In a game, a man wins ₹8 for getting a number greater than 3 and loses ₹3 otherwise, when a fair die is thrown. The man decided to throw a die 4 times but to quit as and when he gets a number greater than 3. If X denotes the amount which the man wins or loses, then which of the following are correct?

(A) All the possible values of X are 8, 5, 2 and -1.

(B) The probability distribution of X is:

(C) The mean value of X is 75/16.

(D) The variance of X is 6615/256.

Choose the correct answer from the options given below:

(B), (C) and (D) only

The correct answer is Option (3) → (B), (C) and (D) only

Outcome of a single throw: number greater than 3 (4,5,6) is success with probability $p=\frac{1}{2}$, failure with probability $q=\frac{1}{2}$.

Payoff on each trial: success gives $+8$, failure gives $-3$. Stopping rule: stop at first success or after 4 throws.

Possible values of $X$ and their probabilities:

$X=8$ when success on 1st throw, $P=\frac{1}{2}$.

$X=5$ when success on 2nd throw, $P=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$.

$X=2$ when success on 3rd throw, $P=\left(\frac{1}{2}\right)^{3}=\frac{1}{8}$.

$X=-1$ when success on 4th throw, $P=\left(\frac{1}{2}\right)^{4}=\frac{1}{16}$.

$X=-12$ when no success in 4 throws, $P=\left(\frac{1}{2}\right)^{4}=\frac{1}{16}$.

Probability distribution:

$\begin{array}{c|ccccc} X & 8 & 5 & 2 & -1 & -12\\\hline P(X) & \frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} & \frac{1}{16} \end{array}$

$E[X]=8\cdot\frac{1}{2}+5\cdot\frac{1}{4}+2\cdot\frac{1}{8}+(-1)\cdot\frac{1}{16}+(-12)\cdot\frac{1}{16}=\frac{75}{16}$

$E[X^{2}]=8^{2}\cdot\frac{1}{2}+5^{2}\cdot\frac{1}{4}+2^{2}\cdot\frac{1}{8}+(-1)^{2}\cdot\frac{1}{16}+(-12)^{2}\cdot\frac{1}{16}=\frac{765}{16}$

$\mathrm{Var}(X)=E[X^{2}]-(E[X])^{2}=\frac{765}{16}-\left(\frac{75}{16}\right)^{2}=\frac{6615}{256}$

Possible values of $X$: $8,5,2,-1,-12$.

Probability distribution as above.

$E[X]=\frac{75}{16}$

$\mathrm{Var}(X)=\frac{6615}{256}$