Question

If the random variable X has the following distribution:X	0	1	2	otherwise
P(X)	k	2k	3k	0

Match List-I with List-II:

Choose the correct answer from the options given below:

• A. (A)- (I), (B)- (II), (C)- (III), (D)- (IV)
• B. (A)- (IV), (B)- (III), (C)- (II), (D)- (I)
• C. (A)- (I), (B)- (II), (C)- (IV), (D)- (III)
• D. (A)- (III), (B)- (IV), (C)- (I), (D)- (II)

Answer 1

Answer: (B)

(A)- (IV), (B)- (III), (C)- (II), (D)- (I)

Answer 2

The given probabilities are $k, 2k, 3k$ for $X = 0, 1, 2$, respectively. Since the sum of probabilities must equal 1:

$k + 2k + 3k = 1 \implies 6k = 1 \implies k = \frac{1}{6}.$

For (A) $k$: From above, $k = \frac{1}{6}$. Match: (A) -> (IV).

For (B) $P(X<2)$: This is the sum of probabilities for $X = 0$ and $X = 1$:

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

Substituting $k = \frac{1}{6}$:

$P(X<2) = 3 \cdot \frac{1}{6} = \frac{1}{2}.$

Match: (B) -> (III).

For (C) $E(X)$: The expected value is:

$E(X) = \sum X \cdot P(X) = 0 \cdot k + 1 \cdot 2k + 2 \cdot 3k = 0 + 2k + 6k = 8k.$

Substituting $k = \frac{1}{6}$:

$E(X) = 8 \cdot \frac{1}{6} = \frac{4}{3}.$

Match: (C) -> (II).

For (D) $P(1 \leq X \leq 2)$: This is the sum of probabilities for $X = 1$ and $X = 2$:

$P(1 \leq X \leq 2) = P(X = 1) + P(X = 2) = 2k + 3k = 5k.$

Substituting $k = \frac{1}{6}$:

$P(1 \leq X \leq 2) = 5 \cdot \frac{1}{6} = \frac{5}{6}.$

Match: (D) -> (I).

Final matching: (A) -> (IV), (B) -> (III), (C) -> (II), (D) -> (I).