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Question

(A) Let $X$ denote the number of queens obtained in two draws with replacement.The probability of drawing a queen in a single draw is $p = \frac{4}{52

Vedclass · Accepted Answer

$X = x$$0$$1$$2$$P(X = x)$$\frac{144}{169}$$\frac{24}{169}$$\frac{1}{169}$

Vedclass · Answer

(A) Let $X$ denote the number of queens obtained in two draws with replacement.The probability of drawing a queen in a single draw is $p = \frac{4}{52} = \frac{1}{13}$.The probability of not drawing a queen is $q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13}$.Since the draws are with replacement,the trials are independent and follow a binomial distribution $B(n, p)$ with $n = 2$ and $p = \frac{1}{13}$.$P(X = 0) = ^2C_0 	imes (\frac{12}{13})^2 = 1 	imes \frac{144}{169} = \frac{144}{169}$.$P(X = 1) = ^2C_1 	imes (\frac{1}{13})^1 	imes (\frac{12}{13})^1 = 2 	imes \frac{12}{169} = \frac{24}{169}$.$P(X = 2) = ^2C_2 	imes (\frac{1}{13})^2 = 1 	imes \frac{1}{169} = \frac{1}{169}$.Thus,the probability distribution is as given in option $A$.

Two cards are drawn successively with replacement from a well-shuffled pack of $52$ cards. Find the probability distribution of the number of queens.

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$X = x$	$0$	$1$	$2$
$P(X = x)$	$\frac{144}{169}$	$\frac{24}{169}$	$\frac{1}{169}$