If the random variable X has the following probability distribution:
|
X |
0 |
1 |
2 |
otherwise |
|
P(X) |
k |
3k |
5k |
0 |
Match List-I with List-II
|
List-I |
List-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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
|
List-I |
List-II |
|
(A) $k$ |
(IV) $\frac{1}{9}$ |
|
(B) $E (X)$ |
(I) $\frac{13}{9}$ |
|
(C) $P (X ≤ 1)$ |
(II) $\frac{4}{9}$ |
|
(D) $P (1 ≤ X ≤ 2)$ |
(III) $\frac{8}{9}$ |
Given probability distribution:
$X: 0, 1, 2$
$P(X): k, 3k, 5k$
Sum of probabilities = 1:
$k + 3k + 5k = 9k = 1 \Rightarrow k = \frac{1}{9}$
(A) $k = \frac{1}{9} \Rightarrow$ (IV)
Expectation: $E(X) = 0*k + 1*3k + 2*5k = 3k + 10k = 13k = 13*\frac{1}{9} = \frac{13}{9}$ → (I)
$P(X \le 1) = P(0) + P(1) = k + 3k = 4k = \frac{4}{9}$ → (II)
$P(1 \le X \le 2) = P(1) + P(2) = 3k + 5k = 8k = \frac{8}{9}$ → (III)