when we need a combined (joint) probability.

We are asked to find the probability that the target is hit by P or Q, but not by R.

First, we consider R’s failure, because the phrase “but not by R” makes R’s missing the target a compulsory condition.

The probability that R misses the target is the complement of the probability that R hits it.

Using the complementary rule $$P(\text{miss}) = 1 - P(\text{hit})$$, we have

$$P(\text{R misses}) = 1 - \dfrac{5}{8} = \dfrac{3}{8}.$$

Next, we need the probability that the target is hit by at least one of P or Q.

Again we use the complementary rule. It is easier to find the probability that neither P nor Q hits, and then subtract that from 1.

The probability that P misses is $$1 - \dfrac{3}{4} = \dfrac{1}{4},$$

and the probability that Q misses is $$1 - \dfrac{1}{2} = \dfrac{1}{2}.$$

Because P and Q act independently, the probability that both miss is the product

$$P(\text{P misses and Q misses}) = \dfrac{1}{4} \times \dfrac{1}{2} = \dfrac{1}{8}.$$

Therefore, the probability that at least one of P or Q hits is

$$P(\text{P or Q (at least one) hits}) = 1 - \dfrac{1}{8} = \dfrac{7}{8}.$$

Now we combine the two independent requirements:

(i) R must miss, which occurs with probability $$\dfrac{3}{8},$$ and

(ii) at least one of P or Q must hit, which occurs with probability $$\dfrac{7}{8}.$$

Multiplying these independent probabilities, we obtain

$$P(\text{desired event}) = \dfrac{3}{8} \times \dfrac{7}{8} = \dfrac{21}{64}.$$

Hence, the correct answer is Option B.

We are told that the experiment “succeeds twice as often as it fails.” Let the probability of one success be $$p$$ and the probability of one failure be $$q$$. The statement “twice as often” translates to the ratio

$$p : q = 2 : 1.$$ So we have the relationship

$$p = 2q.$$

Because every individual trial must result in either success or failure, the two probabilities add up to $$1$$. Hence

$$p + q = 1.$$

Substituting $$p = 2q$$ into this equation gives

$$2q + q = 1 \;\;\Longrightarrow\;\; 3q = 1 \;\;\Longrightarrow\;\; q = \frac{1}{3}.$$

Now we find $$p$$ using $$p = 2q$$:

$$p = 2 \left(\frac{1}{3}\right) = \frac{2}{3}.$$

Thus, for each independent trial,

$$p = \frac{2}{3}, \quad q = \frac{1}{3}.$$

We perform $$n = 6$$ independent trials. Let the random variable $$X$$ denote the number of successes in these six trials. Because each trial is independent and has the same success probability, $$X$$ follows the binomial distribution. The probability mass function of a binomially distributed variable is

$$P(X = r) = \binom{n}{r} p^{\,r} q^{\,n-r},$$

where $$\binom{n}{r}$$ is the binomial coefficient.

The question asks for the probability of “at least 5 successes,” which means either 5 successes or 6 successes. We therefore need

$$P(X \ge 5) = P(X = 5) + P(X = 6).$$

First we compute $$P(X = 5)$$. Using the formula with $$n = 6$$ and $$r = 5$$:

$$P(X = 5) = \binom{6}{5} \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^{6-5}.$$

The binomial coefficient is

$$\binom{6}{5} = 6.$$ Therefore

$$P(X = 5) = 6 \left(\frac{2}{3}\right)^5 \left(\frac{1}{3}\right)^1 = 6 \cdot \frac{2^5}{3^5} \cdot \frac{1}{3} = 6 \cdot \frac{32}{243} \cdot \frac{1}{3} = 6 \cdot \frac{32}{729} = \frac{192}{729}.$$

Next, we compute $$P(X = 6)$$ with $$r = 6$$:

$$P(X = 6) = \binom{6}{6} \left(\frac{2}{3}\right)^6 \left(\frac{1}{3}\right)^{6-6}.$$

Since $$\binom{6}{6} = 1$$ and $$\left(\frac{1}{3}\right)^0 = 1,$$ we get

$$P(X = 6) = 1 \cdot \left(\frac{2}{3}\right)^6 = \frac{2^6}{3^6} = \frac{64}{729}.$$

Now we add the two probabilities:

$$P(X \ge 5) = \frac{192}{729} + \frac{64}{729} = \frac{256}{729}.$$

Hence, the correct answer is Option D.

We are given that $$P(A) = \frac{2}{5}$$ and $$P(A \cap B) = \frac{3}{20}$$. We need to find $$P(A \mid (A' \cup B'))$$, where $$A'$$ and $$B'$$ are the complements of events $$A$$ and $$B$$ respectively.

The conditional probability is defined as $$P(A \mid C) = \frac{P(A \cap C)}{P(C)}$$, where $$C = A' \cup B'$$. So, we have:

$$P(A \mid (A' \cup B')) = \frac{P(A \cap (A' \cup B'))}{P(A' \cup B')}$$

First, simplify the numerator $$P(A \cap (A' \cup B'))$$. Using the distributive law of sets:

$$A \cap (A' \cup B') = (A \cap A') \cup (A \cap B')$$

Since $$A \cap A' = \emptyset$$ (the empty set, as $$A$$ and its complement are disjoint), this simplifies to:

$$A \cap (A' \cup B') = \emptyset \cup (A \cap B') = A \cap B'$$

Thus, $$P(A \cap (A' \cup B')) = P(A \cap B')$$.

Now, $$A \cap B'$$ represents the event that $$A$$ occurs but $$B$$ does not. This can also be written as $$A \setminus (A \cap B)$$. Since $$A \cap B \subseteq A$$, we have:

$$P(A \cap B') = P(A) - P(A \cap B)$$

Substitute the given probabilities:

$$P(A) = \frac{2}{5} = \frac{8}{20}, \quad P(A \cap B) = \frac{3}{20}$$

So,

$$P(A \cap B') = \frac{8}{20} - \frac{3}{20} = \frac{5}{20} = \frac{1}{4}$$

Therefore, the numerator is $$\frac{1}{4}$$.

Next, simplify the denominator $$P(A' \cup B')$$. Using De Morgan's law:

$$A' \cup B' = (A \cap B)'$$

So,

$$P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B)$$

Substitute $$P(A \cap B) = \frac{3}{20}$$:

$$P(A' \cup B') = 1 - \frac{3}{20} = \frac{20}{20} - \frac{3}{20} = \frac{17}{20}$$

Therefore, the denominator is $$\frac{17}{20}$$.

Now, the conditional probability is:

$$P(A \mid (A' \cup B')) = \frac{\frac{1}{4}}{\frac{17}{20}}$$

Dividing fractions is equivalent to multiplying by the reciprocal:

$$\frac{\frac{1}{4}}{\frac{17}{20}} = \frac{1}{4} \times \frac{20}{17} = \frac{1 \times 20}{4 \times 17} = \frac{20}{68}$$

Simplify $$\frac{20}{68}$$ by dividing both numerator and denominator by 4:

$$\frac{20 \div 4}{68 \div 4} = \frac{5}{17}$$

Thus, $$P(A \mid (A' \cup B')) = \frac{5}{17}$$.

Comparing with the options:

• A. $$\frac{11}{20}$$
• B. $$\frac{5}{17}$$
• C. $$\frac{8}{17}$$
• D. $$\frac{1}{4}$$

Hence, the correct answer is Option B.

We begin by noting that each of the two dice is fair and six-faced, so the ordered pairs of outcomes $$\bigl(A,B\bigr)=(1,1),(1,2),\dots,(6,6)$$ are all equally likely. There are in total $$6\times 6=36$$ such pairs, each having probability $$\dfrac1{36}$$.

The three events are defined as follows:

$$E_1:\;A=4,\qquad E_2:\;B=2,\qquad E_3:\;A+B\text{ is odd.}$$

First, we compute the individual probabilities.

For $$E_1$$ we require the first die to show four. The second die may show any of the six numbers, giving $$6$$ favourable pairs: $$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6).$$ Hence $$P(E_1)=\dfrac6{36}=\dfrac16.$$ Exactly the same reasoning applies to $$E_2$$, so $$P(E_2)=\dfrac16.$$ For $$E_3$$, the sum is odd exactly when one die is even and the other odd. There are $$3$$ even numbers and $$3$$ odd numbers on a die, so the number of favourable pairs is $$3\times 3+3\times 3=18.$$ Thus $$P(E_3)=\dfrac{18}{36}=\dfrac12.$$

Now we examine pairwise intersections in order to check pairwise independence.

Intersection of $$E_1$$ and $$E_2$$. Both $$A=4$$ and $$B=2$$ must hold, leaving the single outcome $$(4,2).$$ Therefore $$P(E_1\cap E_2)=\dfrac1{36},\qquad P(E_1)\,P(E_2)=\dfrac16\cdot\dfrac16=\dfrac1{36}.$$ Because these two numbers are equal, $$E_1$$ and $$E_2$$ are independent.

Intersection of $$E_1$$ and $$E_3$$. We already have $$A=4$$ (an even number). For the sum to be odd, $$B$$ must be odd, i.e. $$B=1,3,5.$$ The three favourable outcomes are $$(4,1),(4,3),(4,5).$$ Thus $$P(E_1\cap E_3)=\dfrac3{36}=\dfrac1{12},\qquad P(E_1)\,P(E_3)=\dfrac16\cdot\dfrac12=\dfrac1{12}.$$ The equality confirms that $$E_1$$ and $$E_3$$ are independent.

Intersection of $$E_2$$ and $$E_3$$. Now $$B=2$$ (even), and we need the sum to be odd, so $$A$$ must be odd: $$A=1,3,5.$$ The three favourable pairs are $$(1,2),(3,2),(5,2).$$ Hence $$P(E_2\cap E_3)=\dfrac3{36}=\dfrac1{12},\qquad P(E_2)\,P(E_3)=\dfrac16\cdot\dfrac12=\dfrac1{12}.$$ Again the two values match, so $$E_2$$ and $$E_3$$ are independent.

So far, every pair of events is independent. To see whether all three events are mutually independent, we must consider the triple intersection.

Intersection of $$E_1,$$ $$E_2,$$ and $$E_3$$. Imposing $$A=4$$ and $$B=2$$ fixes the only possible outcome to be $$(4,2).$$ However, for this outcome the sum is $$4+2=6,$$ which is even, not odd. Therefore no single ordered pair satisfies all three conditions simultaneously, and $$P(E_1\cap E_2\cap E_3)=0.$$ On the other hand, $$P(E_1)\,P(E_2)\,P(E_3)=\dfrac16\cdot\dfrac16\cdot\dfrac12=\dfrac1{72}.$$ Because $$0\neq\dfrac1{72},$$ the three events $$E_1,E_2,E_3$$ are not mutually independent.

We have now gathered all the information required to assess the given statements:

• Statement A: “$$E_1$$ and $$E_3$$ are independent.” — True.
• Statement B: “$$E_1$$, $$E_2$$ and $$E_3$$ are independent.” — Not true (they are not mutually independent).
• Statement C: “$$E_1$$ and $$E_2$$ are independent.” — True.
• Statement D: “$$E_2$$ and $$E_3$$ are independent.” — True.

Hence, the correct answer is Option B.