Match List-I with List-II. The probability distribution of number of sixes(X) is when an unbiased die is thrown twice is :
| X | 0 | 1 | 2 |
| P(X) | 25/36 | 5/18 | 1/36 |
| List-I | List-II | ||
| (A) | The value of E(X) | (I) | $\frac{5}{72}$ |
| (B) | Value of Var(X) | (II) | $\frac{1}{8}$ |
| (C) | Value of Var($\frac{x}{2}$) | (III) | $\frac{1}{3}$ |
| (D) | Value of E$(\frac{3}{8}X)$ | (IV) | $\frac{5}{18}$ |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(A) $E(X)=∑XP(X)$
$=0×\frac{25}{36}+1×\frac{5}{18}+2×\frac{1}{18}$
$=\frac{12}{13}=\frac{1}{3}$
(B) $E(X^2)=0^2×\frac{25}{36}+1^2×\frac{5}{18}+2^2×\frac{1}{36}$
$=\frac{7}{18}$
$Var(X)=E(X^2)-[E(X)]^2$
$=\frac{7}{18}-\left(\frac{1}{3}\right)^2=\frac{5}{18}$
(C) $Var(aN)=a^2\,Var(N)$
$⇒Var(\frac{X}{2})=\left(\frac{1}{2}\right)^2Var(X)=\frac{1}{4}×\frac{5}{18}$
$=\frac{5}{72}$
(D) $E(\frac{3}{8}X)=\frac{3}{8}E(X)$
$=\frac{3}{8}×\frac{1}{3}=\frac{1}{8}$