If the random variable x has the following distribution:
|
$X$ |
0 |
1 |
2 |
otherwise |
|
$P(x)$ |
$k$ |
$k$ |
$2k$ |
0 |
Match List-I with List-II
|
List-I |
List-II |
|
(A) $k$ |
(I) $\frac{3}{4}$ |
|
(B) $P(x≥2)$ |
(II) $\frac{1}{4}$ |
|
(C) $P(X ≤2)$ |
(III) $\frac{1}{2}$ |
|
(D) $P(0 < x ≤ 2)$ |
(IV) 1 |
Choose the correct answer from the options given below.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I) **
|
List-I |
List-II |
|
(A) $k$ |
(II) $\frac{1}{4}$ |
|
(B) $P(x≥2)$ |
(III) $\frac{1}{2}$ |
|
(C) $P(X ≤2)$ |
(IV) 1 |
|
(D) $P(0 < x ≤ 2)$ |
(I) $\frac{3}{4}$ |
Given distribution:
X : 0, 1, 2
P(X) : k, k, 2k
Total probability = 1:
$k + k + 2k = 4k = 1 \Rightarrow k = \frac{1}{4}$
Now compute:
$P(x \ge 2) = P(2) = 2k = \frac{1}{2}$
$P(X \le 2) = P(0) + P(1) + P(2) = 1$
$P(0 < x \le 2) = P(1) + P(2) = k + 2k = 3k = \frac{3}{4}$
Final answer: (A)–(II), (B)–(III), (C)–(IV), (D)–(I)