CUET · MATHS · PYQ PAPER 2025
If the random varible 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) \(\frac{13}{9}\) |
| (B) E(X) | (II) \(\frac{4}{9}\) |
| (C) P(X ≤ 1) | (III) \(\frac{8}{9}\) |
| (D) P(1 ≤ X ≤ 2) | (IV) \(\frac{1}{9}\) |
- A (А) - (II), (В) - (I), (C) - (IV), (D) - (III)
- B (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
- C (A) - (IV), (B) - (II), (C) - (I), (D) - (III)
- D (А) - (III), (В) - (II), (C) - (I), (D) - (IV)
Answer & Solution
Correct Answer
(B) (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
Step-by-step Solution
Detailed explanation
\(\sum P(X) = 1\) \(k + 3k + 5k + 0 = 1 \implies 9k = 1 \implies k = \frac{1}{9}\) So, (A) k matches with (IV) \(\frac{1}{9}\). \(E(X) = \sum X P(X)\) \(E(X) = (0 \cdot k) + (1 \cdot 3k) + (2 \cdot 5k) + (3 \cdot 0) = 13k\) \(E(X) = 13 \cdot \frac{1}{9} = \frac{13}{9}\) So, (B)…
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