Suppose that two cards are drawn at random from a deck of cards. Let $X$ be the number of aces obtained. Then the value of $E(X)$ is

(A) $\cfrac{{37}}{{221}}$

(B) $\cfrac{5}{{13}}$

(C) $\cfrac{1}{{13}}$

(D) $\cfrac{2}{{13}}$

Option D is correct

$X$ denotes the random variable which assumes the number of aces obtained

$\therefore$ $X$ can assume values 0, 1 and 2.

$\therefore$ $P(X = 0) = P({\rm{ No ace }}) = \cfrac{{^{48}{C_2}}}{{^{52}{C_2}}} = \cfrac{{48 \times 47}}{{52 \times 51}} = \cfrac{{24 \times 47}}{{26 \times 51}} = \cfrac{{1128}}{{1326}}$

$P(X = 1) = P({\rm{ No ace }}) = 2\cfrac{{{{\rm{ }}^4}{C_1}{ \times ^{48}}{C_1}}}{{^{52}{C_2}}} = \cfrac{{48 \times 4}}{{26 \times 51}} = \cfrac{{192}}{{1326}}$

$P(X = 2) = P({\rm{ No ace }}) = \cfrac{{^4{C_2}}}{{^{52}{C_2}}} = \cfrac{{4 \times 3}}{{52 \times 51}} = \cfrac{6}{{1326}}$

Hence, the probability distribution is :

$\therefore$ $E(X) = \sum X P(X) = (0)\left( {\cfrac{{1128}}{{1326}}} \right) + (1)\left( {\cfrac{{192}}{{1326}}} \right) + (2)\left( {\cfrac{6}{{1326}}} \right)$

$= \cfrac{{192 + 12}}{{1326}} = \cfrac{{204}}{{1326}} = \cfrac{2}{{13}}$

Exercise-13.5