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(C) The total number of ways to draw $2$ cards from $52$ is $^52C_2 = \frac{52 	imes 51}{2} = 1326$. Let $X$ be the number of queens. $X$ can take va

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$\frac{176}{221}$

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(C) The total number of ways to draw $2$ cards from $52$ is $^52C_2 = \frac{52 	imes 51}{2} = 1326$. Let $X$ be the number of queens. $X$ can take values $0, 1, 2$. $P(X=0) = \frac{^{48}C_2}{^{52}C_2} = \frac{48 	imes 47}{52 	imes 51} = \frac{1128}{1326} = \frac{188}{221}$. $P(X=1) = \frac{^4C_1 	imes ^{48}C_1}{^{52}C_2} = \frac{4 	imes 48}{1326} = \frac{192}{1326} = \frac{32}{221}$. $P(X=2) = \frac{^4C_2}{^{52}C_2} = \frac{6}{1326} = \frac{1}{221}$. $E(X) = 0 	imes P(X=0) + 1 	imes P(X=1) + 2 	imes P(X=2) = 0 + \frac{32}{221} + \frac{2}{221} = \frac{34}{221}$. $E(X^2) = 0^2 	imes P(X=0) + 1^2 	imes P(X=1) + 2^2 	imes P(X=2) = 0 + \frac{32}{221} + \frac{4}{221} = \frac{36}{221}$. Now,$2 E(X) + 3 E(X^2) = 2 \left(\frac{34}{221}ight) + 3 \left(\frac{36}{221}ight) = \frac{68 + 108}{221} = \frac{176}{221}$.

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X = x)$	$0.1$	$0.15$	$0.3$	$0.25$	$k$	$k$

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$0.1$	$0.2$	$0.3$	$0.4$

Two cards are drawn simultaneously from a well-shuffled pack of $52$ cards. If $X$ is the random variable representing the number of queens obtained,then the value of $2 E(X) + 3 E(X^2)$ is:

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