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(C) The random variable $X$ represents the number of aces obtained,which can take values $0, 1, 2$. Total number of ways to draw $2$ cards from $52$ i

Vedclass · Accepted Answer

$\frac{2}{13}$

Vedclass · Answer

(C) The random variable $X$ represents the number of aces obtained,which can take values $0, 1, 2$. Total number of ways to draw $2$ cards from $52$ is $^{52}C_2 = \frac{52 	imes 51}{2} = 1326$. Probability of getting $0$ aces: $P(X=0) = \frac{^{48}C_2}{^{52}C_2} = \frac{1128}{1326} = \frac{188}{221}$. Probability of getting $1$ ace: $P(X=1) = \frac{^{4}C_1 	imes ^{48}C_1}{^{52}C_2} = \frac{4 	imes 48}{1326} = \frac{192}{1326} = \frac{32}{221}$. Probability of getting $2$ aces: $P(X=2) = \frac{^{4}C_2}{^{52}C_2} = \frac{6}{1326} = \frac{1}{221}$. Expected value $E(X) = \sum x_i P(x_i) = (0 	imes \frac{188}{221}) + (1 	imes \frac{32}{221}) + (2 	imes \frac{1}{221}) = \frac{32+2}{221} = \frac{34}{221} = \frac{2}{13}$.

Let two cards be drawn at random from a pack of $52$ playing cards. Let $X$ be the number of aces obtained. Then the value of $E(X)$ is

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