A random variable y has the following probability distribution
|
y |
1 |
2 |
3 |
4 |
5 |
|
P(y) |
2k |
3k |
k |
4k |
5k |
Match List-I with List-II
|
List-I |
List-II |
|
(A) $P(y > 2)$ |
(I) 2/5 |
|
(B) $k$ |
(II) 2/3 |
|
(C) $P(y ≤3)$ |
(III) 8/15 |
|
(D) $P(2≤ y ≤ 4)$ |
(IV) 1/15 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
|
List-I |
List-II |
|
(A) $P(y > 2)$ |
(II) 2/3 |
|
(B) $k$ |
(IV) 1/15 |
|
(C) $P(y ≤3)$ |
(I) 2/5 |
|
(D) $P(2≤ y ≤ 4)$ |
(III) 8/15 |
Given probability distribution:
y: 1, 2, 3, 4, 5
P(y): 2k, 3k, k, 4k, 5k
Sum of probabilities = 1 ⇒ 2k + 3k + k + 4k + 5k = 15k = 1 ⇒ k = \frac{1}{15}
Compute required probabilities:
(A)$ P(y > 2) = P(3) + P(4) + P(5) = k + 4k + 5k = 10k = \frac{10}{15} = \frac{2}{3}$
(B) $k = \frac{1}{15}$
(C) $P(y ≤ 3) = P(1) + P(2) + P(3) = 2k + 3k + k = 6k = \frac{6}{15} = \frac{2}{5}$
(D) $P(2 ≤ y ≤ 4) = P(2) + P(3) + P(4) = 3k + k + 4k = 8k = \frac{8}{15}$