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) | \(\displaystyle\frac{5}{6}\) |
| (B) | \(P(X \lt 2)\) | (II) | \(\displaystyle\frac{4}{3}\) |
| (C) | E(X) | (III) | \(\displaystyle\frac{1}{2}\) |
| (D) | \(P(1 \le X \le 2)\) | (IV) | \(\displaystyle\frac{1}{6}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (IV), (B) - (III), (C) - (II), (D) - (I)
(A) k
This matches (IV) \(\frac{1}{6}\).
The probabilities add up to 1:
\(k + 2k + 3k = 1\)
\(6k = 1\)
\( \displaystyle k = \frac{1}{6} \)
(B) \(P(X \lt 2)\)
This matches (III) \(\frac{1}{2}\).
X < 2 means X = 0 or 1:
\( \displaystyle P(X \lt 2) = k + 2k = 3k = \frac{3}{6} = \frac{1}{2} \)
(C) E(X)
This matches (II) \(\frac{4}{3}\).
The mean is \(E(X) = \sum x\,P(X = x)\):
\(E(X) = 0(k) + 1(2k) + 2(3k) = 8k\)
\( \displaystyle = \frac{8}{6} = \frac{4}{3} \)
(D) \(P(1 \le X \le 2)\)
This matches (I) \(\frac{5}{6}\).
\( \displaystyle P(1 \le X \le 2) = 2k + 3k = 5k = \frac{5}{6} \)
So the matching is (A) - (IV), (B) - (III), (C) - (II), (D) - (I).