For a random variable x, probability distribution P(x) is given by $P(x) = \frac{k}{6}(3-x), x = 0,1,2$, then

Match List-I with List-II

Choose the correct answer from the options given below:

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

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

Given: $P(x) = \frac{k}{6}(3 - x)$ for $x = 0, 1, 2$

Use the total probability rule: $\sum P(x) = 1$

$P(0) = \frac{k}{6}(3 - 0) = \frac{3k}{6}$

$P(1) = \frac{k}{6}(3 - 1) = \frac{2k}{6}$

$P(2) = \frac{k}{6}(3 - 2) = \frac{k}{6}$

Total: $\frac{3k + 2k + k}{6} = \frac{6k}{6} = k$

Set $k = 1$ to satisfy total probability = 1

Now compute individual probabilities:

$P(x = 0) = \frac{3}{6} = \frac{1}{2}$

$P(x < 2) = P(0) + P(1) = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$

$P(1 < x \leq 2) = P(2) = \frac{1}{6}$