If a random variable x has the following probability distribution:
|
X |
0 |
1 |
2 |
3 |
4 |
|
P(X) |
$k$ |
$2k$ |
$3k$ |
$k^2$ |
$6k^2$ |
, then
Match List-I with List-II
|
List-I |
List-II |
|
(A) $k$ |
(I) 3/7 |
|
(B) $P(X < 2)$ |
(II) 6/49 |
|
(C) $P(X >3)$ |
(III) 1/7 |
|
(D) $P(2 ≤ X ≤3)$ |
(IV) 22/49 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
|
List-I |
List-II |
|
(A) $k$ |
(III) 1/7 |
|
(B) $P(X < 2)$ |
(I) 3/7 |
|
(C) $P(X >3)$ |
(II) 6/49 |
|
(D) $P(2 ≤ X ≤3)$ |
(IV) 22/49 |
Given probability distribution:
$P(X=0) = k$, $P(X=1) = 2k$, $P(X=2) = 3k$, $P(X=3) = k^2$, $P(X=4) = 6k^2$
Sum of probabilities = 1:
$k + 2k + 3k + k^2 + 6k^2 = 6k + 7k^2 = 1$
Solve quadratic: $7k^2 + 6k - 1 = 0$
Factor: $(7k - 1)(k + 1) = 0 \Rightarrow k = 1/7$ (since probability must be positive)
Compute probabilities:
(A) $k = 1/7 \Rightarrow$ (III)
(B) $P(X<2) = P(X=0) + P(X=1) = k + 2k = 3k = 3/7 \Rightarrow$ (I)
(C) $P(X>3) = P(X=4) = 6k^2 = 6*(1/7)^2 = 6/49 \Rightarrow$ (II)
(D) $P(2 \le X \le 3) = P(X=2) + P(X=3) = 3k + k^2 = 3/7 + 1/49 = 22/49 \Rightarrow$ (IV)