For a random variable x , probability distribution P(x) is given by P…

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

The three probabilities are

    \( \displaystyle P(0) = \frac{3k}{6},\quad P(1) = \frac{2k}{6},\quad P(2) = \frac{k}{6} \)

(A) \(k\) is equal to

This matches (II) 1. The probabilities add up to 1:

    \( \displaystyle \frac{3k + 2k + k}{6} = k = 1 \)

So \(P(0) = \frac{1}{2}\), \(P(1) = \frac{1}{3}\), \(P(2) = \frac{1}{6}\).

(B) \(P(x = 0)\)

This matches (I) \(\frac{1}{2}\), as \(P(0) = \frac{3}{6}\).

(C) \(P(x \lt 2)\)

This matches (IV) \(\frac{5}{6}\). Here x is 0 or 1:

    \( \displaystyle P(0) + P(1) = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \)

(D) \(P(1 \lt x \le 2)\)

This matches (III) \(\frac{1}{6}\). The only value of x with \(1 \lt x \le 2\) is x = 2, so the probability is \(P(2) = \frac{1}{6}\).

So the matching is (A)-(II), (B)-(I), (C)-(IV), (D)-(III).