A random variable y has the following probability distribution

Match List-I with List-II

Choose the correct answer from the options given below:

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

The correct answer is Option (3) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Given probability distribution:

y: 1, 2, 3, 4, 5

P(y): 2k, 3k, k, 4k, 5k

Sum of probabilities = 1 ⇒ 2k + 3k + k + 4k + 5k = 15k = 1 ⇒ k = \frac{1}{15}

Compute required probabilities:

(A)$ P(y > 2) = P(3) + P(4) + P(5) = k + 4k + 5k = 10k = \frac{10}{15} = \frac{2}{3}$

(B) $k = \frac{1}{15}$

(C) $P(y ≤ 3) = P(1) + P(2) + P(3) = 2k + 3k + k = 6k = \frac{6}{15} = \frac{2}{5}$

(D) $P(2 ≤ y ≤ 4) = P(2) + P(3) + P(4) = 3k + k + 4k = 8k = \frac{8}{15}$