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(C) Let $X$ denote the number of aces obtained. Therefore,$X$ can take any of the values $0, 1,$ or $2$.In a deck of $52$ cards,there are $4$ aces and

Vedclass · Accepted Answer

$\frac{2}{13}$

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(C) Let $X$ denote the number of aces obtained. Therefore,$X$ can take any of the values $0, 1,$ or $2$.In a deck of $52$ cards,there are $4$ aces and $48$ non-ace cards.$P(X=0) = \frac{^{4}C_{0} 	imes ^{48}C_{2}}{^{52}C_{2}} = \frac{1 	imes 1128}{1326} = \frac{1128}{1326}$$P(X=1) = \frac{^{4}C_{1} 	imes ^{48}C_{1}}{^{52}C_{2}} = \frac{4 	imes 48}{1326} = \frac{192}{1326}$$P(X=2) = \frac{^{4}C_{2} 	imes ^{48}C_{0}}{^{52}C_{2}} = \frac{6 	imes 1}{1326} = \frac{6}{1326}$The probability distribution is:$x$$P(X)$$0$$\frac{1128}{1326}$$1$$\frac{192}{1326}$$2$$\frac{6}{1326}$$E(X) = \sum x_i p_i = (0 	imes \frac{1128}{1326}) + (1 	imes \frac{192}{1326}) + (2 	imes \frac{6}{1326})$$E(X) = \frac{192 + 12}{1326} = \frac{204}{1326} = \frac{2}{13}$Therefore,the correct answer is $C$.

$x$	$0$	$1$	$2$	$3$
$P(x)$	$0.2$	$k$	$k$	$2k$

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$3k$	$3k^2$	$k^2$	$2k^2$	$7k^2+k$

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

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