The probability distribution of a random variable X is:
$P(X = x)=\left\{\begin{matrix}kx^2,&for\,x=1,2,3\\2kx,&for\,x=4,5,6\\0,&\text{otherwise}\end{matrix}\right.$, where k is a constant.
Match List-I with List-II
|
List-I |
List-II |
|
(A) $k$ |
(I) 7/22 |
|
(B) $P(X ≥4)$ |
(II) 1/44 |
|
(C) $P(X <4)$ |
(III) 95/22 |
|
(D) $E[X]$ |
(IV) 15/22 |
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) $k$ |
(II) 1/44 |
|
(B) $P(X ≥4)$ |
(IV) 15/22 |
|
(C) $P(X <4)$ |
(I) 7/22 |
|
(D) $E[X]$ |
(III) 95/22 |
Given
$P(X=x)=kx^2$ for $x=1,2,3$
$P(X=x)=2kx$ for $x=4,5,6$
Total probability $=1$
$k(1^2+2^2+3^2)+2k(4+5+6)=1$
$k(1+4+9)+2k(15)=1$
$14k+30k=1$
$44k=1$
$k=\frac{1}{44}$
So (A) → (II)
$P(X\ge4)=P(4)+P(5)+P(6)$
$=2k(4+5+6)=30k$
$=30\times\frac{1}{44}=\frac{15}{22}$
So (B) → (IV)
$P(X<4)=1-P(X\ge4)$
$=1-\frac{15}{22}=\frac{7}{22}$
So (C) → (I)
$E(X)=\sum xP(X=x)$
$=k(1^3+2^3+3^3)+2k(4^2+5^2+6^2)$
$=k(1+8+27)+2k(16+25+36)$
$=k(36)+2k(77)$
$=36k+154k=190k$
$=190\times\frac{1}{44}=\frac{95}{22}$
So (D) → (III)
Correct matching: A–II, B–IV, C–I, D–III