CUET · MATHS · PYQ PAPER 2025
If a random variable \(X\) has the following probability distribution :
| X | 0 | 1 | 2 | 3 | 4 |
| P(X) | k | 2k | 3k | \(k^2\) | \(6 k^2\) |
| List-I | List-II |
| (A) k | (I) 3/7 |
| (B) \(P(x<2)\) | (II) \(6 / 49\) |
| (C) \(P(X>3)\) | (III) \(1 / 7\) |
| (D) \(P(2 \leq X \leq 3)\) | (IV) 22/49 |
- A (А) - (Ι), (Β) - (II), (С) - (III), (D) - (IV)
- B (А) - (III), (В) - (I), (C) - (IV), (D) - (II)
- C (А) - (Ι), (Β) - (II), (C) - (IV), (D) - (III)
- D (А) - (III), (В) - (I), (C) - (II), (D) - (IV)
Answer & Solution
Correct Answer
(D) (А) - (III), (В) - (I), (C) - (II), (D) - (IV)
Step-by-step Solution
Detailed explanation
\( \sum P(X) = 1 \) \( k + 2k + 3k + k^2 + 6k^2 = 1 \) \( 7k^2 + 6k - 1 = 0 \) \( (7k - 1)(k + 1) = 0 \) \( k = 1/7 \quad (\text{since } k \ge 0) \) (A) \( k = 1/7 \rightarrow \text{(III)} \) (B) \( P(X(C)…
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