A random variable \(y\) has the following probability distribution
| \(y\) | 1 | 2 | 3 | 4 | 5 |
| \(P(y)\) | \(2k\) | \(3k\) | \(k\) | \(4k\) | \(5k\) |
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(P(y \gt 2)\) | (I) | \(2/5\) |
| (B) | \(k\) | (II) | \(2/3\) |
| (C) | \(P(y \le 3)\) | (III) | \(8/15\) |
| (D) | \(P(2 \le y \le 4)\) | (IV) | \(1/15\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
The probabilities must add up to 1:
\(2k + 3k + k + 4k + 5k = 1\)
\( \displaystyle 15k = 1,\quad k = \frac{1}{15} \)
(A) \(P(y \gt 2)\)
\( \displaystyle P(3) + P(4) + P(5) = k + 4k + 5k = 10k = \frac{10}{15} = \frac{2}{3} \)
This matches (II).
(B) \(k\)
\( \displaystyle k = \frac{1}{15} \)
This matches (IV).
(C) \(P(y \le 3)\)
\( \displaystyle P(1) + P(2) + P(3) = 2k + 3k + k = 6k = \frac{6}{15} = \frac{2}{5} \)
This matches (I).
(D) \(P(2 \le y \le 4)\)
\( \displaystyle P(2) + P(3) + P(4) = 3k + k + 4k = 8k = \frac{8}{15} \)
This matches (III).
So the matching is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).