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