CUET · MATHS · PYQ PAPER 2025
"The probability distribution of a random variable \(x\) is :
\(P(X=x)=\left\{\begin{array}{ll}k x^2, & \text { for } x=1,2,3 \\2 k x, & \text { for } x=4,5,6 \text { where } k \text { is a constant. } \\0, & \text { otherwise }\end{array}\right.\)
Match List-I with List-II
| List I | List II |
| (A) k | (I) 7/22 |
| (B) \(P(X \geq 4)\) | (II) 1/44 |
| (C) \(P(X<4)\) | (III) 95/22 |
| (D) E[X] | (IV) 15/22 |
- A (A) - (I), (B) - (II), (C) - (IV), (D) - (III)
- B (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
- C (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
- D (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
Answer & Solution
Correct Answer
(C) (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
Step-by-step Solution
Detailed explanation
\( \sum P(X=x) = 1 \) \( k(1^2) + k(2^2) + k(3^2) + 2k(4) + 2k(5) + 2k(6) = 1 \) \( k(1+4+9) + k(8+10+12) = 1 \) \( 14k + 30k = 1 \) \( 44k = 1 \) \( k = 1/44 \) (A) - (II) \( P(X \geq 4) = P(X=4) + P(X=5) + P(X=6) = 2k(4) + 2k(5) + 2k(6) \)…
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