ExamBro
ExamBro
CUET · MATHS · PYQ PAPER 2025

If the random varible X has the following probability distribution:
X012otherwise
P(X)K3K5K0
X = 0, P(X) = k; x = 1, P(x) = 3k; X = 2, P(x) = 5k; otherwise, P(X) = 0
Match List - I with List - II
List - IList - 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) \(\frac{1}{9}\)
Choose the correct answer from the options given below:

  1. A (А) - (II), (В) - (I), (C) - (IV), (D) - (III)
  2. B (A) - (IV), (B) - (I), (C) - (II), (D) - (III)
  3. C (A) - (IV), (B) - (II), (C) - (I), (D) - (III)
  4. D (А) - (III), (В) - (II), (C) - (I), (D) - (IV)
Verified Solution

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)…