For a random variable x, probability distribution P(x) is given by $P(x) = \frac{k}{6}(3-x), x = 0,1,2$, then
Match List-I with List-II
|
List-I |
List-II |
|
(A) k is equal to |
(I) $\frac{1}{2}$ |
|
(B) $P(x = 0)$ |
(II) 1 |
|
(C) $P(x <2)$ |
(III) $\frac{1}{6}$ |
|
(D) $P(1 < x ≤2)$ |
(IV) $\frac{5}{6}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
|
List-I |
List-II |
|
(A) k is equal to |
(II) 1 |
|
(B) $P(x = 0)$ |
(I) $\frac{1}{2}$ |
|
(C) $P(x <2)$ |
(IV) $\frac{5}{6}$ |
|
(D) $P(1 < x ≤2)$ |
(III) $\frac{1}{6}$ |
Given: $P(x) = \frac{k}{6}(3 - x)$ for $x = 0, 1, 2$
Use the total probability rule: $\sum P(x) = 1$
$P(0) = \frac{k}{6}(3 - 0) = \frac{3k}{6}$
$P(1) = \frac{k}{6}(3 - 1) = \frac{2k}{6}$
$P(2) = \frac{k}{6}(3 - 2) = \frac{k}{6}$
Total: $\frac{3k + 2k + k}{6} = \frac{6k}{6} = k$
Set $k = 1$ to satisfy total probability = 1
Now compute individual probabilities:
$P(x = 0) = \frac{3}{6} = \frac{1}{2}$
$P(x < 2) = P(0) + P(1) = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$
$P(1 < x \leq 2) = P(2) = \frac{1}{6}$