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(C) Let $p$ be the probability of drawing a ten in a single draw. Since there are $4$ tens in a pack of $52$ cards,$p = \frac{4}{52} = \frac{1}{13}$.T

Vedclass · Accepted Answer

$\frac{2}{13}$

Vedclass · Answer

(C) Let $p$ be the probability of drawing a ten in a single draw. Since there are $4$ tens in a pack of $52$ cards,$p = \frac{4}{52} = \frac{1}{13}$.The probability of not drawing a ten is $q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13}$.Since the cards are drawn with replacement,the number of tens $X$ follows a binomial distribution $B(n, p)$ with $n = 2$ and $p = \frac{1}{13}$.The mean of a binomial distribution is given by $E(X) = np$.Therefore,the mean number of tens is $E(X) = 2 	imes \frac{1}{13} = \frac{2}{13}$.

Two cards are drawn successively with replacement from a well-shuffled pack of $52$ cards. The mean of the number of tens is:

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