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Question

(D) Let $X$ be the random variable representing the number of kings drawn in $2$ trials with replacement. This follows a binomial distribution $B(n, p

Vedclass · Accepted Answer

$\frac{2}{13}$

Vedclass · Answer

(D) Let $X$ be the random variable representing the number of kings drawn in $2$ trials with replacement. This follows a binomial distribution $B(n, p)$ where $n=2$. The probability of drawing a king in a single draw is $p = \frac{4}{52} = \frac{1}{13}$. The probability of not drawing a king is $q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13}$. The mean of a binomial distribution is given by $E(X) = np$. Therefore,the mean is $2 	imes \frac{1}{13} = \frac{2}{13}$.

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

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