If the random variable X has the following probability distribution:

Match List-I with List-II

Choose the correct answer from the options given below:

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

The correct answer is Option (2) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

Given probability distribution:

$X: 0, 1, 2$

$P(X): k, 3k, 5k$

Sum of probabilities = 1:

$k + 3k + 5k = 9k = 1 \Rightarrow k = \frac{1}{9}$

(A) $k = \frac{1}{9} \Rightarrow$ (IV)

Expectation: $E(X) = 0*k + 1*3k + 2*5k = 3k + 10k = 13k = 13*\frac{1}{9} = \frac{13}{9}$ → (I)

$P(X \le 1) = P(0) + P(1) = k + 3k = 4k = \frac{4}{9}$ → (II)

$P(1 \le X \le 2) = P(1) + P(2) = 3k + 5k = 8k = \frac{8}{9}$ → (III)