JEE Probability Questions

P(B₁)=P(B₂)=P(B₃)=1/3

P(W|B₁)=6/10, P(W|B₂)=4/10, P(W|B₃)=5/10

P(W) = (1/3)(6/10+4/10+5/10) = (1/3)(15/10) = 1/2

P(B₂|W) = P(W|B₂)P(B₂)/P(W) = (4/10)(1/3)/(1/2) = (4/30)/(1/2) = 8/30 = 4/15

The correct answer is Option 1: 4/15.

Let the events be:
  $$B_1$$ : bag I is selected,  $$B_2$$ : bag II is selected,  $$B_3$$ : bag III is selected.
  $$R$$ : the ball drawn is Red,  $$G$$ : the ball drawn is Green.

Because the person picks a bag at random, $$P(B_1)=P(B_2)=P(B_3)=\frac13$$.

Conditional probabilities for drawing a Red ball from the three bags are
  $$P(R\mid B_1)=\frac{3}{10},\qquad P(R\mid B_2)=\frac{4}{10},\qquad P(R\mid B_3)=\frac{5}{10}$$

1. Total probability of drawing a Red ball (use the law of total probability):
$$P(R)=P(B_1)P(R\mid B_1)+P(B_2)P(R\mid B_2)+P(B_3)P(R\mid B_3)$$ $$=\frac13\left(\frac{3}{10}+\frac{4}{10}+\frac{5}{10}\right)=\frac13\cdot\frac{12}{10}=\frac{12}{30}=\frac25\;-(1)$$

2. Probability that the Red ball came from bag I (Bayes’ theorem):
$$p=P(B_1\mid R)=\frac{P(B_1)P(R\mid B_1)}{P(R)}=\frac{\frac13\cdot\frac{3}{10}}{\frac25}=\frac{1}{10}\cdot\frac{5}{2}=\frac14\;-(2)$$

Similarly, for Green balls we have the conditional probabilities
  $$P(G\mid B_1)=\frac{5}{10},\qquad P(G\mid B_2)=\frac{3}{10},\qquad P(G\mid B_3)=\frac{4}{10}$$

3. Total probability of drawing a Green ball:
$$P(G)=\frac13\left(\frac{5}{10}+\frac{3}{10}+\frac{4}{10}\right)=\frac13\cdot\frac{12}{10}=\frac{12}{30}=\frac25\;-(3)$$

4. Probability that the Green ball came from bag III:
$$q=P(B_3\mid G)=\frac{P(B_3)P(G\mid B_3)}{P(G)}=\frac{\frac13\cdot\frac{4}{10}}{\frac25}=\frac{4}{30}\cdot\frac{5}{2}=\frac13\;-(4)$$

5. Required expression:
$$\frac1p+\frac1q=\frac1{\frac14}+\frac1{\frac13}=4+3=7$$

Therefore, $$\left(\frac1p+\frac1q\right)=7$$, which corresponds to Option C.

Let the events be defined as follows:
  $$M_1, M_2, M_3$$ : the bulb is produced by unit $$M_1, M_2, M_3$$ respectively.
  $$D$$ : the bulb is defective.

Production proportions (given as $$2:2:1$$) give the prior probabilities
$$P(M_1)=\frac{2}{5},\quad P(M_2)=\frac{2}{5},\quad P(M_3)=\frac{1}{5}$$

Overall defective rate in the factory is $$P(D)=0.20$$.

Defective rates for two of the units are known:
$$P(D\mid M_1)=0.15,\qquad P(M_2\mid D)=\frac{2}{5}=0.40$$

Step 1: Find $$P(D\mid M_2)$$ using Bayes’ theorem.
Bayes’ theorem gives $$P(M_2\mid D)=\frac{P(D\mid M_2)P(M_2)}{P(D)}$$ Therefore $$P(D\mid M_2)=\frac{P(M_2\mid D)\,P(D)}{P(M_2)} =\frac{0.40\times0.20}{\tfrac{2}{5}} =\frac{0.08}{0.40}=0.20$$

Step 2: Use the overall defective rate to solve for $$P(D\mid M_3)$$. The law of total probability gives $$P(D)=P(D\mid M_1)P(M_1)+P(D\mid M_2)P(M_2)+P(D\mid M_3)P(M_3)$$ Substituting the known numbers: $$0.20=0.15\left(\frac{2}{5}\right)+0.20\left(\frac{2}{5}\right)+P(D\mid M_3)\left(\frac{1}{5}\right)$$ Compute the first two terms: $$0.15\left(\frac{2}{5}\right)=0.06,\qquad 0.20\left(\frac{2}{5}\right)=0.08$$ Thus $$0.20=0.06+0.08+\frac{1}{5}P(D\mid M_3)$$ $$0.20-0.14=\frac{1}{5}P(D\mid M_3)$$ $$0.06=\frac{1}{5}P(D\mid M_3)$$ $$P(D\mid M_3)=0.06\times5=0.30$$

Hence, the probability that a bulb produced by $$M_3$$ is defective is $$0.30$$, which lies in the required range $$0.27-0.33$$.

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

Since the total probability must equal 1: $$\displaystyle\sum_{x=0}^{\infty} k(x+1)3^{-x} = 1$$.

We know that $$\displaystyle\sum_{x=0}^{\infty} (x+1)r^x = \dfrac{1}{(1-r)^2}$$ for $$|r| \lt 1$$.

With $$r = 1/3$$: $$\displaystyle\sum_{x=0}^{\infty} (x+1)\left(\dfrac{1}{3}\right)^x = \dfrac{1}{(1 - 1/3)^2} = \dfrac{1}{(2/3)^2} = \dfrac{9}{4}$$.

So $$k \cdot \dfrac{9}{4} = 1$$, giving $$k = \dfrac{4}{9}$$.

Now, $$P(X \ge 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2)$$.

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

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

$$P(X = 2) = \dfrac{4}{9} \cdot 3 \cdot \dfrac{1}{9} = \dfrac{12}{81} = \dfrac{4}{27}$$

$$P(X = 0) + P(X = 1) + P(X = 2) = \dfrac{4}{9} + \dfrac{8}{27} + \dfrac{4}{27} = \dfrac{12}{27} + \dfrac{8}{27} + \dfrac{4}{27} = \dfrac{24}{27} = \dfrac{8}{9}$$

Therefore, $$P(X \ge 3) = 1 - \dfrac{8}{9} = \dfrac{1}{9}$$.

Hence, the correct answer is Option D.

When a pair of dice is thrown, the ways in which we can get a sum of $$5$$ are: $$(1,4)$$, $$(4,1)$$, $$(2,3)$$ and $$(3,2)$$, giving the total probability among all the $$6\times 6$$ possibilities as $$\dfrac{4}{36}$$

When a pair of dice is thrown, the ways in which we can get a sum of $$8$$ are: $$(2,6)$$, $$(6,2)$$, $$(5,3)$$, $$(3,5)$$, and $$(4,4)$$, giving the total probability among all the $$6\times 6$$ possibilities as $$\dfrac{5}{36}$$

The probability that A wins in the very first throw is $$\dfrac{4}{36}$$

The probability that A wins in the second time he throws is $$\dfrac{36-4}{36}\times \dfrac{36-5}{36} \times \dfrac{4}{36}$$

The probability that A wins in the third time he throws is $$\dfrac{36-4}{36}\times \dfrac{36-5}{36} \times \dfrac{36-4}{36}\times \dfrac{36-5}{36} \times \dfrac{4}{36}$$

And so on,

The cases form a GP with the first term $$\dfrac{4}{36}$$ and the common ratio of $$\dfrac{36-4}{36}\times \dfrac{36-5}{36}$$

Thus, the combined probability will be;

$$\dfrac{4/36}{1- \left(\dfrac{36-4}{36}\times \dfrac{36-5}{36}\right)} = \dfrac{144}{304} = \dfrac{36}{76} = \dfrac{9}{19}$$

Option B is the correct answer.

Total ways to choose 2 squares: $$\binom{16}{2} = \frac{16 \times 15}{2} = 120$$.

Number of pairs with a common side (Adjacent pairs):

Horizontal adjacencies: In each of the 4 rows, there are 3 pairs $$\implies 4 \times 3 = 12$$.

Vertical adjacencies: In each of the 4 columns, there are 3 pairs $$\implies 4 \times 3 = 12$$.

Total common side pairs = $$12 + 12 = 24$$.

Pairs with NO common side: $$120 - 24 = 96$$.

Probability: $$P = \frac{96}{120} = \frac{4}{5}$$.

Bag 1: 4 white, 5 black (total 9). Bag 2: n white, 3 black (total n+3).

Case 1: White ball transferred from Bag 1 to Bag 2.

Probability = $$\frac{4}{9}$$. Bag 2 becomes: (n+1) white, 3 black (total n+4).

P(white from Bag 2) = $$\frac{n+1}{n+4}$$

Case 2: Black ball transferred from Bag 1 to Bag 2.

Probability = $$\frac{5}{9}$$. Bag 2 becomes: n white, 4 black (total n+4).

P(white from Bag 2) = $$\frac{n}{n+4}$$

Total probability of drawing white from Bag 2:

$$ P = \frac{4}{9} \cdot \frac{n+1}{n+4} + \frac{5}{9} \cdot \frac{n}{n+4} = \frac{4(n+1) + 5n}{9(n+4)} = \frac{9n+4}{9(n+4)} $$

Setting equal to $$\frac{29}{45}$$:

$$ \frac{9n+4}{9(n+4)} = \frac{29}{45} $$

$$ \frac{9n+4}{9(n+4)} = \frac{29}{45} $$

Cross-multiplying: $$45(9n+4) = 29 \times 9(n+4)$$

$$405n + 180 = 261n + 1044$$

$$144n = 864$$

$$n = 6$$

The correct answer is Option 1: 6.

Let the random variable $$X$$ be the number of defective pens obtained when 2 pens are drawn without replacement from a box containing 10 pens, out of which 3 are defective.

Because the draws are without replacement, $$X$$ follows a hypergeometric distribution with parameters
    Population size $$N = 10$$
    Number of defectives (successes) $$M = 3$$
    Sample size $$n = 2$$.

The variance formula for a hypergeometric distribution is
$$\text{Var}(X)= n\,\frac{M}{N}\left(1-\frac{M}{N}\right)\frac{N-n}{N-1} \quad -(1)$$

Substitute the given numbers into $$(1)$$:

$$\frac{M}{N} = \frac{3}{10}, \quad 1-\frac{M}{N} = \frac{7}{10}, \quad \frac{N-n}{N-1} = \frac{10-2}{10-1} = \frac{8}{9}.$$

Therefore
$$\text{Var}(X)= 2 \times \frac{3}{10} \times \frac{7}{10} \times \frac{8}{9}$$

Simplify step by step:
$$2 \times \frac{3}{10} = \frac{6}{10},$$
$$\frac{6}{10} \times \frac{7}{10} = \frac{42}{100} = \frac{21}{50},$$
$$\frac{21}{50} \times \frac{8}{9} = \frac{168}{450} = \frac{28}{75}.$$

Hence, the variance of $$X$$ equals $$\dfrac{28}{75}$$.

So the correct choice is Option B.

The following are the cases possible when the dies are thrown and the sum of the numbers on the two faces is $$4$$:

1 and 3
First die 1 and the second die 3 = $$\dfrac{2}{6}\times \dfrac{2}{6} = \dfrac{4}{26}$$
First die 3 and the second die 1 = $$\dfrac{1}{6}\times \dfrac{1}{6} = \dfrac{1}{36}$$

2 and 2
First die 2 and the second die 2 = $$\dfrac{2}{6}\times \dfrac{2}{6} = \dfrac{4}{36}$$

The probability of getting a sum of $$4$$ in all: $$\dfrac{4+1+4}{36} = \dfrac{1}{4}$$

The following are the cases possible when the dies are thrown and the sum of the numbers on the two faces is $$5$$:

1 and 4
First die 1 and the second die 4: $$\dfrac{2}{6}\times \dfrac{1}{6} = \dfrac{2}{36}$$
First die 4 and the second die 1: $$\dfrac{1}{6}\times \dfrac{1}{6} = \dfrac{1}{36}$$

2 and 3
First die 2 and the second die 3: $$\dfrac{2}{6}\times \dfrac{2}{6} = \dfrac{4}{36}$$
First die 3 and the second die 2: $$\dfrac{1}{6}\times \dfrac{2}{6} = \dfrac{2}{36}$$

The probability of getting a sum of $$5$$ in all: $$\dfrac{2+1+4+2}{36} = \dfrac{1}{4}$$

Thus, the combined probability of getting a sum of either $$4$$ or $$5$$ when the two dice are thrown is $$\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}$$

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: $$183.$$

Let
A : the lost card is a spade,
B : all the $$n$$ cards drawn from the remaining $$51$$ cards are spades.

We are given $$P(A\mid B)=\dfrac{11}{50}$$ and we have to find $$n$$.

First write Bayes’ theorem:
$$P(A\mid B)=\dfrac{P(B\mid A)\,P(A)}{P(B\mid A)\,P(A)+P(B\mid \overline{A})\,P(\overline{A})}\,\,\,-(1)$$

Initial probabilities before any card is lost:
Total spades in a full pack = $$13$$ out of $$52$$.
Hence $$P(A)=\dfrac{13}{52}=\dfrac14$$ and $$P(\overline{A})=\dfrac{39}{52}=\dfrac34$$.

Case 1 (event A happens): the lost card itself is a spade.
Spades left in the remaining $$51$$ cards = $$12$$.
So
$$P(B\mid A)=\dfrac{{}^{12}C_{n}}{{}^{51}C_{n}}\,\,\,-(2)$$

Case 2 (event $$\overline{A}$$ happens): the lost card is not a spade.
Spades left in the remaining $$51$$ cards = $$13$$.
So
$$P(B\mid \overline{A})=\dfrac{{}^{13}C_{n}}{{}^{51}C_{n}}\,\,\,-(3)$$

Substitute $$(2)$$ and $$(3)$$ into $$(1)$$:

$$P(A\mid B)=\dfrac{\displaystyle \frac{{}^{12}C_{n}}{{}^{51}C_{n}}\cdot \frac14}{\displaystyle \frac{{}^{12}C_{n}}{{}^{51}C_{n}}\cdot \frac14+\frac{{}^{13}C_{n}}{{}^{51}C_{n}}\cdot \frac34}$$

Cancel the common factor $${}^{51}C_{n}$$ and multiply numerator and denominator by $$4$$ to simplify:

$$P(A\mid B)=\dfrac{{}^{12}C_{n}}{{}^{12}C_{n}+3\,{}^{13}C_{n}}\,\,\,-(4)$$

The question tells us that $$(4)=\dfrac{11}{50}$$, therefore

$$\dfrac{{}^{12}C_{n}}{{}^{12}C_{n}+3\,{}^{13}C_{n}}=\dfrac{11}{50}$$

Cross-multiply:

$$50\,{}^{12}C_{n}=11\bigl({}^{12}C_{n}+3\,{}^{13}C_{n}\bigr)$$

$$50\,{}^{12}C_{n}=11\,{}^{12}C_{n}+33\,{}^{13}C_{n}$$

Subtract $$11\,{}^{12}C_{n}$$ from both sides:

$$39\,{}^{12}C_{n}=33\,{}^{13}C_{n}$$

Divide by $$3$$:

$$13\,{}^{12}C_{n}=11\,{}^{13}C_{n}\,\,\,-(5)$$

Use the identity relating the two combinations:
$${}^{13}C_{n}=\dfrac{13}{13-n}\,{}^{12}C_{n}$$

Substitute this into $$(5)$$:

$$13\,{}^{12}C_{n}=11\left(\dfrac{13}{13-n}\,{}^{12}C_{n}\right)$$

Cancel the common factor $${}^{12}C_{n}$$:

$$13=\dfrac{143}{13-n}$$

Multiply both sides by $$(13-n)$$:

$$13(13-n)=143$$

$$169-13n=143$$

$$13n=169-143=26$$

$$n=2$$

Hence the number of cards drawn that turned out to be spades is $$n=2$$.

The problem asks for the probability that three numbers chosen from $\{1, 2, \dots, 40\}$ form an increasing G.P.

1. Total Sample Space

The total number of ways to select 3 distinct numbers from 40 is:

$$n(S) = \binom{40}{3} = \frac{40 \times 39 \times 38}{3 \times 2 \times 1} = 9880$$

2. General Form of the G.P.

For the terms to be integers, they must follow the form $(kp^2, kpq, kq^2)$, where $\text{gcd}(p, q) = 1$ and $q > p$. The constraint is that the largest term $kq^2 \leq 40$.

3. Counting Favorable Cases

We iterate through possible values of $p$ and $q$:

• $p=1$: $q \in \{2, 3, 4, 5, 6\}$ yields $10 + 4 + 2 + 1 + 1 = 18$ cases.
• $p=2$: $q \in \{3, 5\}$ yields $4 + 1 = 5$ cases.
• $p=3$: $q \in \{4, 5\}$ yields $2 + 1 = 3$ cases.
• $p=4$: $q=5$ yields $1$ case.
• $p=5$: $q=6$ yields $1$ case.
• Total Favorable Cases:$$n(E) = 18 + 5 + 3 + 1 + 1 = 28$$

4. Final Probability and Result

The probability $P$ is given by:

$$P = \frac{28}{9880} = \frac{7}{2470} = \frac{m}{n}$$

Since $\text{gcd}(7, 2470) = 1$, we identify $m = 7$ and $n = 2470$.

$$m + n = 7 + 2470 = 2477$$

We need to find $$ P(\text{first black} \mid \text{second black}) $$

We know, for conditional probability, $$ P(A|B) = \frac{P(A \cap B)}{P(B)} $$

Let, $$ A$$ be the event that the first ball is black, and $$ B$$ be that the second ball is black

So, probability both balls are black (without replacement) is $$ P(A \cap B) = \dfrac{6}{10} \cdot \dfrac{5}{9} = \dfrac{1}{3} $$

Probability second ball is black is 

$$ P(B) = P(\text{second black | first black}) \cdot P(\text{first black}) + P(\text{second black | first white}) \cdot P(\text{first white}) $$

$$ = \dfrac{5}{9} \cdot \dfrac{6}{10} + \dfrac{6}{9} \cdot \dfrac{4}{10} = \dfrac{30}{90} + \dfrac{24}{90} = \dfrac{3}{5} $$

Now,  $$ P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{1/3}{3/5} = \dfrac{5}{9} $$

Thus, $$ m+n = 5+9 =14$$ .

We want the conditional probability of $$B$$ given $$A \cup \bar{B}$$, written as $$P\left(B \; \bigl|\; A \cup \bar{B}\right)$$.

By definition of conditional probability,

$$P\left(B \;\bigl|\; A \cup \bar{B}\right)=\dfrac{P\!\left(B \cap (A \cup \bar{B})\right)}{P\!\left(A \cup \bar{B}\right)} \quad -(1)$$

Step 1: Simplify the numerator
Inside the intersection, $$B \cap (A \cup \bar{B}) = (B \cap A)\, \cup\, (B \cap \bar{B})$$.
But $$B \cap \bar{B} = \varnothing$$, so

$$B \cap (A \cup \bar{B}) = A \cap B \quad -(2)$$

Hence the numerator in $$(1)$$ is $$P(A \cap B)$$.

Step 2: Find $$P(A \cap B)$$
Use $$P(A)=P(A \cap B)+P(A \cap \bar{B}) \quad -(3)$$.
Given $$P(A)=0.7$$ and $$P(A \cap \bar{B})=0.5$$, substitute in $$(3)$$:

$$P(A \cap B)=0.7-0.5=0.2 \quad -(4)$$

Step 3: Find $$P(A \cup \bar{B})$$ (the denominator)
The addition theorem states

$$P(A \cup \bar{B}) = P(A) + P(\bar{B}) - P(A \cap \bar{B}) \quad -(5)$$

We already know $$P(A)=0.7$$ and $$P(A \cap \bar{B})=0.5$$. Also, $$P(\bar{B}) = 1 - P(B) = 1 - 0.4 = 0.6$$.
Substitute into $$(5)$$:

$$P(A \cup \bar{B}) = 0.7 + 0.6 - 0.5 = 0.8 \quad -(6)$$

Step 4: Compute the conditional probability
Insert $$(4)$$ and $$(6)$$ into $$(1)$$:

$$P\left(B \; \bigl|\; A \cup \bar{B}\right)=\dfrac{0.2}{0.8}=0.25=\dfrac{1}{4}$$

Therefore, the required probability equals $$\dfrac{1}{4}$$.
Hence, Option A is correct.

Let

    $$U$$ : the drawn coin is unbiased (ordinary coin)
    $$D$$ : the drawn coin is double-headed
    $$H$$ : head turns up on the toss

Bayes’ theorem states

$$P(U\,|\,H)=\dfrac{P(U)\,P(H\,|\,U)}{P(U)\,P(H\,|\,U)+P(D)\,P(H\,|\,D)}$$   $$-(1)$$

Step 1: Prior probabilities of choosing a coin

There are 20 coins in all (19 unbiased + 1 double-headed).

$$P(U)=\frac{19}{20},\qquad P(D)=\frac{1}{20}$$

Step 2: Probabilities of getting head from each type

For an unbiased coin, head appears with probability $$\frac12$$: $$P(H\,|\,U)=\frac12$$.
For a double-headed coin, head always appears: $$P(H\,|\,D)=1$$.

Step 3: Use Bayes’ theorem

Substituting the values into $$(1)$$:

$$P(U\,|\,H)=\dfrac{\frac{19}{20}\times\frac12}{\frac{19}{20}\times\frac12+\frac{1}{20}\times1}$$

Simplify numerator and denominator separately:

Numerator  $$=\frac{19}{20}\times\frac12=\frac{19}{40}$$

Denominator $$=\frac{19}{40}+\frac{1}{20}=\frac{19}{40}+\frac{2}{40}=\frac{21}{40}$$

Hence

$$P(U\,|\,H)=\frac{\frac{19}{40}}{\frac{21}{40}}=\frac{19}{21}$$

Step 4: Identify $$m$$ and $$n$$

The required probability is $$\frac{m}{n}=\frac{19}{21}$$ with $$\gcd(19,21)=1$$, so $$m=19$$ and $$n=21$$.

Step 5: Compute $$n^2-m^2$$

$$n^2-m^2=21^2-19^2=(21-19)(21+19)=2\times40=80$$

Therefore, $$n^2 - m^2 = 80$$.

Option A is correct.

Given that $$A$$ and $$B$$ are two events with $$P(A \cap B) = 0.1$$, and $$P(A \mid B)$$ and $$P(B \mid A)$$ are the roots of the equation $$12x^2 - 7x + 1 = 0$$. We need to find $$\frac{P(\overline{A} \cup \overline{B})}{P(\overline{A} \cap \overline{B})}$$.

First, solve the quadratic equation $$12x^2 - 7x + 1 = 0$$ to find the roots. Using the quadratic formula:

$$x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 12 \cdot 1}}{2 \cdot 12} = \frac{7 \pm \sqrt{49 - 48}}{24} = \frac{7 \pm \sqrt{1}}{24} = \frac{7 \pm 1}{24}$$

So, the roots are $$x = \frac{8}{24} = \frac{1}{3}$$ and $$x = \frac{6}{24} = \frac{1}{4}$$. Therefore, $$P(A \mid B)$$ and $$P(B \mid A)$$ are $$\frac{1}{3}$$ and $$\frac{1}{4}$$, in some order.

Recall the conditional probability formulas:

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)} \quad \text{and} \quad P(B \mid A) = \frac{P(A \cap B)}{P(A)}$$

Given $$P(A \cap B) = 0.1$$, we consider the two possible cases.

Case 1: $$P(A \mid B) = \frac{1}{3}$$ and $$P(B \mid A) = \frac{1}{4}$$

Then,

$$P(B) = \frac{P(A \cap B)}{P(A \mid B)} = \frac{0.1}{\frac{1}{3}} = 0.1 \times 3 = 0.3$$

$$P(A) = \frac{P(A \cap B)}{P(B \mid A)} = \frac{0.1}{\frac{1}{4}} = 0.1 \times 4 = 0.4$$

Case 2: $$P(A \mid B) = \frac{1}{4}$$ and $$P(B \mid A) = \frac{1}{3}$$

Then,

$$P(B) = \frac{P(A \cap B)}{P(A \mid B)} = \frac{0.1}{\frac{1}{4}} = 0.1 \times 4 = 0.4$$

$$P(A) = \frac{P(A \cap B)}{P(B \mid A)} = \frac{0.1}{\frac{1}{3}} = 0.1 \times 3 = 0.3$$

In both cases, $$P(A) + P(B) = 0.4 + 0.3 = 0.7$$ or $$0.3 + 0.4 = 0.7$$, same sum.

Now, find $$P(A \cup B)$$ using the formula:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

In Case 1: $$P(A \cup B) = 0.4 + 0.3 - 0.1 = 0.6$$

In Case 2: $$P(A \cup B) = 0.3 + 0.4 - 0.1 = 0.6$$

So, $$P(A \cup B) = 0.6$$ in both cases.

Using De Morgan's laws:

$$\overline{A} \cup \overline{B} = (A \cap B)^c \quad \text{and} \quad \overline{A} \cap \overline{B} = (A \cup B)^c$$

Therefore,

$$P(\overline{A} \cup \overline{B}) = P((A \cap B)^c) = 1 - P(A \cap B) = 1 - 0.1 = 0.9$$

$$P(\overline{A} \cap \overline{B}) = P((A \cup B)^c) = 1 - P(A \cup B) = 1 - 0.6 = 0.4$$

The ratio is:

$$\frac{P(\overline{A} \cup \overline{B})}{P(\overline{A} \cap \overline{B})} = \frac{0.9}{0.4} = \frac{9}{10} \times \frac{5}{2} = \frac{9}{4}$$

Using fractions for clarity:

$$P(A \cap B) = \frac{1}{10}$$

Case 1: $$P(A) = \frac{2}{5}$$, $$P(B) = \frac{3}{10}$$

$$P(A \cup B) = \frac{2}{5} + \frac{3}{10} - \frac{1}{10} = \frac{4}{10} + \frac{3}{10} - \frac{1}{10} = \frac{6}{10} = \frac{3}{5}$$

$$P(\overline{A} \cup \overline{B}) = 1 - \frac{1}{10} = \frac{9}{10}$$

$$P(\overline{A} \cap \overline{B}) = 1 - \frac{3}{5} = \frac{2}{5}$$

Ratio $$= \frac{\frac{9}{10}}{\frac{2}{5}} = \frac{9}{10} \times \frac{5}{2} = \frac{45}{20} = \frac{9}{4}$$

Same result in Case 2.

Thus, the ratio is $$\frac{9}{4}$$, which corresponds to option D.

Total number of persons available: 4 engineers (E), 2 doctors (D) and 10 professors (P).
Hence total persons = $$4+2+10 = 16$$.

We have to form a committee of 12 persons subject to the two constraints:
  • at least 3 engineers
  • at least 1 doctor.

Probability $$= \dfrac{\text{number of favourable committees}}{\text{total number of possible committees}}$$.

Total committees
Choosing any 12 out of 16 persons gives $$\binom{16}{12} = \binom{16}{4} = 1820$$ committees.

Favourable committees
Let the committee contain $$e$$ engineers, $$d$$ doctors and $$p$$ professors.
We need $$e \ge 3$$, $$d \ge 1$$ and $$e+d+p = 12$$ with the limits $$0 \le p \le 10$$.

Possible integer triples $$(e,d,p)$$ satisfying these conditions are:

• $$e = 3,\; d = 1,\; p = 8$$
• $$e = 3,\; d = 2,\; p = 7$$
• $$e = 4,\; d = 1,\; p = 7$$
• $$e = 4,\; d = 2,\; p = 6$$

For each case count the committees using $$\binom{n}{r}$$.

Case 1: $$(e,d,p) = (3,1,8)$$
$$\binom{4}{3}\binom{2}{1}\binom{10}{8} = 4 \times 2 \times 45 = 360$$

Case 2: $$(e,d,p) = (3,2,7)$$
$$\binom{4}{3}\binom{2}{2}\binom{10}{7} = 4 \times 1 \times 120 = 480$$

Case 3: $$(e,d,p) = (4,1,7)$$
$$\binom{4}{4}\binom{2}{1}\binom{10}{7} = 1 \times 2 \times 120 = 240$$

Case 4: $$(e,d,p) = (4,2,6)$$
$$\binom{4}{4}\binom{2}{2}\binom{10}{6} = 1 \times 1 \times 210 = 210$$

Add the favourable counts:
$$360 + 480 + 240 + 210 = 1290$$.

Probability
$$\text{P} = \dfrac{1290}{1820} = \dfrac{129}{182}$$ (dividing numerator and denominator by 10).

Thus the required probability is $$\dfrac{129}{182}$$, which matches Option A.

We need to find the probability that a randomly selected word from all arrangements of GARDEN does NOT have vowels in alphabetical order.

The word GARDEN has 6 distinct letters: G, A, R, D, E, N. The vowels are A and E.

Total arrangements = $$6! = 720$$.

In any arrangement of the 6 letters, the two vowels A and E occupy 2 of the 6 positions. These two vowels can appear in only 2 orders: A before E (alphabetical) or E before A (non-alphabetical).

By symmetry, exactly half of all arrangements have A before E (vowels in alphabetical order), and exactly half have E before A.

The probability that vowels are NOT in alphabetical order (i.e., E appears before A) is:

$$P = \frac{1}{2}$$

The correct answer is Option (1): $$\frac{1}{2}$$.

The five probabilities are $$P(X = 0), P(X = 1), P(X = 2), P(X = 3), P(X = 4)$$. Because the points $$(x, P(X = x))$$ for $$x = 0,1,2,3,4$$ lie on a straight line, the probability is an affine (linear) function of $$x$$:

$$P(X = x) = a + bx,\qquad x = 0,1,2,3,4$$

Here $$a$$ and $$b$$ are constants to be determined. Outside these five values the probability is zero, so the total probability is

$$\sum_{x=0}^{4} P(X = x) = 1$$ $$\Rightarrow \sum_{x=0}^{4} (a + bx) = 1$$

The useful sums are $$\sum_{x=0}^{4} 1 = 5,\qquad \sum_{x=0}^{4} x = 0+1+2+3+4 = 10$$

Therefore $$5a + 10b = 1 \qquad -(1)$$

Next use the given mean $$E[X] = \dfrac{5}{2} = 2.5$$:

$$E[X] = \sum_{x=0}^{4} x\,P(X = x) = \sum_{x=0}^{4} x(a + bx)$$ Break the sum into two parts:

$$\sum_{x=0}^{4} xa = a\sum_{x=0}^{4} x = a\cdot10$$ $$\sum_{x=0}^{4} bx^{2} = b\sum_{x=0}^{4} x^{2}$$ Since $$\sum_{x=0}^{4} x^{2} = 0^{2}+1^{2}+2^{2}+3^{2}+4^{2} = 30$$,

$$E[X] = 10a + 30b = 2.5 \qquad -(2)$$

Solve the simultaneous equations. Divide $$(1)$$ by $$5$$: $$a + 2b = 0.2$$ Divide $$(2)$$ by $$10$$: $$a + 3b = 0.25$$ Subtract the first from the second:

$$b = 0.25 - 0.2 = 0.05 = \frac{1}{20}$$ Back-substitute into $$a + 2b = 0.2$$:

$$a = 0.2 - 2(0.05) = 0.2 - 0.1 = 0.1$$

Compute $$E[X^{2}]$$ to obtain the variance. First, $$E[X^{2}] = \sum_{x=0}^{4} x^{2}P(X = x) = \sum_{x=0}^{4} x^{2}(a + bx)$$

Break it again: $$\sum_{x=0}^{4} x^{2}a = a\sum_{x=0}^{4} x^{2} = a\cdot30$$ $$\sum_{x=0}^{4} x^{3}b = b\sum_{x=0}^{4} x^{3}$$ Here $$\sum_{x=0}^{4} x^{3} = 0^{3}+1^{3}+2^{3}+3^{3}+4^{3} = 100$$

Thus $$E[X^{2}] = 30a + 100b = 30(0.1) + 100(0.05) = 3 + 5 = 8$$

The variance is $$\alpha = \operatorname{Var}(X) = E[X^{2}] - (E[X])^{2} = 8 - (2.5)^{2} = 8 - 6.25 = 1.75 = \frac{7}{4}$$

Finally, $$24\alpha = 24 \times \frac{7}{4} = 6 \times 7 = 42$$

Therefore, the required value is 42.

Let us define three events for a randomly chosen question:
  $$K$$ : the student knows the answer.
  $$G$$ : the student guesses the answer. (So $$G = K^{\,c}$$)
  $$C$$ : the student’s answer is correct.

Given data in probability notation:
  $$P(C \mid K)=1$$  (if he knows, he is always correct)
  $$P(C \mid G)=\frac12$$  (probability of a correct guess)
  $$P(G \mid C)=\frac16$$  (a correct answer is guessed with probability $$\tfrac16$$)

We want $$P(K)$$, the probability that the student knows a randomly chosen question.

Step 1 - Express $$P(G \cap C)$$ in two ways.
By the definition of conditional probability,
$$P(G \mid C)=\frac{P(G \cap C)}{P(C)}$$ $$-(1)$$
But, for the joint event,
$$P(G \cap C)=P(C \mid G)\,P(G)=\frac12\,P(G)$$ $$-(2)$$

Insert $$(2)$$ into $$(1)$$ and use the given value $$P(G \mid C)=\tfrac16$$:
$$\frac12\,P(G) \;=\; \frac16\,P(C)$$
$$\Rightarrow\; P(G)=\frac{1}{3}\,P(C)$$ $$-(3)$$

Step 2 - Write $$P(C)$$ via the law of total probability.
$$P(C)=P(C \mid K)\,P(K)+P(C \mid G)\,P(G)$$
$$\;\;\; = 1\cdot P(K)+\frac12\,P(G)$$
$$\;\;\; = P(K)+\frac12\,P(G)$$ $$-(4)$$

Step 3 - Substitute $$P(G)=\tfrac13 P(C)$$ from $$(3)$$ into $$(4)$$.
$$P(C)=P(K)+\frac12\left(\frac13 P(C)\right)=P(K)+\frac16\,P(C)$$
Rearrange:
$$P(C)-\frac16 P(C)=P(K)$$
$$\frac56\,P(C)=P(K)$$ $$-(5)$$

Step 4 - Use $$P(K)+P(G)=1$$ with $$(3)$$ and $$(5)$$ to find $$P(C)$$.
$$P(K)+P(G)=1$$
$$\frac56\,P(C)+\frac13\,P(C)=1$$
$$\left(\frac56+\frac26\right)P(C)=1$$
$$\frac76\,P(C)=1$$
$$\Rightarrow\; P(C)=\frac67$$

Step 5 - Obtain the required probability $$P(K)$$.
From $$(5)$$:
$$P(K)=\frac56\,P(C)=\frac56\left(\frac67\right)=\frac57$$

Therefore, the probability that the student knows the answer of a randomly chosen question is $$\frac57$$.

Option C which is: $$\frac{5}{7}$$

Sum of Probabilities = 1.

$$a + 2a + (a+b) + 2b + 3b = 1 \implies 4a + 6b = 1$$.

Use Mean.

$$E(X) = 0(a) + 2(2a) + 4(a+b) + 6(2b) + 8(3b) = 4a + 4a + 4b + 12b + 24b = 8a + 40b$$.

$$8a + 40b = \frac{46}{9} \implies 4a + 20b = \frac{23}{9}$$.

Solve for $$a, b$$.

Subtract $$(4a+6b=1)$$ from $$(4a+20b=23/9)$$:

$$14b = \frac{23}{9} - 1 = \frac{14}{9} \implies \mathbf{b = \frac{1}{9}}$$.

$$4a + 6(1/9) = 1 \implies 4a = 1 - 2/3 = 1/3 \implies \mathbf{a = \frac{1}{12}}$$.

$$Var(X) = E(X^2) - [E(X)]^2$$.

$$E(X^2) = 0^2(a) + 2^2(2a) + 4^2(a+b) + 6^2(2b) + 8^2(3b) = 8a + 16a + 16b + 72b + 192b = 24a + 280b$$.

$$E(X^2) = 24(1/12) + 280(1/9) = 2 + \frac{280}{9} = \frac{298}{9}$$.

$$Var(X) = \frac{298}{9} - (\frac{46}{9})^2 = \frac{2682 - 2116}{81} = \mathbf{\frac{566}{81}}$$ (Option B)

Total marbles = 10+30+20+15=75. P(red)=10/75=2/15. P(white)=30/75=6/15=2/5.

With replacement: P = $$\frac{2}{15}\times\frac{2}{5}=\frac{4}{75}$$.

The answer is Option (4): $$\frac{4}{75}$$.

The bag contains 8 balls, each either white or black. We draw 4 balls without replacement and observe exactly 2 white and 2 black balls. We need to find the probability that the bag originally had an equal number of white and black balls, i.e., 4 white and 4 black balls.

Let $$W$$ be the number of white balls in the bag. Since we drew 2 white and 2 black balls, $$W$$ must be at least 2 and at most 6 (so that there are at least 2 black balls, i.e., $$8 - W \geq 2$$). Thus, $$W$$ can be 2, 3, 4, 5, or 6.

We assume that all possible values of $$W$$ from 0 to 8 are equally likely initially. Therefore, the prior probability $$P(W = k) = \frac{1}{9}$$ for $$k = 0, 1, \dots, 8$$. However, since the event $$A$$ (drawing 2 white and 2 black balls) can only occur if $$W$$ is between 2 and 6, we restrict our calculation to these values.

By Bayes' theorem, the conditional probability is:

$$P(W=4 \mid A) = \frac{P(A \mid W=4) \cdot P(W=4)}{\sum_{k=2}^{6} P(A \mid W=k) \cdot P(W=k)}$$

Substituting $$P(W=k) = \frac{1}{9}$$ for all $$k$$:

$$P(W=4 \mid A) = \frac{P(A \mid W=4) \cdot \frac{1}{9}}{\sum_{k=2}^{6} P(A \mid W=k) \cdot \frac{1}{9}} = \frac{P(A \mid W=4)}{\sum_{k=2}^{6} P(A \mid W=k)}$$

Now, $$P(A \mid W=k)$$ is the probability of drawing exactly 2 white and 2 black balls from a bag with $$k$$ white and $$8-k$$ black balls, without replacement. This follows the hypergeometric distribution:

$$P(A \mid W=k) = \frac{\binom{k}{2} \binom{8-k}{2}}{\binom{8}{4}}$$

First, compute $$\binom{8}{4}$$:

$$\binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$$

Now, compute for each $$k$$:

• For $$k=2$$: $$\binom{2}{2} = 1$$, $$\binom{6}{2} = \frac{6 \times 5}{2} = 15$$, so $$P(A \mid W=2) = \frac{1 \times 15}{70} = \frac{15}{70}$$
• For $$k=3$$: $$\binom{3}{2} = 3$$, $$\binom{5}{2} = \frac{5 \times 4}{2} = 10$$, so $$P(A \mid W=3) = \frac{3 \times 10}{70} = \frac{30}{70}$$
• For $$k=4$$: $$\binom{4}{2} = \frac{4 \times 3}{2} = 6$$, $$\binom{4}{2} = 6$$, so $$P(A \mid W=4) = \frac{6 \times 6}{70} = \frac{36}{70}$$
• For $$k=5$$: $$\binom{5}{2} = \frac{5 \times 4}{2} = 10$$, $$\binom{3}{2} = 3$$, so $$P(A \mid W=5) = \frac{10 \times 3}{70} = \frac{30}{70}$$
• For $$k=6$$: $$\binom{6}{2} = \frac{6 \times 5}{2} = 15$$, $$\binom{2}{2} = 1$$, so $$P(A \mid W=6) = \frac{15 \times 1}{70} = \frac{15}{70}$$

The denominator is the sum:

$$\sum_{k=2}^{6} P(A \mid W=k) = \frac{15}{70} + \frac{30}{70} + \frac{36}{70} + \frac{30}{70} + \frac{15}{70} = \frac{15 + 30 + 36 + 30 + 15}{70} = \frac{126}{70}$$

Now, the probability is:

$$P(W=4 \mid A) = \frac{\frac{36}{70}}{\frac{126}{70}} = \frac{36}{70} \times \frac{70}{126} = \frac{36}{126}$$

Simplify $$\frac{36}{126}$$ by dividing both numerator and denominator by 18:

$$\frac{36 \div 18}{126 \div 18} = \frac{2}{7}$$

Thus, the probability that the bag contains an equal number of white and black balls is $$\frac{2}{7}$$.

The correct option is B: $$\frac{2}{7}$$.

We are told the coin is biased so that a head is twice as likely as a tail. Let us find the probability of getting exactly two tails and one head in 3 tosses.

To begin, we determine the individual probabilities. Let $$P(\text{Tail})=p$$ and $$P(\text{Head})=2p$$. Since the probabilities must sum to 1:

$$ p + 2p = 1 \implies 3p = 1 \implies p = \frac{1}{3} $$

So $$P(\text{Tail}) = \frac{1}{3}$$ and $$P(\text{Head}) = \frac{2}{3}$$.

Next, we use the binomial probability formula, which states that the probability of obtaining exactly $$k$$ tails in $$n$$ tosses is:

$$ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} $$

In this case, $$n = 3$$, $$k = 2$$, $$p = \frac{1}{3}$$, and $$1-p = \frac{2}{3}$$.

We now calculate the binomial coefficient $$\binom{3}{2}$$, which counts the number of ways to choose which two of the three tosses will be tails. The three possible arrangements are TTH, THT, and HTT.

$$ \binom{3}{2} = \frac{3!}{2! \cdot 1!} = 3 $$

Each such sequence with exactly two tails and one head has probability:

$$ \left(\frac{1}{3}\right)^2 \times \frac{2}{3} = \frac{1}{9} \times \frac{2}{3} = \frac{2}{27} $$

Multiplying by the number of arrangements gives:

$$ P(X = 2) = 3 \times \frac{2}{27} = \frac{6}{27} = \frac{2}{9} $$

The correct answer is Option A: $$\frac{2}{9}$$.

P(A) = 0.6, P(B) = 0.4. P(Quality|A) = 0.8, P(Quality|B) = 0.9.

P(Quality) = 0.6×0.8 + 0.4×0.9 = 0.48 + 0.36 = 0.84.

p = P(B|Quality) = 0.36/0.84 = 36/84 = 3/7.

126p = 126 × 3/7 = 54.

The correct answer is Option (1): 54.

Let $$p$$ be the probability of getting a $$2$$ in a single throw: $$p = \frac{1}{6}$$.

 Let $$q$$ be the probability of not getting a $$2$$: $$q = 1 - p = \frac{5}{6}$$.

The event "2 appears in an even number of throws" happens if it occurs on the 2nd, 4th, 6th... throw.

The total probability $$P$$ is an infinite geometric series:

$$P = (q \cdot p) + (q^3 \cdot p) + (q^5 \cdot p) + \dots$$

$$P = qp(1 + q^2 + q^4 + \dots)$$

Using the sum of an infinite GP $$S = \frac{a}{1-r}$$:

$$P = \frac{qp}{1 - q^2} = \frac{\frac{5}{6} \cdot \frac{1}{6}}{1 - (\frac{5}{6})^2} = \frac{\frac{5}{36}}{1 - \frac{25}{36}} = \frac{\frac{5}{36}}{\frac{11}{36}} = \frac{5}{11}$$

We use the Principle of Inclusion-Exclusion for three sets $$A$$ (multiples of 4), $$B$$ (multiples of 6), and $$C$$ (multiples of 7).

• $$n(A) = \lfloor 50/4 \rfloor = 12$$

• $$n(B) = \lfloor 50/6 \rfloor = 8$$

• $$n(C) = \lfloor 50/7 \rfloor = 7$$

• $$n(A \cap B) = \text{multiples of } \text{lcm}(4, 6) = 12 \implies \lfloor 50/12 \rfloor = 4$$

• $$n(B \cap C) = \text{multiples of } \text{lcm}(6, 7) = 42 \implies \lfloor 50/42 \rfloor = 1$$

• $$n(A \cap C) = \text{multiples of } \text{lcm}(4, 7) = 28 \implies \lfloor 50/28 \rfloor = 1$$

• $$n(A \cap B \cap C) = \text{multiples of } \text{lcm}(4, 6, 7) = 84 \implies 0$$

Total count: $$12 + 8 + 7 - (4 + 1 + 1) + 0 = 21$$.

Probability: $$P = \frac{21}{50}$$

Total balls: 6 white + 9 black = 15.

First draw: 4 white balls from 6 white: $$\binom{6}{4} / \binom{15}{4}$$.

After first draw: 2 white + 9 black = 11 balls remain.

Second draw: 4 black balls from 9: $$\binom{9}{4} / \binom{11}{4}$$.

$$P = \frac{\binom{6}{4}}{\binom{15}{4}} \times \frac{\binom{9}{4}}{\binom{11}{4}} = \frac{15}{1365} \times \frac{126}{330} = \frac{15 \times 126}{1365 \times 330}$$

$$= \frac{1890}{450450} = \frac{3}{715}$$

The answer corresponds to Option (3).

$$P(A) = P(B) = 1/2$$. $$P(W|A) = 3/10$$, $$P(W|B) = 3/5$$.

$$P(W) = P(A) \cdot P(W|A) + P(B) \cdot P(W|B) = \frac{1}{2} \cdot \frac{3}{10} + \frac{1}{2} \cdot \frac{3}{5} = \frac{3}{20} + \frac{3}{10} = \frac{3}{20} + \frac{6}{20} = \frac{9}{20}$$

By Bayes' theorem: $$P(A|W) = \frac{P(A) \cdot P(W|A)}{P(W)} = \frac{3/20}{9/20} = \frac{3}{9} = \frac{1}{3}$$

The answer is Option (3): $$\boxed{\frac{1}{3}}$$.

We are looking for the probability that the three rolls $$(x_1, x_2, x_3)$$ satisfy the condition $$x_1 < x_2 < x_3$$.

Total Outcomes: Since a die has 6 faces and is rolled 3 times, total outcomes = $$6 \times 6 \times 6 = 216$$.

Favourable Outcomes: We need to choose 3 distinct numbers from the set $$\{1, 2, 3, 4, 5, 6\}$$. Once 3 distinct numbers are chosen, there is only one way to arrange them in strictly increasing order.

Number of ways to choose 3 numbers = $$\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$.

Probability:

$$P = \frac{20}{216}$$

Dividing both by 4:

$$P = \frac{5}{54}$$

We need to find the probability that three letters posted to 5 addresses go to exactly two addresses.

We start by counting the total outcomes. Each letter can go to any of 5 addresses independently, so total ways = $$5^3 = 125$$.

Next, we count the favorable outcomes where exactly two addresses are used. We first choose which 2 addresses: $$\binom{5}{2} = 10$$ ways.

Then we count surjective maps from 3 letters to 2 addresses. Total ways to post 3 letters to 2 addresses = $$2^3 = 8$$. From these, we subtract the cases where all letters go to the same address, of which there are 2, giving $$8 - 2 = 6$$.

Therefore, the total favorable outcomes = $$10 \times 6 = 60$$.

Substituting these values into the probability formula gives $$P = \frac{60}{125} = \frac{12}{25}$$.

The correct answer is Option B: $$\frac{12}{25}$$.

Let $$P(\text{Ajay not appear}) = p = \frac{2}{7}$$.

So $$P(\text{Ajay appears}) = 1 - \frac{2}{7} = \frac{5}{7}$$.

$$P(\text{Both Ajay and Vijay appear}) = q = \frac{1}{5}$$.

We need: $$P(\text{Ajay appears and Vijay does not appear})$$.

$$P(\text{Ajay appears and Vijay does not}) = P(\text{Ajay appears}) - P(\text{Both appear})$$

$$= \frac{5}{7} - \frac{1}{5} = \frac{25 - 7}{35} = \frac{18}{35}$$

The answer is $$\boxed{\dfrac{18}{35}}$$, which corresponds to Option (2).

Two positive integers sum to 24. Find the probability their product is at least $$\frac{3}{4}$$ of the greatest possible product.

Since $$x + y = 24$$ with $$x, y \in \mathbb{Z}^+$$, the product can be written as $$P = xy = x(24 - x)$$, which is a quadratic in $$x$$ that attains its maximum when $$x = 12$$. This gives $$P_{\max} = 144$$.

From the condition $$xy \geq \tfrac{3}{4}\cdot144 = 108$$, we require $$x(24 - x) \geq 108$$, so that $$24x - x^2 \geq 108$$ and hence $$x^2 - 24x + 108 \leq 0$$. The roots of the equation $$x^2 - 24x + 108 = 0$$ are found by $$x = \frac{24 \pm \sqrt{576 - 432}}{2} = \frac{24 \pm 12}{2}$$, yielding $$x = 6$$ or $$x = 18$$, and therefore $$6 \leq x \leq 18$$.

Since $$x$$ can be any integer from 1 to 23 (so that $$y = 24 - x > 0$$), there are 23 total equally likely outcomes. Substituting the range for $$x$$, the favorable values are $$x = 6, 7, 8, \dots, 18$$, giving $$18 - 6 + 1 = 13$$ outcomes.

This gives the probability $$\frac{13}{23}$$. Because $$\gcd(13,23) = 1$$, we have $$m = 13$$ and $$n = 23$$, so that $$n - m = 23 - 13 = 10$$.

The correct answer is Option (1): 10

$$D > 0$$.

$$D = b^2 - 4ac > 0 \implies b^2 > 4ac$$

Count pairs $$(a, c)$$ such that $$4ac < b^2$$ for each $$b \in \{1...6\}$$:

o $$b=1, 2$$: $$b^2 \le 4$$, no cases.

o $$b=3$$: $$b^2=9 > 4ac \implies ac < 2.25$$. Pairs: $$(1,1), (1,2), (2,1)$$ (3 cases)

o $$b=4$$: $$b^2=16 > 4ac \implies ac < 4$$. Pairs: $$(1,1), (1,2), (1,3), (2,1), (3,1)$$ (5 cases)

o $$b=5$$: $$b^2=25 > 4ac \implies ac < 6.25$$. Pairs: $$(1,1..6), (2,1..3), (3,1,2), (4,1), (5,1), (6,1)$$ (14 cases)

o $$b=6$$: $$b^2=36 > 4ac \implies ac < 9$$. Pairs: $$(1,1..6), (2,1..4), (3,1,2), (4,1,2), (5,1), (6,1), (8 \text{ is not possible})$$

Total for $$b=6$$ is 16 cases.

Total cases $$= 3 + 5 + 14 + 16 = 38$$.

Total sample space $$= 6^3 = 216$$.

$$p = 38/216$$. Therefore, $$216p = 38$$.

Repeated roots requires $$b^2 = 4ac$$ with $$a, b, c \in \{1,...,8\}$$. Since $$b^2$$ must be divisible by 4, $$b$$ must be even.

$$b = 2$$: $$ac = 1$$. Pairs: $$(1,1)$$. Count = 1.

$$b = 4$$: $$ac = 4$$. Pairs: $$(1,4),(2,2),(4,1)$$. Count = 3.

$$b = 6$$: $$ac = 9$$. Pairs from $$\{1,...,8\}$$: $$(3,3)$$ only. Count = 1.

$$b = 8$$: $$ac = 16$$. Pairs: $$(2,8),(4,4),(8,2)$$. Count = 3.

Total favorable = 1 + 3 + 1 + 3 = 8. Total outcomes = $$8^3 = 512$$.

Probability = $$\frac{8}{512} = \frac{1}{64}$$.

The correct answer is Option 2: $$\frac{1}{64}$$.

Using Bayes' theorem to find the probability that the one-rupee coin came from Bag Y.

Let $$P(X) = P(Y) = P(Z) = \frac{1}{3}$$ (each bag equally likely).

$$P(\text{1-rupee}|X) = \frac{5}{9}$$, $$P(\text{1-rupee}|Y) = \frac{4}{9}$$, $$P(\text{1-rupee}|Z) = \frac{3}{9} = \frac{1}{3}$$.

By Bayes' theorem:

$$P(Y|\text{1-rupee}) = \frac{P(\text{1-rupee}|Y) \cdot P(Y)}{P(\text{1-rupee}|X)P(X) + P(\text{1-rupee}|Y)P(Y) + P(\text{1-rupee}|Z)P(Z)}$$

$$= \frac{\frac{4}{9} \cdot \frac{1}{3}}{\frac{5}{9} \cdot \frac{1}{3} + \frac{4}{9} \cdot \frac{1}{3} + \frac{3}{9} \cdot \frac{1}{3}} = \frac{\frac{4}{27}}{\frac{5+4+3}{27}} = \frac{4}{12} = \frac{1}{3}$$

The correct answer is Option 4: $$\frac{1}{3}$$.

18 apples total (3 rotten, 15 good). X = number of rotten in draw of 2.

$$P(X=0)=\frac{\binom{15}{2}}{\binom{18}{2}}=\frac{105}{153}$$. $$P(X=1)=\frac{\binom{3}{1}\binom{15}{1}}{\binom{18}{2}}=\frac{45}{153}$$. $$P(X=2)=\frac{\binom{3}{2}}{\binom{18}{2}}=\frac{3}{153}$$.

$$E(X)=0+45/153+6/153=51/153=1/3$$. $$E(X^2)=0+45/153+12/153=57/153$$.

$$Var=E(X^2)-[E(X)]^2=\frac{57}{153}-\frac{1}{9}=\frac{57}{153}-\frac{17}{153}=\frac{40}{153}$$.

The answer is Option (4): $$\frac{40}{153}$$.

We can use the Bayes Theorem:

$$E_1$$: Selecting Urn A

$$E_2$$: Selecting Urn B

$$E_3$$: Selecting Urn C

Because the urn is chosen completely at random, the probability of selecting any single urn is equal:

$$P(E_1) = P(E_2) = P(E_3) = \dfrac{1}{3}$$

Let $$B$$ be the event of drawing a black ball. We have to find the probability of drawing a black ball from each specific urn.

Each urn contains exactly 12 balls in total (7+5 = 12, 5+7 = 12, 6+6 = 12).

Urn A has 5 black balls: $$P(B|E_1) = \dfrac{5}{12}$$

Urn B has 7 black balls: $$P(B|E_2) = \dfrac{7}{12}$$

Urn C has 6 black balls: $$P(B|E_3) = \dfrac{6}{12}$$

 Bayes Theorem: $$P(E_1|B) = \dfrac{P(E_1) \cdot P(B|E_1)}{P(E_1) \cdot P(B|E_1) + P(E_2) \cdot P(B|E_2) + P(E_3) \cdot P(B|E_3)}$$

$$P(E_1|B) = \dfrac{\left(\dfrac{1}{3}\right) \cdot \left(\dfrac{5}{12}\right)}{\left(\dfrac{1}{3}\right) \cdot \left(\dfrac{5}{12}\right) + \left(\dfrac{1}{3}\right) \cdot \left(\dfrac{7}{12}\right) + \left(\dfrac{1}{3}\right) \cdot \left(\dfrac{6}{12}\right)}$$

$$P(E_1|B) = \dfrac{5}{5 + 7 + 6}$$

$$P(E_1|B) = \dfrac{5}{18}$$

Two integers $$x, y$$ are chosen from $$\{0, 1, 2, ..., 10\}$$. Total outcomes = $$11 \times 11 = 121$$.

We need $$P(|x - y| > 5)$$, i.e., $$|x - y| \geq 6$$.

Count pairs where $$x - y \geq 6$$: For each difference $$d = 6, 7, 8, 9, 10$$:

$$d = 6$$: $$(6,0),(7,1),(8,2),(9,3),(10,4)$$ → 5 pairs

$$d = 7$$: 4 pairs

$$d = 8$$: 3 pairs

$$d = 9$$: 2 pairs

$$d = 10$$: 1 pair

Total with $$x - y \geq 6$$: $$5 + 4 + 3 + 2 + 1 = 15$$.

By symmetry, pairs with $$y - x \geq 6$$ = 15.

Total favorable = $$15 + 15 = 30$$.

$$P = \frac{30}{121}$$.

The answer is Option (1): $$\boxed{\frac{30}{121}}$$.

Let the total number of balls be $$N$$.
Numbers of each colour:
    White   = $$3$$
    Green  = $$6$$
    Blue   = $$N-9$$

Because the draws are without replacement, multiply the successive probabilities.

Step 1 : Compute $$P(W_1 \cap G_2 \cap B_3)$$
$$P(W_1)=\frac{3}{N}$$
After removing one white ball, total balls become $$N-1$$ and green balls remain $$6$$, so
$$P(G_2 \mid W_1)=\frac{6}{N-1}$$
After removing a white and a green ball, total balls become $$N-2$$ and blue balls are still $$N-9$$, so
$$P(B_3 \mid W_1 \cap G_2)=\frac{N-9}{N-2}$$
Therefore
$$P(W_1 \cap G_2 \cap B_3)=\frac{3}{N}\cdot\frac{6}{N-1}\cdot\frac{N-9}{N-2}=\frac{18(N-9)}{N(N-1)(N-2)}$$

This probability is given to be $$\dfrac{2}{5N}$$, hence
$$\frac{18(N-9)}{N(N-1)(N-2)}=\frac{2}{5N} \quad -(1)$$

Step 2 : Compute the conditional probability $$P(B_3 \mid W_1 \cap G_2)$$
First find $$P(W_1 \cap G_2)$$:
$$P(W_1 \cap G_2)=\frac{3}{N}\cdot\frac{6}{N-1}=\frac{18}{N(N-1)}$$
Thus
$$P(B_3 \mid W_1 \cap G_2)=\frac{P(W_1 \cap G_2 \cap B_3)}{P(W_1 \cap G_2)} =\frac{\dfrac{18(N-9)}{N(N-1)(N-2)}}{\dfrac{18}{N(N-1)}} =\frac{N-9}{N-2}$$
This conditional probability is given to be $$\dfrac{2}{9}$$, so
$$\frac{N-9}{N-2}=\frac{2}{9} \quad -(2)$$

Step 3 : Solve equation $$(2)$$ for $$N$$
Cross-multiplying:
$$9(N-9)=2(N-2)$$
$$9N-81=2N-4$$
$$7N=77$$
$$N=11$$

Step 4 : Verification with equation $$(1)$$ (optional but reassuring)
Substitute $$N=11$$ in $$(1)$$:
Left side $$=\dfrac{18(11-9)}{11\cdot10\cdot9}=\dfrac{18\cdot2}{990}=\dfrac{36}{990}=\dfrac{2}{55}$$
Right side $$=\dfrac{2}{5\cdot11}=\dfrac{2}{55}$$
Both sides match, confirming the value.

Hence the total number of balls is
11

A fair die: $$P(\text{six}) = 1/6$$, $$P(\text{not six}) = 5/6$$.

$$a = P(X = 3) = (5/6)^2 \cdot (1/6) = 25/216$$

$$b = P(X \geq 3) = (5/6)^2 = 25/36$$ (first two tosses are not six)

$$c = P(X \geq 6 \mid X > 3)$$: Given that first 3 tosses are not six, the probability that we need at least 6 tosses means tosses 4 and 5 are also not six.

$$c = (5/6)^2 = 25/36$$

$$\frac{b + c}{a} = \frac{25/36 + 25/36}{25/216} = \frac{50/36}{25/216} = \frac{50}{36} \times \frac{216}{25} = \frac{50 \times 6}{25} = 12$$

The answer is $$\boxed{12}$$.

We need the variance of $$X$$ (number of defectives in a sample of 5 from 12 items with 3 defectives, drawn without replacement). Here, $$X$$ follows a Hypergeometric distribution with parameters $$N=12$$, $$K=3$$ (defectives), and $$n=5$$ (sample size). The variance for a Hypergeometric distribution is given by $$\text{Var}(X) = n \cdot \frac{K}{N} \cdot \frac{N-K}{N} \cdot \frac{N-n}{N-1}$$.

Substituting the values yields $$= 5 \cdot \frac{3}{12} \cdot \frac{9}{12} \cdot \frac{7}{11} = \frac{5 \times 3 \times 9 \times 7}{12 \times 12 \times 11} = \frac{945}{1584}.$$ Since $$\gcd(945, 1584) = 9$$, we have $$\frac{945}{1584} = \frac{105}{176}.$$ Verifying the factors, $$105 = 3\times5\times7$$ and $$176 = 2^4\times11$$, confirms that $$\gcd(105,176)=1$$.

Setting $$m=105$$ and $$n=176$$ gives $$n - m = 176 - 105 = 71$$.

The correct answer is 71.

A team plays 10 matches with probability of winning each match $$p = \frac{1}{3}$$ and losing $$q = \frac{2}{3}$$. Let $$x$$ denote the number of wins and $$y$$ the number of losses. Since $$x + y = 10$$, it follows that $$y = 10 - x$$.

Substituting into the condition $$|x - y| \le 2$$ gives $$|x - (10 - x)| \le 2 \implies |2x - 10| \le 2 \implies -2 \le 2x - 10 \le 2 \implies 4 \le x \le 6$$, so the possible values of $$x$$ are $$4, 5, 6$$.

Since $$X \sim \mathrm{Binomial}(10,1/3)$$, the probability mass function is $$P(X = r) = \binom{10}{r} \left(\frac{1}{3}\right)^r \left(\frac{2}{3}\right)^{10-r} = \frac{\binom{10}{r}\cdot2^{10-r}}{3^{10}}\,. $$

For $$r = 4$$ this yields $$P(X=4) = \frac{\binom{10}{4}\cdot2^6}{3^{10}} = \frac{210 \times 64}{3^{10}} = \frac{13440}{3^{10}}\,. $$

For $$r = 5$$ one finds $$P(X=5) = \frac{\binom{10}{5}\cdot2^5}{3^{10}} = \frac{252 \times 32}{3^{10}} = \frac{8064}{3^{10}}\,. $$

For $$r = 6$$ we have $$P(X=6) = \frac{\binom{10}{6}\cdot2^4}{3^{10}} = \frac{210 \times 16}{3^{10}} = \frac{3360}{3^{10}}\,. $$

Adding these probabilities gives $$p = P(|x - y| \le 2) = \frac{13440 + 8064 + 3360}{3^{10}} = \frac{24864}{3^{10}}\,. $$

Multiplying by $$3^9$$ yields $$3^9 p = 3^9 \cdot \frac{24864}{3^{10}} = \frac{24864}{3} = 8288\,. $$

Therefore, the final answer is $$\boxed{8288}$$.

Fair tetrahedral die with faces 1,2,3,4. For $$ax^2+bx+c=0$$ to have all real roots, we need $$b^2 - 4ac \geq 0$$.

Total outcomes = $$4^3 = 64$$.

We need $$b^2 \geq 4ac$$. Enumerate favorable cases:

b=1: $$1 \geq 4ac$$, impossible for $$a,c \geq 1$$.

b=2: $$4 \geq 4ac \implies ac \leq 1 \implies a=c=1$$. 1 case.

b=3: $$9 \geq 4ac \implies ac \leq 2$$. Cases: $$(1,1),(1,2),(2,1)$$. 3 cases.

b=4: $$16 \geq 4ac \implies ac \leq 4$$. Cases: $$(1$$, $$1)$$, $$(1$$, $$2)$$, $$(1$$, $$3)$$, $$(1$$, $$4)$$, $$(2$$, $$1)$$, $$(2$$, $$2)$$, $$(3$$, $$1)$$, $$(4$$, $$1)$$. 8 cases.

Total favorable = $$0+1+3+8 = 12$$.

Probability = $$\frac{12}{64} = \frac{3}{16}$$.

$$\gcd(3,16) = 1$$, so $$m = 3, n = 16$$.

$$m + n = 19$$.

The answer is $$\boxed{19}$$.

We consider drawing 3 balls from a bag containing 5 blue and 4 yellow balls. Let X and Y be the number of blue and yellow balls drawn, respectively, and we seek to evaluate 7$$\bar{X}$$ + 4$$\bar{Y}$$.

Since the total number of balls is 9 and we draw 3 without replacement, by the hypergeometric distribution we have $$\bar{X}=E[X]=3\times\frac{5}{9}=\frac{15}{9}=\frac{5}{3}$$ and $$\bar{Y}=E[Y]=3\times\frac{4}{9}=\frac{12}{9}=\frac{4}{3}$$. Notice that X+Y=3 so $$\bar{X}+\bar{Y}=3$$, which checks as $$\tfrac{5}{3}+\tfrac{4}{3}=3$$.

Substituting these into the expression gives $$7\bar{X}+4\bar{Y}=7\times\frac{5}{3}+4\times\frac{4}{3}=\frac{35}{3}+\frac{16}{3}=\frac{51}{3}=17$$.

Therefore, the final answer is 17.

Let the probability of getting Head on a single toss be $$p=\frac13$$. Hence, the probability of getting Tail is $$q=1-p=\frac23$$.

When the experiment ends, it must end either with the pair $$HH$$ or with the pair $$TT$$. We want the probability that it ends with $$HH$$.

Define the following conditional probabilities:
  • $$P_H$$ : probability that the experiment finally stops with $$HH$$, given that the most recent toss is a Head.
  • $$P_T$$ : probability that the experiment finally stops with $$HH$$, given that the most recent toss is a Tail.

1. Recurrence for $$P_H$$ If the last toss is H, then on the next toss:
  • With probability $$p$$ we get another H → the pair $$HH$$ appears and we stop; this contributes $$p\times1$$.
  • With probability $$q$$ we get T → no stop; the new “last toss” is now T, so the chance of ending with $$HH$$ becomes $$P_T$$. Hence $$P_H = p + qP_T \quad -(1)$$

2. Recurrence for $$P_T$$ If the last toss is T, then on the next toss:
  • With probability $$p$$ we get H → no stop; the new “last toss” is H, so the chance becomes $$P_H$$.
  • With probability $$q$$ we get another T → the pair $$TT$$ appears; we stop, but this is a failure for our requirement, so contributes 0. Thus $$P_T = pP_H \quad -(2)$$

3. Solve the two equations Substitute $$P_T$$ from $$(2)$$ into $$(1)$$:
$$P_H = p + q(pP_H) = p + pqP_H$$ $$P_H(1 - pq) = p$$ $$P_H = \dfrac{p}{1-pq}$$

With $$p=\frac13,\; q=\frac23$$ we get $$pq = \frac13\cdot\frac23 = \frac29,\quad 1-pq = 1-\frac29 = \frac79$$ $$P_H = \frac{\tfrac13}{\tfrac79} = \frac13 \times \frac97 = \frac37$$

4. Overall probability of stopping with $$HH$$ The very first toss can be H or T:
  • First toss H (probability $$p$$): we are in the “last toss = H” state, so eventual success chance is $$P_H$$.
  • First toss T (probability $$q$$): we are in the “last toss = T” state, so eventual success chance is $$P_T = pP_H$$ (from $$(2)$$).
Therefore $$P=\;pP_H + qP_T = pP_H + q(pP_H) = pP_H(1+q)$$

Substituting the values: $$1+q = 1+\frac23 = \frac53$$ $$P = \frac13 \times \frac37 \times \frac53 = \frac{15}{63} = \frac{5}{21}$$

Hence the probability that the experiment stops with Head is $$\dfrac{5}{21}$$.

Option B which is: $$\frac{5}{21}$$

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: $$((\sim P) \vee Q) \wedge ((\sim Q) \vee P)$$.

A total of 100 persons includes 75 English speakers and 40 Hindi speakers, each speaking at least one language. By the inclusion-exclusion principle, the number of persons speaking both languages is 75 + 40 - 100 = 15, so the number speaking only English is $$\alpha$$ = 75 - 15 = 60 and the number speaking only Hindi is $$\beta$$ = 40 - 15 = 25.

The equation of the ellipse can be written as $$25(\beta^2 x^2 + \alpha^2 y^2) = \alpha^2\beta^2$$
which yields $$\frac{x^2}{\alpha^2/25} + \frac{y^2}{\beta^2/25} = 1 \implies \frac{x^2}{144} + \frac{y^2}{25} = 1\ .$$

In this form, $$a^2 = 144$$ and $$b^2 = 25$$ since $$\alpha^2/25 = 3600/25 = 144$$ and $$\beta^2/25 = 625/25 = 25$$.

The eccentricity of the ellipse is given by $$ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{25}{144}} = \sqrt{\frac{119}{144}} = \frac{\sqrt{119}}{12}\ .$$

The correct answer is $$\dfrac{\sqrt{119}}{12}$$.

We consider two positive integers summing to 66 and let the maximum product be $$M$$. We then set $$S = \{x \in \mathbb{Z} : x(66-x) \geq \frac{5}{9}M\}$$ and $$A = \{x \in S : x$$ is multiple of 3$$\}$$.

Since the arithmetic mean-geometric mean inequality shows that the product is maximized when the integers are equal, each integer is 33 and hence $$M = 33^2 = 1089$$.

Now we determine the set $$S$$ by solving the inequality $$x(66-x) \geq \frac{5}{9} \times 1089 = 605$$, which simplifies to $$66x - x^2 \geq 605$$ or equivalently $$x^2 - 66x + 605 \leq 0$$. Solving the corresponding quadratic equation gives $$x = \frac{66 \pm \sqrt{4356 - 2420}}{2} = \frac{66 \pm \sqrt{1936}}{2} = \frac{66 \pm 44}{2}$$, so that $$x = 11$$ or $$x = 55$$. Therefore, $$S = \{11, 12, 13, \ldots, 55\}$$, which contains $$55 - 11 + 1 = 45$$ elements.

Subsequently, the multiples of 3 in the interval $$[11, 55]$$ are $$12, 15, 18, \ldots, 54$$, and their count is given by $$\frac{54 - 12}{3} + 1 = 15$$.

Therefore, the probability is $$P(A) = \frac{15}{45} = \frac{1}{3}$$.

The correct answer is Option B: $$\boxed{\frac{1}{3}}$$.

Expected is Option 9 (likely encoding). Saving for review.

Two dice are thrown. Let us enumerate the events.

Event A (1st die < 2nd die): 15 outcomes

$$\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)\}$$

Event B (1st even, 2nd odd): 9 outcomes

$$\{(2,1),(2,3),(2,5),(4,1),(4,3),(4,5),(6,1),(6,3),(6,5)\}$$

Event C (1st odd, 2nd even): 9 outcomes

$$\{(1,2),(1,4),(1,6),(3,2),(3,4),(3,6),(5,2),(5,4),(5,6)\}$$

Now check each option:

Option A: Favourable cases of $$(A \cup B) \cap C$$.

$$A \cap C = \{(1,2),(1,4),(1,6),(3,4),(3,6),(5,6)\}$$ — 6 cases (where 1st < 2nd AND 1st odd, 2nd even)

$$B \cap C = \emptyset$$ (1st die cannot be both even and odd)

$$(A \cup B) \cap C = (A \cap C) \cup (B \cap C) = 6 + 0 = 6$$ ✓

Option B: A and B mutually exclusive?

$$A \cap B$$ includes $$(2,3), (2,5), (4,5)$$ — not empty ✗

Option C: Favourable cases of A, B, C are 15, 6, 6?

A has 15 ✓, but B has 9 ✗ and C has 9 ✗

Option D: B and C independent?

$$P(B) = 9/36 = 1/4$$, $$P(C) = 9/36 = 1/4$$

$$P(B \cap C) = 0 \neq P(B) \cdot P(C) = 1/16$$ ✗

The answer is Option A: The number of favourable cases of $$(A \cup B) \cap C$$ is 6.

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

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

A bag contains 6 balls. Two balls are drawn and both are black. We need to find $$P(\text{at least 5 black} | \text{2 drawn are black})$$.

Assuming each possible composition of black balls (0 to 6) is equally likely a priori, we have for each $$k=0,1,2,\ldots,6$$ that $$P(B=k)=\frac{1}{7}$$.

The likelihood of drawing two black balls given there are $$k$$ black balls in the bag is $$P(\text{2 black drawn}|B=k)=\frac{\binom{k}{2}}{\binom{6}{2}}=\frac{k(k-1)}{30},$$ which equals 0 for $$k=0,1$$, $$\frac{2}{30}$$ for $$k=2$$, $$\frac{6}{30}$$ for $$k=3$$, $$\frac{12}{30}$$ for $$k=4$$, $$\frac{20}{30}$$ for $$k=5$$, and $$\frac{30}{30}$$ for $$k=6$$.

By the law of total probability we compute $$P(\text{2 black})=\frac{1}{7}\cdot\frac{1}{30}(0+0+2+6+12+20+30)=\frac{70}{210}=\frac{1}{3}\,.$$

Applying Bayes' theorem to find the probability of at least five black balls given two black draws yields $$P(B\ge5|\text{2 black})=\frac{P(\text{2 black}|B=5)P(B=5)+P(\text{2 black}|B=6)P(B=6)}{P(\text{2 black})} =\frac{\frac{1}{7}\bigl(\frac{20}{30}+\frac{30}{30}\bigr)}{\frac{1}{3}} =\frac{\frac{50}{210}}{\frac{1}{3}} =\frac{150}{210} =\frac{5}{7}\,.$$

Hence the correct answer is Option (1): $$\boxed{\frac{5}{7}}$$.

Total balls: 6 white + 4 black = 10. A die is rolled, and the number shown determines how many balls are drawn.

P(all white) = $$\sum_{k=1}^{6} P(\text{die shows } k) \times P(\text{all } k \text{ balls are white})$$

$$= \frac{1}{6}\left[\frac{\binom{6}{1}}{\binom{10}{1}} + \frac{\binom{6}{2}}{\binom{10}{2}} + \frac{\binom{6}{3}}{\binom{10}{3}} + \frac{\binom{6}{4}}{\binom{10}{4}} + \frac{\binom{6}{5}}{\binom{10}{5}} + \frac{\binom{6}{6}}{\binom{10}{6}}\right]$$

$$= \frac{1}{6}\left[\frac{6}{10} + \frac{15}{45} + \frac{20}{120} + \frac{15}{210} + \frac{6}{252} + \frac{1}{210}\right]$$

$$= \frac{1}{6}\left[\frac{3}{5} + \frac{1}{3} + \frac{1}{6} + \frac{1}{14} + \frac{1}{42} + \frac{1}{210}\right]$$

Finding common denominator (210):

$$= \frac{1}{6}\left[\frac{126}{210} + \frac{70}{210} + \frac{35}{210} + \frac{15}{210} + \frac{5}{210} + \frac{1}{210}\right]$$

$$= \frac{1}{6} \times \frac{252}{210} = \frac{1}{6} \times \frac{6}{5} = \frac{1}{5}$$

This matches option 3: $$\frac{1}{5}$$.

For throwing a pair of dice, the ways to get a total of 5 are: (1,4), (2,3), (3,2), (4,1) = 4 outcomes out of 36.

Probability of success: $$p = \frac{4}{36} = \frac{1}{9}$$

Probability of failure: $$q = 1 - \frac{1}{9} = \frac{8}{9}$$

The dice are thrown 5 times. We need P(at least 4 successes) = P(4) + P(5).

$$ P(5) = \binom{5}{5}\left(\frac{1}{9}\right)^5 = \frac{1}{9^5} $$

$$ P(4) = \binom{5}{4}\left(\frac{1}{9}\right)^4\left(\frac{8}{9}\right) = \frac{5 \times 8}{9^5} = \frac{40}{9^5} $$

$$ P(\geq 4) = \frac{41}{9^5} = \frac{41}{59049} $$

Converting to the form $$\frac{k}{3^{11}}$$: Since $$9^5 = 3^{10}$$, we have:

$$ \frac{41}{3^{10}} = \frac{41 \times 3}{3^{11}} = \frac{123}{3^{11}} $$

Therefore $$k = \mathbf{123}$$.

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: $$\frac{7}{27}$$.

We begin by using Bayes’ theorem and letting D denote the event that a bolt is defective.

Next, we have the probabilities $$P(A) = 0.20$$, $$P(B) = 0.30$$, and $$P(C) = 0.50$$.

Similarly, the conditional probabilities are $$P(D|A) = 0.03$$, $$P(D|B) = 0.04$$, and $$P(D|C) = 0.02$$.

To find the total probability of a defective bolt, we apply the law of total probability and compute $$ P(D) = P(A) \cdot P(D|A) + P(B) \cdot P(D|B) + P(C) \cdot P(D|C) $$.
Substituting the values gives $$ = 0.20 \times 0.03 + 0.30 \times 0.04 + 0.50 \times 0.02 $$.
This gives $$ = 0.006 + 0.012 + 0.010 = 0.028 $$.

Then, by Bayes’ theorem, the probability that a defective bolt was manufactured by C is given by $$ P(C|D) = \frac{P(D|C) \cdot P(C)}{P(D)} = \frac{0.02 \times 0.50}{0.028} = \frac{0.01}{0.028} = \frac{10}{28} = \frac{5}{14} $$.

Therefore, the correct answer is $$\dfrac{5}{14}$$.

Each roll of a fair die gives an odd number (1, 3, 5) with probability $$\tfrac12$$ and an even number (2, 4, 6) with probability $$\tfrac12$$.
Hence, when the die is rolled $$n$$ times, the number of odd outcomes, say $$X$$, follows the binomial distribution $$X \sim \text{Bin}(n,\tfrac12)$$.

The question says the probability of getting odd numbers exactly seven times equals the probability of getting them exactly nine times.
Using the binomial formula,

$$P(X=7)=\binom{n}{7}\left(\tfrac12\right)^{n}, \qquad P(X=9)=\binom{n}{9}\left(\tfrac12\right)^{n}$$

Equality of these probabilities gives

$$\binom{n}{7}=\binom{n}{9}$$

Divide the two combinations to remove the common factorial terms:

$$\frac{\binom{n}{9}}{\binom{n}{7}} =\frac{7!\,(n-7)!}{9!\,(n-9)!} =\frac{1}{9\cdot8}\,(n-7)(n-8)$$

Setting this ratio equal to 1 (because the two combinations are equal):

$$\frac{1}{9\cdot8}\,(n-7)(n-8)=1 \;\;\Longrightarrow\;\; (n-7)(n-8)=72$$

Expanding and solving the quadratic:

$$n^{2}-15n+56-72=0 \;\;\Longrightarrow\;\; n^{2}-15n-16=0$$

The positive root is

$$n=\frac{15+\sqrt{15^{2}+64}}{2} =\frac{15+17}{2} =16$$

Therefore, the die is rolled $$n=16$$ times.

Let $$Y$$ be the number of even outcomes. Because every roll is either odd or even, $$Y = n - X$$.
We need the probability of getting even numbers exactly twice, i.e. $$Y=2$$ when $$n=16$$.

Again using the binomial distribution (with probability $$\tfrac12$$ for an even number),

$$P(Y=2)=\binom{16}{2}\left(\tfrac12\right)^{16}$$

Calculate the combination:

$$\binom{16}{2}= \frac{16\cdot15}{2}=120$$

Thus

$$P(Y=2)=\frac{120}{2^{16}} =\frac{60}{2^{15}}$$

The given form of the probability is $$\dfrac{k}{2^{15}}$$, so $$k=60$$.

Hence, the required value of $$k$$ is 60.
Option A is correct.

1. Analyze the condition $$2^N < N!$$

For the sum $$N$$ of two dice, $$N$$ can range from $$2$$ to $$12$$.

• If $$N=2$$: $$2^2 = 4$$, $$2! = 2 \implies 4 \nless 2$$ (False)
• If $$N=3$$: $$2^3 = 8$$, $$3! = 6 \implies 8 \nless 6$$ (False)
• If $$N=4$$: $$2^4 = 16$$, $$4! = 24 \implies 16 < 24$$ (True)

The condition $$2^N < N!$$ is true for all $$N \geq 4$$.

2. Calculate the Probability

The total number of outcomes when rolling two dice is $$36$$.

We need $$P(N \geq 4)$$, which is easier to calculate as $$1 - P(N < 4)$$.

• Outcomes for $$N=2$$: $$(1,1)$$ — (1 outcome)
• Outcomes for $$N=3$$: $$(1,2), (2,1)$$ — (2 outcomes)

Total outcomes for $$N < 4$$ is $$1 + 2 = 3$$.

$$P(N \geq 4) = 1 - \frac{3}{36} = 1 - \frac{1}{12} = \frac{11}{12}$$

Here, $$m = 11$$ and $$n = 12$$ (they are coprime).

3. Final Calculation

Substitute $$m$$ and $$n$$ into the expression $$4m - 3n$$:

$$4(11) - 3(12) = 44 - 36 = \mathbf{8}$$

Correct Option: (D)

We need to analyze the two statements about probability.

(S1): If $$P(A) = 0$$, then $$A = \phi$$.

This is false. In a continuous sample space (e.g., choosing a real number uniformly from $$[0, 1]$$), any singleton event $$\{x\}$$ has probability 0 but is not the empty set. Even in discrete cases with countably infinite sample spaces, events can have probability 0 without being empty.

(S2): If $$P(A) = 1$$, then $$A = \Omega$$.

This is also false. For example, if $$\Omega = [0, 1]$$ with uniform probability, then $$A = (0, 1)$$ (open interval) has $$P(A) = 1$$ but $$A \neq \Omega$$ since it doesn't contain the endpoints.

Both statements are false.

The answer is Option 4: both (S1) and (S2) are false.

We need to find the probability that a randomly chosen 2x2 matrix with entries from $$\{0, 1, 2\}$$ is invertible.

Determine the sample space.

A 2x2 matrix $$M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ has 4 entries, each chosen from $$\{0, 1, 2\}$$. Total number of matrices = $$3^4 = 81$$.

Determine the condition for invertibility.

A 2x2 matrix is invertible if and only if its determinant is non-zero:

$$ \det(M) = ad - bc \neq 0 $$

So we need to count matrices where $$ad = bc$$ (non-invertible), then subtract from 81.

Count the number of ways each product value can occur.

For two numbers chosen from $$\{0, 1, 2\}$$, the possible products and the number of ordered pairs $$(x, y)$$ giving each product are:

- Product = 0: pairs where at least one is 0. These are: $$(0,0), (0,1), (0,2), (1,0), (2,0)$$ = 5 ways

- Product = 1: $$(1,1)$$ = 1 way

- Product = 2: $$(1,2), (2,1)$$ = 2 ways

- Product = 4: $$(2,2)$$ = 1 way

Verification: $$5 + 1 + 2 + 1 = 9 = 3^2$$ .

Count matrices with $$ad = bc$$ (determinant = 0).

For $$ad = bc$$, both products must equal the same value. The number of such matrices is:

$$ N(\det = 0) = 5 \times 5 + 1 \times 1 + 2 \times 2 + 1 \times 1 $$

$$ = 25 + 1 + 4 + 1 = 31 $$

Here, for each possible product value $$p$$, we multiply the number of ways to get $$ad = p$$ by the number of ways to get $$bc = p$$.

Calculate the number of invertible matrices and the probability.

Number of invertible matrices = $$81 - 31 = 50$$

$$ P(A) = \frac{50}{81} $$

The correct answer is Option 4: $$\dfrac{50}{81}$$.

We begin by noting that three dice are rolled, so the total number of outcomes is $$6^3 = 216$$.

Next, the favorable outcomes, where all three dice show different numbers, can be counted by considering that the first die has 6 choices, the second die has 5 choices, and the third die has 4 choices, which gives $$\text{Favorable} = 6 \times 5 \times 4 = 120$$.

Therefore, the probability can be expressed as $$\frac{p}{q} = \frac{120}{216} = \frac{5}{9}$$, and here $$p = 5$$ and $$q = 9$$ are coprime.

Subtracting these values yields $$q - p = 9 - 5 = 4$$.

Hence, the correct answer is 4.

Given: Two dice A and B are rolled. $$\alpha$$ and $$\beta$$ are the numbers obtained. Find the variance of $$\alpha - \beta$$.

Use properties of variance.

Since $$\alpha$$ and $$\beta$$ are independent: $$\text{Var}(\alpha - \beta) = \text{Var}(\alpha) + \text{Var}(\beta)$$

For a fair die: $$E[X] = \frac{7}{2}$$, $$E[X^2] = \frac{1+4+9+16+25+36}{6} = \frac{91}{6}$$

$$\text{Var}(X) = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12}$$

$$\text{Var}(\alpha - \beta) = \frac{35}{12} + \frac{35}{12} = \frac{35}{6}$$

So $$\frac{p}{q} = \frac{35}{6}$$, where $$\gcd(35, 6) = 1$$. Thus $$p = 35$$.

Find the sum of positive divisors of $$p = 35$$.

$$35 = 5 \times 7$$

Divisors: $$1, 5, 7, 35$$

Sum = $$1 + 5 + 7 + 35 = 48$$

The correct answer is Option C: $$48$$.

Some couples participated in a mixed doubles badminton tournament with no couple playing in a match. Total matches = 840. Find total persons.

Let there be $$n$$ couples ($$2n$$ persons total). In mixed doubles, each team has one male and one female. For one match, choose 2 males in $$\binom{n}{2}$$ ways, then choose 2 females who are not the wives of the chosen males in $$\binom{n-2}{2}$$ ways, and pair them into two teams in 2 ways. Since each pairing corresponds to one distinct match, the total number of matches is $$\binom{n}{2}\binom{n-2}{2} \times 2 = 840$$. Substituting the binomial coefficients gives $$\dfrac{n(n-1)}{2} \times \dfrac{(n-2)(n-3)}{2} \times 2 = 840$$, which simplifies to $$\dfrac{n(n-1)(n-2)(n-3)}{2} = 840$$ and hence $$n(n-1)(n-2)(n-3) = 1680$$. Testing $$n = 8$$: $$8 \times 7 \times 6 \times 5 = 1680$$ $$\checkmark$$, so $$2n = 16$$.

16

Find the value of $$k$$ where the probability that $$N-2, \sqrt{3N}, N+2$$ are in geometric progression is $$\frac{k}{48}$$.

For three terms in GP, the square of the middle term equals the product of the other two: $$(\sqrt{3N})^2 = (N-2)(N+2)$$. This gives $$3N = N^2 - 4$$ and hence $$N^2 - 3N - 4 = 0$$.

The equation factors as $$(N-4)(N+1) = 0$$, giving $$N = 4$$ or $$N = -1$$; since $$N$$ is the sum of two dice, $$N \geq 2$$, we take $$N = 4$$.

For sum 4, the favorable outcomes are $$(1,3), (2,2), (3,1)$$, yielding 3 outcomes out of a total of $$6 \times 6 = 36$$. Hence $$P(N=4) = \frac{3}{36} = \frac{1}{12} = \frac{4}{48}$$.

Equating $$\frac{k}{48} = \frac{4}{48}$$ gives $$k = 4$$, so the correct answer is Option B: $$4$$.

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 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.

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 $$24$$.

For each round of the game let $$x$$ and $$y$$ be the numbers shown by $$P_1$$ and $$P_2$$, respectively.
If $$x\gt y$$ the score pair is $$(5,0)$$; if $$x=y$$ it is $$(2,2)$$; if $$x\lt y$$ it is $$(0,5)$$.

Because the dice are fair and independent, among the $$36$$ ordered pairs $$(x,y)$$:
• $$x\gt y$$ occurs in $$15$$ pairs  $$\Rightarrow$$ probability $$\dfrac{15}{36}=\dfrac{5}{12}$$.
• $$x=y$$ occurs in $$6$$ pairs    $$\Rightarrow$$ probability $$\dfrac{6}{36}=\dfrac{1}{6}$$.
• $$x\lt y$$ also occurs with probability $$\dfrac{5}{12}$$ (symmetry).

Define the “score difference” for one round as $$D=({\text{score of }}P_1)-({\text{score of }}P_2).$$ Hence $$D$$ takes three values

$$D=+5\;(\text{prob. }5/12),\quad D=0\;(\text{prob. }1/6),\quad D=-5\;(\text{prob. }5/12).$$

Totals after $$n$$ rounds satisfy $$X_n-Y_n=D_1+D_2+\dots +D_n,$$ with $$D_1,D_2,\dots$$ i.i.d. as above.

Case 1: Two rounds (distribution of $$S_2=D_1+D_2$$)

All ordered pairs $$(D_1,D_2)$$ and their probabilities:

$$\begin{array}{c|c} S_2 & \text{Probability} \\ \hline +10 & (5/12)^2 = 25/144\\ +5 & 2\,(5/12)(1/6)=20/144\\ 0 & 2\,(5/12)^2 + (1/6)^2 = 50/144+4/144 = 54/144\\ -5 & 2\,(1/6)(5/12)=20/144\\ -10 & (5/12)^2 = 25/144 \end{array}$$

Probability $$X_2\ge Y_2$$ (i.e. $$S_2\ge0$$): $$\dfrac{25+20+54}{144}=\dfrac{99}{144}=\dfrac{11}{16}.$$

Probability $$X_2\gt Y_2$$ (i.e. $$S_2\gt0$$): $$\dfrac{25+20}{144}=\dfrac{45}{144}=\dfrac{5}{16}.$$

Case 2: Three rounds (distribution of $$S_3=D_1+D_2+D_3$$)

Write $$(W,T,L)$$ for the counts of wins, ties, losses (so $$W+T+L=3$$ and $$S_3=5(W-L)$$).

For $$S_3=0$$ we need $$W=L$$. Two possibilities:

1. $$(0,3,0)$$ - three ties.   Probability $$\left(\dfrac{1}{6}\right)^3=\dfrac{1}{216}.$$

2. $$(1,1,1)$$ - one win, one tie, one loss.
  Number of permutations $$=3!/(1!1!1!)=6.$$
  Probability of each permutation $$\;(5/12)(1/6)(5/12)=\dfrac{25}{864}.$$
  Total probability $$6\times\dfrac{25}{864}=\dfrac{150}{864}=\dfrac{25}{144}.$$

Hence $$P(S_3=0)=\dfrac{1}{216}+\dfrac{25}{144}= \dfrac{4}{864}+\dfrac{150}{864}= \dfrac{154}{864}= \dfrac{77}{432}.$$

Because the game is symmetric, $$P(S_3\gt0)=P(S_3\lt0).$$ Therefore

$$P(S_3\gt0)=\dfrac{1-P(S_3=0)}{2}= \dfrac{1-\dfrac{77}{432}}{2}= \dfrac{355}{432}\times\dfrac{1}{2}= \dfrac{355}{864}.$$

We now list the required probabilities:

(I) $$P(X_2\ge Y_2)=\dfrac{11}{16} \; (Q)$$
(II) $$P(X_2\gt Y_2)=\dfrac{5}{16} \; (R)$$
(III) $$P(X_3=Y_3)=\dfrac{77}{432} \; (T)$$
(IV) $$P(X_3\gt Y_3)=\dfrac{355}{864} \; (S)$$

Thus the correct matching is:
(I) → (Q), (II) → (R), (III) → (T), (IV) → (S).

Hence the correct option is:
Option A which is: (I) → (Q); (II) → (R); (III) → (T); (IV) → (S).

Let the two events be defined as follows:
  • $$A$$ : “one of the two chosen balls is white’’
  • $$B$$ : “at least one of the two chosen balls is green’’

We must evaluate the conditional probability $$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$$.

Step 1 : Probability of each colour chosen from Box I
Box I has 8 red, 3 blue and 5 green balls; total $$16$$ balls.

$$P(R_1)=\frac{8}{16}=\frac12,\quad P(B_1)=\frac{3}{16},\quad P(G_1)=\frac{5}{16}$$

(Here the subscript ‘1’ marks the first ball.)

Step 2 : Composition of the other boxes
Box II : 24 R, 9 B, 15 G  (total 48)
Box III: 1 B, 12 G, 3 Y (total 16) (no white)
Box IV : 10 G, 16 O, 6 W (total 32)

Step 3 : Compute $$P(A\cap B)$$
The event $$A\cap B$$ requires that a white ball is obtained and at least one of the two balls is green.

Case 1 : first ball red
Probability $$\dfrac12$$. The second ball then comes from Box II, which contains no white balls, so $$A$$ cannot occur. Hence this case contributes 0 to $$P(A\cap B)$$.

Case 2 : first ball blue
Probability $$\dfrac{3}{16}$$. The second ball comes from Box III, which also has no white balls, so again $$A$$ is impossible. Contribution = 0.

Case 3 : first ball green
Probability $$\dfrac{5}{16}$$. Here $$B$$ is already satisfied because the first ball is green. The second ball is drawn from Box IV, which contains 6 white balls out of 32.
$$P(\text{second ball white}\mid G_1)=\frac{6}{32}=\frac{3}{16}$$

Therefore the probability of both $$A$$ and $$B$$ occurring is
$$P(A\cap B)=\frac{5}{16}\times\frac{3}{16}=\frac{15}{256}$$

Step 4 : Compute $$P(B)$$

Case 1 : first ball red
We need the second ball (from Box II) to be green.
$$P(G_2\mid R_1)=\frac{15}{48}=\frac{5}{16}$$
Contribution: $$\frac12\times\frac{5}{16}=\frac{5}{32}$$

Case 2 : first ball blue
Second ball is from Box III; it must be green.
$$P(G_2\mid B_1)=\frac{12}{16}=\frac34$$
Contribution: $$\frac{3}{16}\times\frac34=\frac{9}{64}$$

Case 3 : first ball green
Event $$B$$ is automatically satisfied regardless of the second draw.
Contribution: $$\frac{5}{16}\times 1=\frac{5}{16}$$

Add the three contributions:
$$P(B)=\frac{5}{32}+\frac{9}{64}+\frac{5}{16}$$

Convert to a common denominator 64:
$$P(B)=\frac{10}{64}+\frac{9}{64}+\frac{20}{64}=\frac{39}{64}$$

Step 5 : Conditional probability
$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}= \frac{\dfrac{15}{256}}{\dfrac{39}{64}}= \frac{15}{256}\times\frac{64}{39}$$

Simplify: $$\frac{64}{256}=\frac14\;,\quad P(A\mid B)=\frac{15}{4\times39}=\frac{15}{156}=\frac{5}{52}$$

Hence the required conditional probability is $$\dfrac{5}{52}$$.

Option C which is: $$\dfrac{5}{52}$$.

Let the sets be: $$F$$ = persons with fever, $$C$$ = persons with cough, $$B$$ = persons with breathing problem.

The given data are:
$$|F| = 190,\; |C| = 220,\; |B| = 220$$
$$|F \cup C| = 330,\; |C \cup B| = 350,\; |F \cup B| = 340$$
$$|F \cap C \cap B| = 30$$

Denote the pair-wise intersections (each includes the triple intersection) by
$$x = |F \cap C|,\; y = |C \cap B|,\; z = |F \cap B|.$$ Using the union formula $$|A \cup B| = |A| + |B| - |A \cap B|$$ we get:

For $$F$$ and $$C$$:
$$190 + 220 - x = 330 \;\Rightarrow\; x = 80.$$

For $$C$$ and $$B$$:
$$220 + 220 - y = 350 \;\Rightarrow\; y = 90.$$

For $$F$$ and $$B$$:
$$190 + 220 - z = 340 \;\Rightarrow\; z = 70.$$

Split every region of the Venn diagram:
Case 1: exactly two symptoms (excluding the third):
$$|F \cap C \text{ only}| = 80 - 30 = 50,$$
$$|C \cap B \text{ only}| = 90 - 30 = 60,$$
$$|F \cap B \text{ only}| = 70 - 30 = 40.$$

Case 2: exactly one symptom. Let these counts be $$a,b,c$$ respectively:

From $$|F| = a + 50 + 40 + 30$$ $$190 = a + 120 \;\Rightarrow\; a = 70.$$ From $$|C| = b + 50 + 60 + 30$$ $$220 = b + 140 \;\Rightarrow\; b = 80.$$ From $$|B| = c + 60 + 40 + 30$$ $$220 = c + 130 \;\Rightarrow\; c = 90.$$

Case 3: all three symptoms: $$g = 30.$$

Total counted in Venn regions:
$$70 + 80 + 90 + 50 + 60 + 40 + 30 = 420.$$

Hence the number having none of the symptoms is
$$h = 900 - 420 = 480.$$

People with at most one symptom = (exactly one) + (none)
$$= (70 + 80 + 90) + 480 = 240 + 480 = 720.$$

Required probability $$= \frac{720}{900} = 0.80.$$

Therefore, the probability that a randomly chosen person has at most one symptom is 0.80.

We need to find the probability that a randomly chosen one-one function $$f: \{a, b, c, d\} \to \{1, 2, 3, 4, 5\}$$ satisfies $$f(a) + 2f(b) - f(c) = f(d)$$.

Count total one-one functions:

$$\text{Total} = 5 \times 4 \times 3 \times 2 = 120$$

Find all injections satisfying the equation:

We need four distinct values from $$\{1, 2, 3, 4, 5\}$$ assigned to $$a, b, c, d$$ such that $$f(a) + 2f(b) - f(c) = f(d)$$.

Equivalently: $$f(a) + 2f(b) = f(c) + f(d)$$, where $$f(a), f(b), f(c), f(d)$$ are all distinct.

Let us denote the values as $$p = f(a), q = f(b), r = f(c), s = f(d)$$, all distinct, with $$p + 2q - r = s$$ and each in $$\{1, 2, 3, 4, 5\}$$.

We systematically check all possible values of $$q$$ (since it has the largest coefficient):

Case $$q = 1$$: $$s = p + 2 - r$$. Try all distinct $$(p, r)$$ from remaining values $$\{2,3,4,5\}$$:

• $$(p, r) = (3, 4)$$: $$s = 3 + 2 - 4 = 1 = q$$ $$\times$$
• $$(p, r) = (4, 3)$$: $$s = 4 + 2 - 3 = 3 = r$$ $$\times$$
• $$(p, r) = (3, 5)$$: $$s = 0 \notin \{1,...,5\}$$ $$\times$$
• $$(p, r) = (5, 3)$$: $$s = 4$$, all distinct: $$\{5, 1, 3, 4\}$$ $$\checkmark$$
• $$(p, r) = (4, 5)$$: $$s = 1 = q$$ $$\times$$
• $$(p, r) = (5, 4)$$: $$s = 3$$, all distinct: $$\{5, 1, 4, 3\}$$ $$\checkmark$$
• $$(p, r) = (2, 3)$$: $$s = 1 = q$$ $$\times$$
• $$(p, r) = (3, 2)$$: $$s = 3 = p$$ $$\times$$
• $$(p, r) = (2, 4)$$: $$s = 0$$ $$\times$$
• $$(p, r) = (4, 2)$$: $$s = 4 = p$$ $$\times$$
• $$(p, r) = (2, 5)$$: $$s = -1$$ $$\times$$
• $$(p, r) = (5, 2)$$: $$s = 5 = p$$ $$\times$$

2 valid assignments for $$q = 1$$.

Case $$q = 2$$: $$s = p + 4 - r$$. Try $$(p, r)$$ from $$\{1,3,4,5\}$$:

• $$(1, 4)$$: $$s = 1$$, $$= p$$ $$\times$$
• $$(4, 1)$$: $$s = 7$$ $$\times$$
• $$(1, 5)$$: $$s = 0$$ $$\times$$
• $$(5, 1)$$: $$s = 8$$ $$\times$$
• $$(3, 4)$$: $$s = 3 = p$$ $$\times$$
• $$(4, 3)$$: $$s = 5$$, all distinct: $$\{4, 2, 3, 5\}$$ $$\checkmark$$
• $$(3, 5)$$: $$s = 2 = q$$ $$\times$$
• $$(5, 3)$$: $$s = 6$$ $$\times$$
• $$(1, 3)$$: $$s = 2 = q$$ $$\times$$
• $$(3, 1)$$: $$s = 6$$ $$\times$$
• $$(4, 5)$$: $$s = 3$$, all distinct: $$\{4, 2, 5, 3\}$$ $$\checkmark$$
• $$(5, 4)$$: $$s = 5 = p$$ $$\times$$

2 valid assignments for $$q = 2$$.

Case $$q = 3$$: $$s = p + 6 - r$$. Try $$(p, r)$$ from $$\{1,2,4,5\}$$:

• $$(1, 2)$$: $$s = 5$$, all distinct: $$\{1, 3, 2, 5\}$$ $$\checkmark$$
• $$(2, 1)$$: $$s = 7$$ $$\times$$
• $$(1, 4)$$: $$s = 3 = q$$ $$\times$$
• $$(4, 1)$$: $$s = 9$$ $$\times$$
• $$(1, 5)$$: $$s = 2$$, all distinct: $$\{1, 3, 5, 2\}$$ $$\checkmark$$
• $$(5, 1)$$: $$s = 10$$ $$\times$$
• $$(2, 4)$$: $$s = 4 = r$$ $$\times$$
• $$(4, 2)$$: $$s = 8$$ $$\times$$
• $$(2, 5)$$: $$s = 3 = q$$ $$\times$$
• $$(5, 2)$$: $$s = 9$$ $$\times$$
• $$(4, 5)$$: $$s = 5 = r$$ $$\times$$
• $$(5, 4)$$: $$s = 7$$ $$\times$$

2 valid assignments for $$q = 3$$.

Case $$q = 4$$: $$s = p + 8 - r$$. Since the minimum of $$p + 8 - r$$ with $$p \geq 1$$ and $$r \leq 5$$ is $$1 + 8 - 5 = 4$$, most values will be too large or equal to $$q = 4$$. Checking all:

• All combinations yield $$s \geq 4$$, and $$s$$ must differ from $$p, q, r$$ and be $$\leq 5$$. No valid assignments.

Case $$q = 5$$: $$s = p + 10 - r \geq 1 + 10 - 4 = 7 > 5$$. No valid assignments.

Compute the probability:

Total favourable assignments: $$2 + 2 + 2 + 0 + 0 = 6$$

$$\text{Probability} = \frac{6}{120} = \frac{1}{20}$$

The correct answer is Option D: $$\dfrac{1}{20}$$.

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)

For a binomial distribution with parameters $$n$$ and $$p$$, the mean is $$np$$ and the variance is $$npq$$ where $$q = 1 - p$$.

Since the mean plus variance equals 24 and their product is 128, we have

To simplify these equations, introduce the notation $$m = np$$, which transforms the relationships into

From (1), it follows that

Substituting this expression for $$m$$ into equation (2) yields:

Since $$0 \le q \le 1$$, the valid solution is $$q = \dfrac{1}{2}$$ and hence $$p = \dfrac{1}{2}$$.

Having determined $$p$$, the relation $$m = np$$ gives

Finally, the probability of observing one or two successes is computed as

Thus, the required probability is $$\dfrac{33}{2^{28}}$$, which corresponds to Option C.

We have $$X \sim B(n, p)$$ with mean $$\mu = np$$ and variance $$\sigma^2 = npq$$ (where $$q = 1 - p$$). We are given:

$$\mu + \sigma^2 = 24$$ and $$\mu \cdot \sigma^2 = 128$$.

From the first equation: $$np = \frac{24}{1+q} = \frac{24}{2-p}$$.

From the second: $$(np)^2 \cdot q = 128$$.

Substituting: $$\left(\frac{24}{2-p}\right)^2 (1-p) = 128$$.

So $$p = \frac{1}{2}$$ (rejecting $$p = -1$$).

Then $$q = \frac{1}{2}$$, and $$np(1 + q) = np \cdot \frac{3}{2} = 24$$, so $$np = 16$$, giving $$n = 32$$.

Since $$P(X > n - 3) = \frac{k}{2^n} = \frac{k}{2^{32}}$$, we get $$k = 529$$.

The correct answer is Option B: 529.

We are given a binomial distribution with mean $$\alpha$$ and variance $$\dfrac{\alpha}{3}$$, and $$P(X = 1) = \dfrac{4}{243}$$.

First, to find $$p$$ and $$q$$, note that Mean $$= np = \alpha$$ and Variance $$= npq = \dfrac{\alpha}{3}$$. From this, dividing gives $$q = \dfrac{1}{3}$$, so $$p = 1 - q = \dfrac{2}{3}$$.

Next, to find $$n$$, use $$P(X = 1) = \binom{n}{1} p^1 q^{n-1} = n \cdot \dfrac{2}{3} \cdot \left(\dfrac{1}{3}\right)^{n-1} = \dfrac{2n}{3^n} = \dfrac{4}{243} = \dfrac{4}{3^5}$$. Hence, $$\dfrac{2n}{3^n} = \dfrac{4}{3^5}$$, which implies $$n = \dfrac{2 \cdot 3^n}{3^5} = 2 \cdot 3^{n-5}$$. Testing $$n = 6$$ gives $$2 \cdot 3^1 = 6 = n$$. $$\checkmark$$

Then compute $$P(X = 4 \text{ or } 5)$$. With $$n = 6$$, $$p = \dfrac{2}{3}$$, $$q = \dfrac{1}{3}$$: $$P(X = 4) = \binom{6}{4}\left(\dfrac{2}{3}\right)^4\left(\dfrac{1}{3}\right)^2 = 15 \cdot \dfrac{16}{81} \cdot \dfrac{1}{9} = \dfrac{240}{729}$$ and $$P(X = 5) = \binom{6}{5}\left(\dfrac{2}{3}\right)^5\left(\dfrac{1}{3}\right)^1 = 6 \cdot \dfrac{32}{243} \cdot \dfrac{1}{3} = \dfrac{192}{729}$$. From this, $$P(X = 4 \text{ or } 5) = \dfrac{240 + 192}{729} = \dfrac{432}{729} = \dfrac{16}{27}$$.

The correct answer is Option C: $$\dfrac{16}{27}$$.

A biased die has faces marked $$2, 4, 8, 16, 32, 32$$ with $$P(\text{face } n) = \frac{1}{n}$$.

First, we find the individual probabilities: $$P(2) = \frac{1}{2}, \quad P(4) = \frac{1}{4}, \quad P(8) = \frac{1}{8}, \quad P(16) = \frac{1}{16}$$. Since $$32$$ appears on two faces, its probability is $$P(32) = \frac{1}{32} + \frac{1}{32} = \frac{1}{16}$$. Verification shows $$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{16} = 1$$ ✓

Next, we look for all ways to obtain a sum of 48 in three throws by selecting numbers from $$\{2,4,8,16,32\}$$.

One way is $$(16,16,16)$$ since $$16 + 16 + 16 = 48$$. The probability of this sequence is $$\left(\frac{1}{16}\right)^3 = \frac{1}{4096}$$.

Another way is $$(32,8,8)$$ in any order, because $$32 + 8 + 8 = 48$$. There are $$\frac{3!}{2!} = 3$$ arrangements, giving a combined probability of $$3 \times \frac{1}{16} \times \frac{1}{8} \times \frac{1}{8} = \frac{3}{1024} = \frac{12}{4096}$$.

No other combinations work (for example, $$32 + 4 + 12$$ fails since 12 is not a face, and $$32 + 16 + 0$$ fails since 0 is not a face).

Therefore, the total probability is $$P = \frac{1}{4096} + \frac{12}{4096} = \frac{13}{4096} = \frac{13}{2^{12}}$$.

The answer is Option D: $$\frac{13}{2^{12}}$$.

We are given a random variable $$X$$ with the probability distribution:

$$P(X=0) = k, \quad P(X=1) = 2k, \quad P(X=2) = 4k, \quad P(X=3) = 6k, \quad P(X=4) = 8k$$

Since the total probability must equal 1:

$$k + 2k + 4k + 6k + 8k = 1$$

$$21k = 1$$, so $$k = \frac{1}{21}$$

We need to find $$P\left(\frac{1 < X < 4}{X \leq 2}\right)$$, which is the conditional probability $$P(1 < X < 4 \mid X \leq 2)$$.

The event $$\{1 < X < 4\} = \{X = 2, X = 3\}$$

The event $$\{X \leq 2\} = \{X = 0, X = 1, X = 2\}$$

The intersection $$\{1 < X < 4\} \cap \{X \leq 2\} = \{X = 2\}$$

$$P(X = 2) = 4k = \frac{4}{21}$$

$$P(X \leq 2) = k + 2k + 4k = 7k = \frac{7}{21} = \frac{1}{3}$$

Therefore:

$$P(1 < X < 4 \mid X \leq 2) = \frac{P(\{1 < X < 4\} \cap \{X \leq 2\})}{P(X \leq 2)} = \frac{4/21}{7/21} = \frac{4}{7}$$

The correct answer is Option A.

A die has faces {1, 2, 3, 4, 5, 6}. The primes are {2, 3, 5}, the composites are {4, 6}, and {1} is neither.

Let $$P(\text{each prime face}) = p$$, $$P(\text{each composite face}) = q$$, and $$P(1) = r$$.

The given condition is: $$3p = 6q = 2r$$.

Here $$3 \times P(\text{a prime number})$$ means $$3$$ times the probability of getting any one specific prime. Since each prime face has probability $$p$$, we interpret the condition as $$3p = 6q = 2r$$.

From $$3p = 6q$$: $$p = 2q$$.

From $$6q = 2r$$: $$r = 3q$$.

Total probability must equal 1:

Therefore: $$p = \frac{2}{11}$$, $$q = \frac{1}{11}$$, $$r = \frac{3}{11}$$.

Perfect squares on a die: {1, 4}.

X counts perfect squares in 2 throws, so $$X \sim \text{Binomial}\left(2, \frac{4}{11}\right)$$.

The correct answer is Option D: $$\frac{8}{11}$$.

Bag I contains 3 red, 4 black, and 3 white balls (total 10). Bag II contains 2 red, 5 black, and 2 white balls (total 9). One ball is transferred from Bag I to Bag II, then a ball is drawn from Bag II and found to be black. We need the probability that the transferred ball was red.

We use Bayes' theorem. Let $$R, B, W$$ denote the events that a red, black, or white ball was transferred, respectively.

The prior probabilities are $$P(R) = \frac{3}{10}$$, $$P(B) = \frac{4}{10}$$, $$P(W) = \frac{3}{10}$$.

After transfer, Bag II has 10 balls. The probability of drawing a black ball from Bag II given each case:
Case 1: If red was transferred, Bag II has 5 black out of 10, so $$P(\text{black}|R) = \frac{5}{10} = \frac{1}{2}$$.
Case 2: If black was transferred, Bag II has 6 black out of 10, so $$P(\text{black}|B) = \frac{6}{10} = \frac{3}{5}$$.
Case 3: If white was transferred, Bag II has 5 black out of 10, so $$P(\text{black}|W) = \frac{5}{10} = \frac{1}{2}$$.

By the law of total probability: $$P(\text{black}) = \frac{3}{10} \cdot \frac{1}{2} + \frac{4}{10} \cdot \frac{3}{5} + \frac{3}{10} \cdot \frac{1}{2} = \frac{3}{20} + \frac{12}{50} + \frac{3}{20} = \frac{3}{20} + \frac{6}{25} + \frac{3}{20}$$.

Converting to a common denominator of 100: $$= \frac{15}{100} + \frac{24}{100} + \frac{15}{100} = \frac{54}{100} = \frac{27}{50}$$.

By Bayes' theorem: $$P(R|\text{black}) = \frac{P(\text{black}|R) \cdot P(R)}{P(\text{black})} = \frac{\frac{1}{2} \cdot \frac{3}{10}}{\frac{27}{50}} = \frac{\frac{3}{20}}{\frac{27}{50}} = \frac{3}{20} \cdot \frac{50}{27} = \frac{150}{540} = \frac{5}{18}$$.

Hence, the correct answer is Option B.

We are given $$P(A) = \dfrac{1}{3}$$, $$P(B) = \dfrac{1}{5}$$, and $$P(A \cup B) = \dfrac{1}{2}$$.

First, we find $$P(A \cap B)$$ using the addition rule: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$.

$$\dfrac{1}{2} = \dfrac{1}{3} + \dfrac{1}{5} - P(A \cap B)$$

$$\dfrac{1}{2} = \dfrac{5 + 3}{15} - P(A \cap B) = \dfrac{8}{15} - P(A \cap B)$$

$$P(A \cap B) = \dfrac{8}{15} - \dfrac{1}{2} = \dfrac{16 - 15}{30} = \dfrac{1}{30}$$

We also need $$P(A') = 1 - \dfrac{1}{3} = \dfrac{2}{3}$$ and $$P(B') = 1 - \dfrac{1}{5} = \dfrac{4}{5}$$.

Now we compute $$P(A \mid B')$$. By definition, $$P(A \mid B') = \dfrac{P(A \cap B')}{P(B')}$$.

$$P(A \cap B') = P(A) - P(A \cap B) = \dfrac{1}{3} - \dfrac{1}{30} = \dfrac{10 - 1}{30} = \dfrac{9}{30} = \dfrac{3}{10}$$

$$P(A \mid B') = \dfrac{3/10}{4/5} = \dfrac{3}{10} \times \dfrac{5}{4} = \dfrac{15}{40} = \dfrac{3}{8}$$

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

$$P(B \cap A') = P(B) - P(A \cap B) = \dfrac{1}{5} - \dfrac{1}{30} = \dfrac{6 - 1}{30} = \dfrac{5}{30} = \dfrac{1}{6}$$

$$P(B \mid A') = \dfrac{1/6}{2/3} = \dfrac{1}{6} \times \dfrac{3}{2} = \dfrac{3}{12} = \dfrac{1}{4}$$

Therefore, $$P(A \mid B') + P(B \mid A') = \dfrac{3}{8} + \dfrac{1}{4} = \dfrac{3}{8} + \dfrac{2}{8} = \dfrac{5}{8}$$.

The answer is Option B: $$\dfrac{5}{8}$$.

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.

We are given: $$P(E_1 | E_2) = \frac{1}{2}$$, $$P(E_2 | E_1) = \frac{3}{4}$$, and $$P(E_1 \cap E_2) = \frac{1}{8}$$.

We use the relation $$ P(E_1 \cap E_2) = P(E_1 | E_2) \cdot P(E_2) \implies \frac{1}{8} = \frac{1}{2} \cdot P(E_2) \implies P(E_2) = \frac{1}{4} $$ and similarly apply $$ P(E_1 \cap E_2) = P(E_2 | E_1) \cdot P(E_1) \implies \frac{1}{8} = \frac{3}{4} \cdot P(E_1) \implies P(E_1) = \frac{1}{6} $$.

Next, we check each option.

Option A asks whether $$P(E_1 \cap E_2) = P(E_1) \cdot P(E_2)$$. The left-hand side is $$\frac{1}{8}$$, while the right-hand side is $$\frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}$$, so they are not equal.

Option B considers whether $$P(E_1' \cap E_2') = P(E_1') \cdot P(E_2)$$. We note that
$$P(E_1' \cap E_2') = 1 - P(E_1 \cup E_2) = 1 - (P(E_1) + P(E_2) - P(E_1 \cap E_2))$$
$$= 1 - \frac{1}{6} - \frac{1}{4} + \frac{1}{8} = 1 - \frac{4 + 6 - 3}{24} = 1 - \frac{7}{24} = \frac{17}{24}$$
and
$$P(E_1') \cdot P(E_2) = \frac{5}{6} \cdot \frac{1}{4} = \frac{5}{24}$$, which are not equal.

Option C checks whether $$P(E_1 \cap E_2') = P(E_1) \cdot P(E_2)$$. We compute
$$P(E_1 \cap E_2') = P(E_1) - P(E_1 \cap E_2) = \frac{1}{6} - \frac{1}{8} = \frac{4-3}{24} = \frac{1}{24}$$
and
$$P(E_1) \cdot P(E_2) = \frac{1}{6} \cdot \frac{1}{4} = \frac{1}{24}$$, showing that they are equal.

Option D asks if $$P(E_1 \cup E_2) = P(E_1) \cdot P(E_2)$$. Since $$P(E_1 \cup E_2) = \frac{7}{24}$$ and $$P(E_1) \cdot P(E_2) = \frac{1}{24}$$, they do not match.

Option C: $$P(E_1 \cap E_2') = P(E_1) \cdot P(E_2)$$.

We need to find the range of $$p$$ such that $$E_1, E_2, E_3$$ are valid mutually exclusive events.

First, impose the non-negativity conditions. Each probability must be $$\ge 0$$:

$$P(E_1) = \dfrac{2 + 3p}{6} \ge 0 \implies p \ge -\dfrac{2}{3}$$

$$P(E_2) = \dfrac{2 - p}{8} \ge 0 \implies p \le 2$$

$$P(E_3) = \dfrac{1 - p}{2} \ge 0 \implies p \le 1$$

Next, consider the sum condition since the events are mutually exclusive:

$$\dfrac{2 + 3p}{6} + \dfrac{2 - p}{8} + \dfrac{1 - p}{2} \le 1$$

Taking LCM = 24:

$$4(2 + 3p) + 3(2 - p) + 12(1 - p) \le 24$$

$$8 + 12p + 6 - 3p + 12 - 12p \le 24$$

$$26 - 3p \le 24$$

$$p \ge \dfrac{2}{3}$$

From the non-negativity conditions we have $$-\dfrac{2}{3} \le p \le 1$$ and from the sum condition $$p \ge \dfrac{2}{3}$$. This gives $$\dfrac{2}{3} \le p \le 1$$.

Finally, to find $$p_1 + p_2$$, note that the maximum value $$p_1 = 1$$ and minimum value $$p_2 = \dfrac{2}{3}$$. Hence $$p_1 + p_2 = 1 + \dfrac{2}{3} = \dfrac{5}{3}$$.

The correct answer is Option B: $$\dfrac{5}{3}$$.

We need to find the probability that a randomly chosen number $$n$$ from $$S = \{1, 2, 3, \ldots, 2022\}$$ satisfies $$\gcd(n, 2022) = 1$$.

We first factorise 2022: $$2022 = 2 \times 1011 = 2 \times 3 \times 337$$. We verify that 337 is prime: it is not divisible by 2, 3, 5, 7, 11, 13, 17, or 19 (and $$19^2 = 361 > 337$$), so 337 is indeed prime.

The count of numbers from 1 to $$N$$ that are coprime to $$N$$ is given by Euler's totient function $$\phi(N)$$. Since $$2022 = 2 \times 3 \times 337$$:

$$\phi(2022) = 2022\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{337}\right) = 2022 \times \frac{1}{2} \times \frac{2}{3} \times \frac{336}{337}$$

We compute step by step: $$2022 \times \frac{1}{2} = 1011$$, then $$1011 \times \frac{2}{3} = 674$$, then $$674 \times \frac{336}{337} = \frac{674 \times 336}{337} = \frac{2 \times 337 \times 336}{337} = 2 \times 336 = 672$$.

So $$\phi(2022) = 672$$, and the required probability is $$\frac{672}{2022}$$. We simplify: $$\gcd(672, 2022) = 6$$ (since $$672 = 6 \times 112$$ and $$2022 = 6 \times 337$$), giving $$\frac{672}{2022} = \frac{112}{337}$$.

Hence, the correct answer is Option D.

We need to find $$54P(X \leq 2)$$ where $$X \sim \text{Binomial}(n, p)$$ with mean $$4$$ and variance $$\dfrac{4}{3}$$.

Mean: $$np = 4$$

Variance: $$npq = \dfrac{4}{3}$$, where $$q = 1 - p$$.

Dividing: $$q = \dfrac{4/3}{4} = \dfrac{1}{3}$$, so $$p = \dfrac{2}{3}$$.

From $$np = 4$$: $$n = \dfrac{4}{2/3} = 6$$.

$$P(X = k) = \binom{6}{k}\left(\dfrac{2}{3}\right)^k\left(\dfrac{1}{3}\right)^{6-k}$$

$$P(X = 0) = \binom{6}{0}\left(\dfrac{1}{3}\right)^6 = \dfrac{1}{729}$$

$$P(X = 1) = \binom{6}{1}\left(\dfrac{2}{3}\right)\left(\dfrac{1}{3}\right)^5 = 6 \cdot \dfrac{2}{729} = \dfrac{12}{729}$$

$$P(X = 2) = \binom{6}{2}\left(\dfrac{2}{3}\right)^2\left(\dfrac{1}{3}\right)^4 = 15 \cdot \dfrac{4}{729} = \dfrac{60}{729}$$

$$P(X \leq 2) = \dfrac{1 + 12 + 60}{729} = \dfrac{73}{729}$$

$$54 \cdot \dfrac{73}{729} = \dfrac{54 \cdot 73}{729} = \dfrac{73}{729/54} = \dfrac{73}{13.5} = \dfrac{146}{27}$$

The correct answer is Option B: $$\dfrac{146}{27}$$.

$$X \sim B(7, p)$$ 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 need to find the probability that a relation $$R$$ from $$\{x, y\}$$ to $$\{x, y\}$$ is both symmetric and transitive. Such a relation is a subset of $$\{(x,x), (x,y), (y,x), (y,y)\}$$, so there are $$2^4 = 16$$ possible relations.

Symmetry requires that if $$(a,b)\in R$$ then $$(b,a)\in R$$, and transitivity requires that if $$(a,b)\in R$$ and $$(b,c)\in R$$ then $$(a,c)\in R$$. The symmetric subsets are those where $$(x,y)$$ and $$(y,x)$$ either both appear or both don’t. There are 2 choices for including or excluding $$(x,x)$$, 2 choices for $$(y,y)$$, and 2 choices for the pair $$\{(x,y),(y,x)\}$$, giving $$2^3 = 8$$ symmetric relations.

Checking these for transitivity shows that the empty set, $$\{(x,x)\}$$, $$\{(y,y)\}$$, $$\{(x,x),(y,y)\}$$, and the full set $$\{(x,x),(y,y),(x,y),(y,x)\}$$ are transitive, while the other three fail because having $$(x,y)$$ and $$(y,x)$$ demands inclusion of $$(x,x)$$ or $$(y,y)$$, which they lack. Thus there are 5 relations that are both symmetric and transitive.

Therefore, the probability is $$\f\frac{5}{16}$$, and the answer is $$\boldsymbol{\f\frac{5}{16}}$$.

A 3-digit number ranges from 100 to 999. Total count = 900.

We need P(at least 2 digits are odd). It is easier to compute:

$$P(\text{at least 2 odd}) = 1 - P(\text{0 odd}) - P(\text{exactly 1 odd})$$

Case 1: No odd digits (all even)

Even digits: {0, 2, 4, 6, 8}. The first digit cannot be 0, so first digit has 4 choices (2, 4, 6, 8). Second and third digits each have 5 choices.

Count = $$4 \times 5 \times 5 = 100$$

Case 2: Exactly 1 odd digit

Odd digits: {1, 3, 5, 7, 9} (5 choices). Even digits: {0, 2, 4, 6, 8}.

Sub-case 2a: First digit is odd, rest even

First digit: 5 choices. Second: 5 choices. Third: 5 choices.

Count = $$5 \times 5 \times 5 = 125$$

Sub-case 2b: Second digit is odd, rest even

First digit: 4 choices (even, non-zero). Second: 5 choices. Third: 5 choices.

Count = $$4 \times 5 \times 5 = 100$$

Sub-case 2c: Third digit is odd, rest even

First digit: 4 choices. Second: 5 choices. Third: 5 choices.

Count = $$4 \times 5 \times 5 = 100$$

Total for exactly 1 odd = $$125 + 100 + 100 = 325$$

At least 2 odd digits:

$$900 - 100 - 325 = 475$$

$$P = \frac{475}{900} = \frac{19}{36}$$

Hence the correct answer is Option A: $$\dfrac{19}{36}$$.

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 $$33$$.

There are 10 true-false questions. For 4 of them, the student guesses correctly with probability $$\frac{3}{4}$$, and for the remaining 6, the probability of a correct guess is $$\frac{1}{4}$$.

We need $$P(\text{exactly 8 correct out of 10})$$.

Let $$X$$ = number of correct answers among the 4 questions (probability $$\frac{3}{4}$$ each).

Let $$Y$$ = number of correct answers among the 6 questions (probability $$\frac{1}{4}$$ each).

We need $$P(X + Y = 8) = \sum_{i} P(X = i) \cdot P(Y = 8-i)$$.

Since $$X \leq 4$$ and $$Y \leq 6$$, we need $$i \geq 2$$ and $$8 - i \leq 6$$, so $$i \geq 2$$. Also $$i \leq 4$$.

Case $$i = 2$$, $$Y = 6$$:

$$P(X=2) = \binom{4}{2}\left(\frac{3}{4}\right)^2\left(\frac{1}{4}\right)^2 = 6 \cdot \frac{9}{16} \cdot \frac{1}{16} = \frac{54}{256}$$

$$P(Y=6) = \binom{6}{6}\left(\frac{1}{4}\right)^6 = \frac{1}{4096}$$

Product: $$\frac{54}{256} \cdot \frac{1}{4096} = \frac{54}{4^{10}}$$

Case $$i = 3$$, $$Y = 5$$:

$$P(X=3) = \binom{4}{3}\left(\frac{3}{4}\right)^3\left(\frac{1}{4}\right)^1 = 4 \cdot \frac{27}{64} \cdot \frac{1}{4} = \frac{108}{256}$$

$$P(Y=5) = \binom{6}{5}\left(\frac{1}{4}\right)^5\left(\frac{3}{4}\right)^1 = 6 \cdot \frac{1}{1024} \cdot \frac{3}{4} = \frac{18}{4096}$$

Product: $$\frac{108}{256} \cdot \frac{18}{4096} = \frac{1944}{4^{10}}$$

Case $$i = 4$$, $$Y = 4$$:

$$P(X=4) = \left(\frac{3}{4}\right)^4 = \frac{81}{256}$$

$$P(Y=4) = \binom{6}{4}\left(\frac{1}{4}\right)^4\left(\frac{3}{4}\right)^2 = 15 \cdot \frac{1}{256} \cdot \frac{9}{16} = \frac{135}{4096}$$

Product: $$\frac{81}{256} \cdot \frac{135}{4096} = \frac{10935}{4^{10}}$$

Total probability: $$\frac{54 + 1944 + 10935}{4^{10}} = \frac{12933}{4^{10}}$$

We are told this equals $$\frac{27k}{4^{10}}$$.

$$27k = 12933$$

$$k = \frac{12933}{27} = 479$$

The correct answer is $$479$$.

Given $$S = \{E_1, E_2, \ldots, E_8\}$$ with $$P(E_n) = \frac{n}{36}$$, and noting that $$\sum_{n=1}^{8} \frac{n}{36} = \frac{36}{36} = 1$$, for a subset $$A$$ we have $$P(A) = \frac{1}{36}\sum_{E_n \in A} n$$, so the condition $$P(A) \geq \frac{4}{5}$$ translates into $$\sum_{E_n \in A} n \geq \frac{4}{5}\times 36 = 28.8$$, i.e., $$\sum_{E_n \in A} n \geq 29$$.

If $$A^c$$ denotes the complement, then $$\sum_{A} n + \sum_{A^c} n = 36$$, so $$\sum_{A} n \geq 29 \iff \sum_{A^c} n \leq 7$$. We count subsets $$B \subseteq \{1,2,\ldots,8\}$$ with $$\text{sum}(B) \leq 7$$.

For $$|B| = 0$$ there is $$\emptyset$$ with sum 0 (1 subset). For $$|B| = 1$$ the subsets are $$\{1\}$$, $$\{2\}$$, $$\{3\}$$, $$\{4\}$$, $$\{5\}$$, $$\{6\}$$, $$\{7\}$$ (7 subsets).

For $$|B| = 2$$ the pairs with sum $$\leq 7$$ are $$\{1,2\}$$, $$\{1,3\}$$, $$\{1,4\}$$, $$\{1,5\}$$, $$\{1,6\}$$, $$\{2,3\}$$, $$\{2,4\}$$, $$\{2,5\}$$, $$\{3,4\}$$ (9 subsets). For $$|B| = 3$$ the triples with sum $$\leq 7$$ are $$\{1,2,3\}, \{1,2,4\}$$ (2 subsets). For $$|B| \geq 4$$ the minimum sum is $$1+2+3+4 = 10 > 7$$, so there are none.

$$1 + 7 + 9 + 2 = 19$$ so the answer is 19.

The line $$L$$ is defined by the intersection of two planes:

Plane 1: $$l x - y + (31 - l) z = 1$$, with normal vector $$\vec{n_1} = (l, -1, 31 - l)$$

Plane 2: $$x + 2y - z = 2$$, with normal vector $$\vec{n_2} = (1, 2, -1)$$

The direction vector $$\vec{d}$$ of line $$L$$ is given by the cross product $$\vec{n_1} \times \vec{n_2}$$:

$$$\vec{d} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ l & -1 & 31 - l \\ 1 & 2 & -1 \\ \end{vmatrix} = \vec{i}((-1)(-1) - (31 - l)(2)) - \vec{j}((l)(-1) - (31 - l)(1)) + \vec{k}((l)(2) - (-1)(1))$$$

Simplifying each component:

$$\vec{i}$$-component: $$1 - 2(31 - l) = 1 - 62 + 2l = 2l - 61$$

$$\vec{j}$$-component: $$-\left(-l - (31 - l)\right) = -\left(-31\right) = 31$$

$$\vec{k}$$-component: $$2l - (-1) = 2l + 1$$

Thus, $$\vec{d} = (2l - 61, 31, 2l + 1)$$.

The plane $$3x - 8y + 7z = 4$$ passes through line $$L$$ and is perpendicular to the plane $$3x + 2y + z = 6$$. The normal vector of the given perpendicular plane is $$\vec{n} = (3, -8, 7)$$. Since the plane contains line $$L$$, the direction vector $$\vec{d}$$ must be perpendicular to $$\vec{n}$$, so $$\vec{d} \cdot \vec{n} = 0$$:

$$$(2l - 61)(3) + (31)(-8) + (2l + 1)(7) = 0$$$

Expanding:

$$$6l - 183 - 248 + 14l + 7 = 0$$$

Combining like terms:

$$$20l - 424 = 0$$$

Solving for $$l$$:

$$$20l = 424 \implies l = \frac{424}{20} = \frac{106}{5}$$$

Substitute $$l = \frac{106}{5}$$ into $$\vec{d}$$:

$$$2l - 61 = 2\left(\frac{106}{5}\right) - 61 = \frac{212}{5} - \frac{305}{5} = -\frac{93}{5}$$$

$$$2l + 1 = 2\left(\frac{106}{5}\right) + 1 = \frac{212}{5} + \frac{5}{5} = \frac{217}{5}$$$

So $$\vec{d} = \left(-\frac{93}{5}, 31, \frac{217}{5}\right)$$. Factor out $$\frac{1}{5}$$:

$$$\vec{d} = \frac{1}{5}(-93, 155, 217)$$$

Notice that $$-93 = -3 \times 31$$, $$155 = 5 \times 31$$, and $$217 = 7 \times 31$$, so:

$$$\vec{d} = \frac{31}{5}(-3, 5, 7)$$$

The direction vector can be simplified to $$\vec{d} = (-3, 5, 7)$$ by scaling.

The acute angle $$\theta$$ between line $$L$$ and the $$y$$-axis (direction vector $$\vec{j} = (0, 1, 0)$$) is given by:

$$$\cos \theta = \frac{|\vec{d} \cdot \vec{j}|}{|\vec{d}| \cdot |\vec{j}|}$$$

Since $$|\vec{j}| = 1$$:

$$$\vec{d} \cdot \vec{j} = (-3)(0) + (5)(1) + (7)(0) = 5$$$

$$$|\vec{d}| = \sqrt{(-3)^2 + 5^2 + 7^2} = \sqrt{9 + 25 + 49} = \sqrt{83}$$$

Thus:

$$$\cos \theta = \frac{|5|}{\sqrt{83}} = \frac{5}{\sqrt{83}}$$$

Then:

$$$\cos^2 \theta = \left(\frac{5}{\sqrt{83}}\right)^2 = \frac{25}{83}$$$

Now compute $$415 \cos^2 \theta$$:

$$$415 \times \frac{25}{83} = \frac{415 \times 25}{83}$$$

Since $$415 \div 83 = 5$$:

$$$\frac{415}{83} \times 25 = 5 \times 25 = 125$$$

Therefore, $$415 \cos^2 \theta = 125$$.

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.

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 $$\dfrac{97}{297}$$.

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: $$\frac{17}{36}$$.

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)  &rightarrow;  impossible, reject.

B. $$37\%$$    (which is < 39)  &rightarrow;  impossible, reject.

C. $$65\%$$    (which is > 63)  &rightarrow;  impossible, reject.

D. $$55\%$$    (which lies between 39 and 63)  &rightarrow;  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 $$30$$:

$$2,\,3,\,5,\,7,\,11,\,13,\,17,\,19,\,23,\,29.$$