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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

Vedclass · Accepted Answer

$\frac{176}{221}$

Vedclass · Answer

(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,$2 E(X) + 3 E(X^2) = 2 \left(\frac{34}{221}ight) + 3 \left(\frac{36}{221}ight) = \frac{68 + 108}{221} = \frac{176}{221}$.

$X$	$4k$	$\frac{30}{7}k$	$\frac{32}{7}k$	$\frac{34}{7}k$	$\frac{36}{7}k$	$\frac{38}{7}k$	$\frac{40}{7}k$	$6k$
$P(X)$	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$

Outcome	$\omega_{1}$	$\omega_{2}$	$\omega_{3}$	$\omega_{4}$	$\omega_{5}$	$\omega_{6}$	$\omega_{7}$
Probability	$-0.1$	$0.2$	$0.3$	$0.4$	$-0.2$	$0.1$	$0.3$

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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