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 $$n=2$$.

We are selecting 3 distinct numbers randomly from the set $$ \{1, 2, 3, \ldots, 40\} $$.

Step 1: Calculate the Total Number of Outcomes

The total number of ways to select 3 distinct numbers from a set of 40 numbers is given by the combination formula:

$$ \text{Total Ways} = \binom{40}{3} = \frac{40 \times 39 \times 38}{3 \times 2 \times 1} = 9880 $$

Step 2: Find the Favorable Outcomes (Increasing G.P.)

Let the three selected numbers be $$ a, b, c $$ such that they form an increasing geometric progression ($$ a < b < c $$).

Any three integers forming a geometric progression can be written in the form:

$$ a = x \cdot p^2 $$

$$ b = x \cdot p \cdot q $$

$$ c = x \cdot q^2 $$

where $$ x, p, q $$ are positive integers such that $$ \gcd(p, q) = 1 $$ and $$ q > p \ge 1 $$.

Since the numbers belong to the set, the maximum number $$ c $$ must satisfy:

$$ c = x \cdot q^2 \le 40 $$

We can find the number of valid combinations by iterating through the possible values of $$ q $$ (where $$ q^2 \le 40 $$):

1. For $$ q = 2 $$:

The only coprime value for $$ p $$ is 1.

$$ x \cdot 2^2 \le 40 \implies 4x \le 40 \implies x \in \{1, 2, \ldots, 10\} $$ (10 ways)

2. For $$ q = 3 $$:

The coprime values for $$ p $$ are 1 and 2.

$$ x \cdot 3^2 \le 40 \implies 9x \le 40 \implies x \in \{1, 2, 3, 4\} $$ (4 ways per $$ p $$, so $$ 2 \times 4 = 8 $$ ways)

3. For $$ q = 4 $$:

The coprime values for $$ p $$ are 1 and 3 (since $$ \gcd(2,4) \neq 1 $$).

$$ x \cdot 4^2 \le 40 \implies 16x \le 40 \implies x \in \{1, 2\} $$ (2 ways per $$ p $$, so $$ 2 \times 2 = 4 $$ ways)

4. For $$ q = 5 $$:

The coprime values for $$ p $$ are 1, 2, 3, and 4.

$$ x \cdot 5^2 \le 40 \implies 25x \le 40 \implies x \in \{1\} $$ (1 way per $$ p $$, so $$ 4 \times 1 = 4 $$ ways)

5. For $$ q = 6 $$:

The coprime values for $$ p $$ are 1 and 5.

$$ x \cdot 6^2 \le 40 \implies 36x \le 40 \implies x \in \{1\} $$ (1 way per $$ p $$, so $$ 2 \times 1 = 2 $$ ways)

Summing all the favorable ways together:

$$ \text{Favorable Ways} = 10 + 8 + 4 + 4 + 2 = 28 $$

Step 3: Compute the Probability $$ \frac{m}{n} $$

The probability of selecting numbers that form an increasing G.P. is:

$$ \text{Probability} = \frac{\text{Favorable Ways}}{\text{Total Ways}} = \frac{28}{9880} $$

Simplifying this fraction by dividing the numerator and denominator by their greatest common divisor, which is 4:

$$ \frac{m}{n} = \frac{7}{2470} $$

Since 7 is a prime number and does not divide 2470, the fraction is in its lowest terms ($$ \gcd(m, n) = 1 $$).

Thus, we have:

$$ m = 7 $$

$$ n = 2470 $$

Step 4: Calculate $$ m + n $$

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

Therefore, the final value of $$ m + n $$ is 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 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: $$\frac{2}{7}$$.

 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:

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

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.

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

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

$$P(X=2)=\frac{\binom{3}{2}}{\binom{18}{2}}=\frac{3}{153}$$.

$$E(X)=0+45/153+6/153=51/153=1/3$$. 

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

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

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.

To find the probability that at least 3 players pick their correct T-shirt, we use the concept of derangements. A derangement is a permutation where no element appears in its original position.

Let $$X$$ be the number of players who pick their correct T-shirt. We want to find $$P(X \ge 3)$$, which can be calculated using the complement rule:

$$P(X \ge 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))$$

Step 1: Formula for exactly $$k$$ matches

The number of ways exactly $$k$$ players pick their correct T-shirts is found by choosing the $$k$$ players, and then finding the number of derangements for the remaining $$15 - k$$ players.

Ways = $$\binom{15}{k} D_{15-k}$$

Where $$D_n$$is the number of derangements of $$n$$ items, given by:

$$D_n = n! \sum_{i=0}^{n} \frac{(-1)^i}{i!}$$

The total number of ways the 15 players can pick the T-shirts is $$15!$$.

Thus, the probability of exactly $$k$$ matches is:

$$P(X = k) = \frac{\binom{15}{k} D_{15-k}}{15!}$$

Expanding the combination $$\binom{15}{k} = \frac{15!}{k!(15-k)!}$$:

$$P(X = k) = \frac{\frac{15!}{k!(15-k)!} (15-k)! \sum_{i=0}^{15-k} \frac{(-1)^i}{i!}}{15!}$$

$$P(X = k) = \frac{1}{k!} \sum_{i=0}^{15-k} \frac{(-1)^i}{i!}$$

Step 2: Calculate $$P(X=0)$$, $$P(X=1)$$, and $$P(X=2)$$

For $$k = 0$$:

$$P(X = 0) = \frac{1}{0!} \sum_{i=0}^{15} \frac{(-1)^i}{i!} = \sum_{i=0}^{15} \frac{(-1)^i}{i!}$$

For $$k = 1$$:

$$P(X = 1) = \frac{1}{1!} \sum_{i=0}^{14} \frac{(-1)^i}{i!} = \sum_{i=0}^{14} \frac{(-1)^i}{i!}$$

For $$k = 2$$:

$$P(X = 2) = \frac{1}{2!} \sum_{i=0}^{13} \frac{(-1)^i}{i!} = \frac{1}{2} \sum_{i=0}^{13} \frac{(-1)^i}{i!}$$

Step 3: Approximate the probabilities using the Maclaurin series

For large values of $$n$$, the summation part closely approximates the expansion of $$e^{-1}$$:

$$e^{-1} = \sum_{i=0}^{\infty} \frac{(-1)^i}{i!} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$$

Since 15, 14, and 13 are sufficiently large, we can approximate the sums as:

$$\sum_{i=0}^{15} \frac{(-1)^i}{i!} \approx e^{-1}$$

$$\sum_{i=0}^{14} \frac{(-1)^i}{i!} \approx e^{-1}$$

$$\sum_{i=0}^{13} \frac{(-1)^i}{i!} \approx e^{-1}$$

Substituting these approximations:

$$P(X = 0) \approx e^{-1}$$

$$P(X = 1) \approx e^{-1}$$

$$P(X = 2) \approx \frac{1}{2} e^{-1}$$

Step 4: Sum the approximate probabilities

$$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$$

$$P(X < 3) \approx e^{-1} + e^{-1} + \frac{1}{2} e^{-1}$$

$$P(X < 3) \approx \frac{5}{2} e^{-1}$$

Step 5: Apply the series truncation for fractional options

To match the given fractional options, we truncate the series for $$e^{-1}$$ after the first four terms:

$$e^{-1} \approx 1 - 1 + \frac{1}{2!} - \frac{1}{3!}$$

$$e^{-1} \approx \frac{1}{2} - \frac{1}{6}$$

$$e^{-1} \approx \frac{3}{6} - \frac{1}{6}$$

$$e^{-1} \approx \frac{2}{6} = \frac{1}{3}$$

Now, substitute this approximated value of 1/3 back into our probability equation:

$$P(X < 3) \approx \frac{5}{2} \left( \frac{1}{3} \right)$$

$$P(X < 3) \approx \frac{5}{6}$$

Final Answer

Subtract this sum from 1 to find the probability that at least 3 players pick the correct T-shirt:

$$P(X \ge 3) = 1 - P(X < 3)$$

$$P(X \ge 3) \approx 1 - \frac{5}{6}$$

$$P(X \ge 3) \approx \frac{1}{6}$$

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

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

Step 1: Determine the probabilities of the individual sample points

We are given that $$S = \{w_1, w_2, w_3, \ldots\}$$ is the sample space of a random experiment. The relation between successive probabilities is:

$$P(w_n) = \frac{P(w_{n-1})}{2} \quad \text{for} \quad n \geq 2$$

This forms a geometric sequence with a common ratio of $$\frac{1}{2}$$:

$$P(w_1) = P(w_1)$$

$$P(w_2) = \frac{P(w_1)}{2}$$

$$P(w_3) = \frac{P(w_1)}{2^2}$$

$$P(w_n) = \frac{P(w_1)}{2^{n-1}}$$

Since the sum of all probabilities in a sample space must equal 1:

$$\sum_{n=1}^{\infty} P(w_n) = 1$$

$$P(w_1) \left(1 + \frac{1}{2} + \frac{1}{2^2} + \dots\right) = 1$$

Using the sum of an infinite geometric progression formula $$S_{\infty} = \frac{a}{1-r}$$:

$$P(w_1) \cdot \left(\frac{1}{1 - \frac{1}{2}}\right) = 1$$

$$2 P(w_1) = 1 \implies P(w_1) = \frac{1}{2}$$

Substituting $$P(w_1)$$ back into the general term gives:

$$P(w_n) = \frac{1}{2^n}$$

Step 2: Identify the elements of set A

The set $$A$$ is defined for natural numbers $$k, l \in \mathbb{N}$$ ($$k, l \geq 1$$) as:

$$A = \{2k + 3l; \, k, l \in \mathbb{N}\}$$

Let us find the possible values generated by this linear combination starting from the minimum values:

- For $$k = 1, l = 1 \implies 2(1) + 3(1) = 5$$

- For $$k = 2, l = 1 \implies 2(2) + 3(1) = 7$$

- For $$k = 1, l = 2 \implies 2(1) + 3(2) = 8$$

- For $$k = 3, l = 1 \implies 2(3) + 3(1) = 9$$

- For $$k = 2, l = 2 \implies 2(2) + 3(2) = 10$$

By analyzing the values, we observe that:

- The numbers $$1, 2, 3, 4$$ cannot be formed because the minimum value is 5.

- The number $$6$$ cannot be formed because if $$2k + 3l = 6$$, since $$l \geq 1$$, the only options are $$l=1 \implies 2k=3$$ (no integer solution) or $$l \geq 2 \implies 2k \leq 0$$ (not a natural number).

- All other natural numbers greater than or equal to 5 can be formed.

Thus, the set $$A$$ contains all natural numbers except $$\{1, 2, 3, 4, 6\}$$:

$$A = \{5, 7, 8, 9, 10, 11, \ldots\}$$

Step 3: Calculate the probability of event B

The event $$B$$ is defined as $$B = \{w_n; \, n \in A\}$$. The total probability $$P(B)$$ is the sum of probabilities of $$w_n$$ for all $$n \in A$$:

$$P(B) = \sum_{n \in A} P(w_n) = \sum_{n \in A} \frac{1}{2^n}$$

We can compute this by taking the sum of the entire infinite series from $$n = 1$$ to $$\infty$$ and subtracting the probabilities of the missing indices $$\{1, 2, 3, 4, 6\}$$:

$$P(B) = \sum_{n=1}^{\infty} \frac{1}{2^n} - \left(\frac{1}{2^1} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \frac{1}{2^6}\right)$$

We know that the full sum evaluates to:

$$\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1$$

Now, calculate the sum of the excluded terms:

$$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{64} = \frac{32 + 16 + 8 + 4 + 1}{64} = \frac{61}{64}$$

Subtracting this from 1:

$$P(B) = 1 - \frac{61}{64} = \frac{3}{64}$$

Conclusion:

The value of $$P(B)$$ is equal to $$\frac{3}{64}$$.

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

Let mean $$M = np$$ and variance $$V = npq$$.

$$M + V = 5$$

$$M \cdot V = 6$$

Since variance must be less than the mean ($$V < M$$): $$V = 2, \quad M = 3$$

$$npq = 2, \quad np = 3$$

$$q = \frac{npq}{np} = \frac{2}{3}$$

$$p = 1 - \frac{2}{3} = \frac{1}{3}$$

$$n\left(\frac{1}{3}\right) = 3 \implies n = 9$$

$$6(n + p - q) = 2 \times 26 = 52$$

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.

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

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

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

The quadratic equation is

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

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

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

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

$$25N^2<256$$

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

$$N^2<10.24$$

Since $$N$$ is a positive integer,

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

Therefore, the required probability is

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

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

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

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

Hence,

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

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

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

Thus,

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

giving

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

Therefore,

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

Hence, the correct answer is

$$27$$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Thus the required value is $$24$$.

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: $$\dfrac{1}{20}$$.

Given curves are

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

and

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

First rewrite them.

Parabola:

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

Hence,

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

Circle:

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

Completing the square,

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

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

So the circle has centre

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

and radius

$$4$$

Now find points of intersection.

Substitute

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

in the circle:

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

Solving gives

$$y=\pm2$$

and corresponding

$$x=0$$

Hence, intersection points are

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

Now the smaller enclosed region lies between:

- parabola on the right

- left arc of the circle on the left

For the circle,

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

Hence required area is

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

Using symmetry,

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

Now,

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

Also,

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

Now evaluate

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

Using standard formula,

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

with

$$a=4,$$

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

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

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

Therefore,

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

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

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

Rearranging,

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

Since area is positive,

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

Hence, the required area is

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

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

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

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

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

Apply Bayes' theorem.

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

Simplify and solve.

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

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

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

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

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

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

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

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

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

Answer: Option C (n = 4)

Total number of ways to select $$5$$ numbers from $$\{1,2,3,\ldots,18\}$$ is $${18\choose5}=8568$$

Given,

$$x_2=7,\qquad x_4=11$$

Since

$$x_1<x_2<x_3<x_4<x_5$$

we must have:

$$x_1\in\{1,2,3,4,5,6\}$$

So, number of choices for $$x_1$$ is

$$6$$

Also,

$$x_3\in\{8,9,10\}$$

So, number of choices for $$x_3$$ is

$$3$$

Further,

$$x_5\in\{12,13,14,15,16,17,18\}$$

So, number of choices for $$x_5$$ is

$$7$$

Hence, favourable outcomes are

$$6\times3\times7=126$$

Therefore, required probability is

$$\frac{126}{8568}=\frac1{68}$$

Hence,

$$\boxed{\frac1{68}}$$

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

$$np + npq = 24 \implies np(1 + q) = 24$$

$$np \cdot npq = 128 \implies n^2p^2q = 128$$

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

$$m(1 + q) = 24 \quad \ldots (1)$$

$$m^2 q = 128 \quad \ldots (2)$$

From (1), it follows that

$$m = \dfrac{24}{1 + q}$$

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

$$\dfrac{576}{(1+q)^2} \cdot q = 128$$

$$576q = 128(1+q)^2 = 128(1 + 2q + q^2)$$

$$576q = 128 + 256q + 128q^2$$

$$128q^2 - 320q + 128 = 0$$

$$q^2 - 2.5q + 1 = 0 \implies 2q^2 - 5q + 2 = 0$$

$$(2q - 1)(q - 2) = 0$$

$$q = \dfrac{1}{2} \text{ or } q = 2$$

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

$$m = np = \dfrac{24}{1 + \frac{1}{2}} = \dfrac{24}{\frac{3}{2}} = 16$$

$$n = \dfrac{16}{p} = \dfrac{16}{1/2} = 32$$

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

$$P(X = 1) + P(X = 2) = \binom{32}{1}\left(\dfrac{1}{2}\right)^{32} + \binom{32}{2}\left(\dfrac{1}{2}\right)^{32}$$

$$= \dfrac{1}{2^{32}}\left(32 + \dfrac{32 \cdot 31}{2}\right) = \dfrac{1}{2^{32}}(32 + 496) = \dfrac{528}{2^{32}}$$

$$= \dfrac{528}{2^{32}} = \dfrac{33 \times 16}{2^{32}} = \dfrac{33 \times 2^4}{2^{32}} = \dfrac{33}{2^{28}}$$

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 $$\mu \cdot \sigma^2 = 128$$.

$$np + npq = 24 \implies np(1 + q) = 24$$

$$np \cdot npq = 128 \implies n^2p^2q = 128$$

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

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

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

$$\frac{576(1-p)}{(2-p)^2} = 128$$

$$576(1-p) = 128(2-p)^2$$

$$576 - 576p = 128(4 - 4p + p^2)$$

$$576 - 576p = 512 - 512p + 128p^2$$

$$64 - 64p = 128p^2$$

$$1 - p = 2p^2$$

$$2p^2 + p - 1 = 0$$

$$(2p - 1)(p + 1) = 0$$

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 $$n = 32$$.

$$P(X = r) = \binom{32}{r}\left(\frac{1}{2}\right)^{32}$$

$$P(X > 29) = \frac{1}{2^{32}}\left[\binom{32}{30} + \binom{32}{31} + \binom{32}{32}\right]$$

$$\binom{32}{30} = \binom{32}{2} = \frac{32 \times 31}{2} = 496$$

$$\binom{32}{31} = 32$$

$$\binom{32}{32} = 1$$

$$P(X > 29) = \frac{496 + 32 + 1}{2^{32}} = \frac{529}{2^{32}}$$

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:

$$3p + 2q + r = 1$$

$$3(2q) + 2q + 3q = 1$$

$$6q + 2q + 3q = 1$$

$$11q = 1 \implies q = \frac{1}{11}$$

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

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

$$P(\text{perfect square}) = P(1) + P(4) = \frac{3}{11} + \frac{1}{11} = \frac{4}{11}$$

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

$$E(X) = np = 2 \times \frac{4}{11} = \frac{8}{11}$$

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: $$\dfrac{5}{8}$$.

The given region is bounded by

$$x=0$$

$$x+2y=6$$

and

$$5x-6y=30$$

The vertices of the triangle are obtained as follows:

From

$$x=0,\quad x+2y=6$$

we get

$$y=3$$

So, one vertex is

$$(0,3)$$

From

$$x=0,\quad 5x-6y=30$$

we get

$$y=-5$$

So, another vertex is

$$(0,-5)$$

Now, solving

$$x+2y=6$$

and

$$5x-6y=30$$

we get

$$x=6,\qquad y=0$$

Hence, the third vertex is

$$(6,0)$$

Therefore, total area of the triangle is

$$\frac12\times8\times6=24$$

Now, the line

$$y=1$$

intersects

$$x+2y=6$$

at

$$x=4$$

So, the region where

$$y\ge1$$

is a triangle with vertices

$$(0,3),\ (0,1),\ (4,1)$$

Its area is

$$\frac12\times2\times4=4$$

Hence, area where

$$y<1$$

is

$$24-4=20$$

Therefore, required probability is

$$\frac{20}{24}=\frac56$$

Hence, the correct answer is

$$\boxed{\text{Option B }\frac56}$$

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)

Given quadratic expression

$$x^2+\alpha x+\beta$$

For

$$x^2+\alpha x+\beta>0\quad \forall x\in\mathbb R,$$

the quadratic must have a positive leading coefficient and no real roots.

Since the coefficient of $$x^2$$ is already positive, we only require

$$\Delta<0$$

Therefore,

$$\alpha^2-4\beta<0$$

$$\alpha^2<4\beta$$

Now,

$$\alpha,\beta\in\{1,2,3,4,5,6\}$$

We count favorable cases.

For $$\alpha=1,$$

$$1<4\beta$$

which is true for all

$$\beta=1,2,3,4,5,6$$

Hence,

$$6$$ cases.

For $$\alpha=2,$$

$$4<4\beta$$

$$\beta>1$$

Hence,

$$\beta=2,3,4,5,6$$

giving

$$5$$ cases.

For $$\alpha=3,$$

$$9<4\beta$$

$$\beta\ge3$$

Hence,

$$4$$ cases.

For $$\alpha=4,$$

$$16<4\beta$$

$$\beta>4$$

Hence,

$$\beta=5,6$$

giving

$$2$$ cases.

For $$\alpha=5,$$

$$25<4\beta$$

is impossible.

Hence,

$$0$$ cases.

For $$\alpha=6,$$

$$36<4\beta$$

is impossible.

Hence,

$$0$$ cases.

Therefore, total favorable outcomes are

$$6+5+4+2=17$$

Total possible outcomes are

$$6\times6=36$$

Hence, the required probability is

$$\boxed{\frac{17}{36}}$$

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 the intersection of the planes

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

Any plane passing through $$L$$ is

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

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

Given plane is

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

Comparing coefficients,

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

Using

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

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

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

$$24\lambda=-4$$

$$\lambda=-\frac16$$

Now,

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

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

$$l-\frac16=\frac12$$

$$l=\frac23$$

Hence, normals of the two planes are

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

and

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

Therefore, direction vector of the line is

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

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

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

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

Thus,

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

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

Then,

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

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

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

Therefore,

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

Hence,

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

$$=5\cdot25$$

$$=125$$

Therefore, the required value is

$$\boxed{125}$$.

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

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

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

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

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

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

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

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

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

Hence $$n = 96$$.

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

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

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

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

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

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

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

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

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

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

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

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

This corresponds to Option 2.

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

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

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

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

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

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

Combining these two facts, we write

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

Substituting the given values,

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

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

Now we simplify step by step:

$$187-K\le 100,$$

so

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

hence

$$-K\le -87,$$

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

$$K\ge 87.$$

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

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

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

$$K\le 89.$$

Putting the two inequalities together, we get the precise interval

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

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

Now we inspect each given option:

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

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

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

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

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

Hence, the correct answer is Option A.

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

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

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

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

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

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