CUET Preparation Today
If a random variable X has the following probability distribution:
then, Match List-I with List-II
Choose the correct answer from the options given below: |
(A)-(I), (B)-(II), (C)-(III), (D)-(IV) (A)-(II), (B)-(III), (C)-(I), (D)-(IV) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (A)-(IV), (B)-(I), (C)-(III), (D)-(II) |
(A)-(III), (B)-(IV), (C)-(I), (D)-(II) |
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Given: Probability distribution of random variable $X$:
Total probability = 1: $K + \frac{K}{2} + \frac{K}{4} + \frac{K}{8} = 1$ LCM = 8: $\frac{8K + 4K + 2K + K}{8} = \frac{15K}{8} = 1$ $\Rightarrow K = \frac{8}{15}$ (A) The value of K ⟹ $\frac{8}{15}$ (B) $P(0 < X < 2)$ = $P(X = 1)$ = $\frac{K}{2} = \frac{8}{15} \cdot \frac{1}{2} = \frac{4}{15}$ (C) $P(1 < X < 3)$ = $P(X = 2)$ = $\frac{K}{4} = \frac{8}{15} \cdot \frac{1}{4} = \frac{2}{15}$ (D) $P(X > 2)$ = $P(X = 3)$ = $\frac{K}{8} = \frac{8}{15} \cdot \frac{1}{8} = \frac{1}{15}$ |
