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

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$\frac{25}{169}$

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(D) The total number of cards is $52$,and the number of aces is $4$. Since the cards are drawn with replacement,the probability of drawing an ace in a single draw is $p = \frac{4}{52} = \frac{1}{13}$,and the probability of not drawing an ace is $q = 1 - \frac{1}{13} = \frac{12}{13}$. This follows a binomial distribution $B(n, p)$ where $n = 2$ and $p = \frac{1}{13}$. $P(X = 1) = \binom{2}{1} 	imes p^1 	imes q^1 = 2 	imes \frac{1}{13} 	imes \frac{12}{13} = \frac{24}{169}$. $P(X = 2) = \binom{2}{2} 	imes p^2 	imes q^0 = 1 	imes \left(\frac{1}{13}ight)^2 	imes 1 = \frac{1}{169}$. Therefore,$P(X = 1) + P(X = 2) = \frac{24}{169} + \frac{1}{169} = \frac{25}{169}$.

Two cards are drawn successively with replacement from a well-shuffled deck of $52$ cards. Let $X$ denote the random variable of the number of aces obtained in the two drawn cards. Then $P(X = 1) + P(X = 2)$ equals

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