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(D) The total number of cards is $52$. The number of kings in the deck is $4$. Since the cards are drawn with replacement,the probability of drawing a

Vedclass · Accepted Answer

$\frac{28}{169}$

Vedclass · Answer

(D) The total number of cards is $52$. The number of kings in the deck is $4$. Since the cards are drawn with replacement,the probability of drawing a king in a single trial is $p = \frac{4}{52} = \frac{1}{13}$. The probability of not drawing a king is $q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13}$. Let $X$ be the random variable representing the number of kings in $n = 2$ trials. $X$ follows a binomial distribution $B(n, p) = B(2, \frac{1}{13})$. The probability distribution of $X$ is: $P(X=0) = \binom{2}{0} p^0 q^2 = 1 	imes 1 	imes (\frac{12}{13})^2 = \frac{144}{169}$ $P(X=1) = \binom{2}{1} p^1 q^1 = 2 	imes \frac{1}{13} 	imes \frac{12}{13} = \frac{24}{169}$ $P(X=2) = \binom{2}{2} p^2 q^0 = 1 	imes (\frac{1}{13})^2 	imes 1 = \frac{1}{169}$ We need to find $E(X^2) = \sum x^2 P(X=x)$. $E(X^2) = (0^2 	imes \frac{144}{169}) + (1^2 	imes \frac{24}{169}) + (2^2 	imes \frac{1}{169})$ $E(X^2) = 0 + \frac{24}{169} + \frac{4}{169} = \frac{28}{169}$.

Two cards are drawn successively with replacement from a fair deck of $52$ cards. Let $X$ denote the number of kings obtained when two cards are drawn. Then $E(X^2) = $

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