If the random variable X has the following probability distribution:
| X | 0 | 1 | 2 | otherwise |
| P(X) | k | 3k | 5k | 0 |
Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(k\) | (I) | \(\displaystyle\frac{13}{9}\) |
| (B) | \(E(X)\) | (II) | \(\displaystyle\frac{4}{9}\) |
| (C) | \(P(X \le 1)\) | (III) | \(\displaystyle\frac{8}{9}\) |
| (D) | \(P(1 \le X \le 2)\) | (IV) | \(\displaystyle\frac{1}{9}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(A) \(k\)
This matches (IV). The probabilities add up to 1:
\(k + 3k + 5k = 1\)
\( \displaystyle 9k = 1,\quad k = \frac{1}{9} \)
(B) \(E(X)\)
This matches (I). The mean is \(E(X) = \sum x\,P(x)\):
\(E(X) = 0(k) + 1(3k) + 2(5k) = 13k\)
\( \displaystyle E(X) = \frac{13}{9} \)
(C) \(P(X \le 1)\)
This matches (II):
\( \displaystyle P(X \le 1) = P(0) + P(1) = k + 3k = \frac{4}{9} \)
(D) \(P(1 \le X \le 2)\)
This matches (III):
\( \displaystyle P(1 \le X \le 2) = P(1) + P(2) = 3k + 5k = \frac{8}{9} \)
So the matching is (A)-(IV), (B)-(I), (C)-(II), (D)-(III).