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: | List-I | List-II | | --- | --- | | (A) k | (I) 5/6 | | (B) P(X < 2) | (II) 4/3 | | (C) E(X) | (III) 1/2 | | (D) P(1 ≤ X ≤ 2) | (IV) 1/6 | Option 1: (A) - (I), (B) - (II), (C) - (III), (D) - (IV) Option 2: (A) - (IV), (B) - (III), (C) - (II), (D) - (I) Option 3: (A) - (I), (B) - (II), (C) - (IV), (D) - (III) Option 4: (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

The answer is '(A) - (IV), (B) - (III), (C) - (II), (D) - (I)'. View full solution:

Algebra > Probability

Easy

core

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: | List-I | List-II | | --- | --- | | (A) $ k $ | (I) $ \frac{5}{6} $ | | (B) $ P(X < 2) $ | (II) $ \frac{4}{3} $ | | (C) $ E(X) $ | (III) $ \frac{1}{2} $ | | (D) $ P(1 \leq X \leq 2) $ | (IV) $ \frac{1}{6} $ |

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

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

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

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

✅ Correct Option: 2

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