The probability distribution of a random variable X is
|
$X$ |
0 |
1 |
2 |
3 |
4 |
|
$P(X)$ |
0.2 |
$k$ |
$k$ |
$2k$ |
$k$ |
Match List-I with List-II
|
List-I |
List-II |
|
(A) value of $k$ |
(I) $\frac{16}{25}$ |
|
(B) $P(x≥2)$ |
(II) $\frac{9}{25}$ |
|
(C) $P(x=3)$ |
(III) $\frac{4}{25}$ |
|
(D) $P(x <2)$ |
(IV) $\frac{8}{25}$ |
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)-(IV), (D)-(II) **
|
List-I |
List-II |
|
(A) value of $k$ |
(III) $\frac{4}{25}$ |
|
(B) $P(x≥2)$ |
(I) $\frac{16}{25}$ |
|
(C) $P(x=3)$ |
(IV) $\frac{8}{25}$ |
|
(D) $P(x <2)$ |
(II) $\frac{9}{25}$ |
Given probability distribution:
Total probability = 1
$0.2 + k + k + 2k + k = 1$
$0.2 + 5k = 1 \Rightarrow 5k = 0.8 \Rightarrow k = 0.16$
(A) value of k = 0.16 = $\frac{4}{25}$ → (III)
(B) $P(x \ge 2) = P(2)+P(3)+P(4)=k+2k+k=4k=4(0.16)=0.64=\frac{16}{25}$ → (I)
(C) $P(x=3)=2k=0.32=\frac{8}{25}$ → (IV)
(D) $P(x<2)=P(0)+P(1)=0.2+k=0.36=\frac{9}{25}$ → (II)
Matching:
(A) → (III), (B) → (I), (C) → (IV), (D) → (II)