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:

X	8	5	2	-1	-12
P(X)	1/2	1/4	1/8	1/16	1/16

(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:

The probability of getting a number greater than 3 (i.e., 4, 5, or 6) is $\frac{3}{6} = \frac{1}{2}$.

The probability of getting a number less than or equal to 3 (i.e., 1, 2, or 3) is $\frac{3}{6} = \frac{1}{2}$.

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Case 1: First throw shows a number > 3

The man wins ₹8 and quits.

$X = 8$

$P(X = 8) = \frac{1}{2}$

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Case 2: First throw ≤ 3, second throw > 3

The man loses ₹3, then wins ₹8 and quits.

$X = -3 + 8 = 5$

$P(X = 5) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

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Case 3: First two throws ≤ 3, third throw > 3

The man loses ₹3 twice, then wins ₹8 and quits.

$X = -3 - 3 + 8 = 2$

$P(X = 2) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$

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Case 4: First three throws ≤ 3, fourth throw > 3

The man loses ₹3 thrice, then wins ₹8 and quits.

$X = -3 - 3 - 3 + 8 = -1$

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

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Case 5: All four throws ≤ 3

The man loses ₹3 four times and must quit after 4 throws.

$X = -3 - 3 - 3 - 3 = -12$

$P(X = -12) = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$

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The possible values of X are: 8, 5, 2, -1, and -12.

Statement (A) is incorrect as it does not include -12.

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The probability distribution matches statement (B):

X	8	5	2	-1	-12
P(X)	1/2	1/4	1/8	1/16	1/16

Statement (B) is correct.

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The mean value of X:

$E(X) = 8 \times \frac{1}{2} + 5 \times \frac{1}{4} + 2 \times \frac{1}{8} + (-1) \times \frac{1}{16} + (-12) \times \frac{1}{16}$

$E(X) = 4 + \frac{5}{4} + \frac{1}{4} - \frac{1}{16} - \frac{12}{16}$

$E(X) = \frac{64}{16} + \frac{20}{16} + \frac{4}{16} - \frac{1}{16} - \frac{12}{16}$

$E(X) = \frac{75}{16}$

Statement (C) is correct.

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For variance, calculate $E(X^2)$:

$E(X^2) = 64 \times \frac{1}{2} + 25 \times \frac{1}{4} + 4 \times \frac{1}{8} + 1 \times \frac{1}{16} + 144 \times \frac{1}{16}$

$E(X^2) = 32 + \frac{25}{4} + \frac{1}{2} + \frac{1}{16} + 9$

$E(X^2) = \frac{512}{16} + \frac{100}{16} + \frac{8}{16} + \frac{1}{16} + \frac{144}{16}$

$E(X^2) = \frac{765}{16}$

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The variance:

$\text{Var}(X) = E(X^2) - [E(X)]^2$

$\text{Var}(X) = \frac{765}{16} - \left(\frac{75}{16}\right)^2$

$\text{Var}(X) = \frac{765}{16} - \frac{5625}{256}$

$\text{Var}(X) = \frac{12240}{256} - \frac{5625}{256}$

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

Statement (D) is correct.

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Therefore, statements (B), (C), and (D) are correct.