CUET · MATHS · PYQ PAPER 2025
A random variable y has the following probability distribution
| y | 1 | 2 | 3 | 4 | 5 |
| P(y) | 2k | 3k | k | 4k | 5k |
| List - I | List - II |
| (A) P(y > 2) | (I) \(\frac{2}{5}\) |
| (B) k | (II) \(\frac{2}{3}\) |
| (C) \(P(y \leq 3)\) | (II) \(\frac{8}{15}\) |
| (D) \(P(2 \leq y \leq 4)\) | (II) \(\frac{1}{15}\) |
- A (А)-(III), (В)-(IV), (C)-(I), (D)-(II)
- B (А)-(II), (B)-(I), (C)-(IV), (D)-(III)
- C (А)-(II), (B)-(IV), (C)-(I), (D)-(III)
- D (A)-(IV), (B)-(III), (С)-(II), (D)-(I)
Answer & Solution
Correct Answer
(C) (А)-(II), (B)-(IV), (C)-(I), (D)-(III)
Step-by-step Solution
Detailed explanation
\( \sum P(y) = 1 \) \( 2k + 3k + k + 4k + 5k = 1 \) \( 15k = 1 \) k = \( \frac{1}{15} \) (B) → (IV) \( P(y > 2) = P(y=3) + P(y=4) + P(y=5) = k + 4k + 5k = 10k \) \( P(y > 2) = 10 \times \frac{1}{15} = \frac{10}{15} = \frac{2}{3} \) (A) → (II)…
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