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(D) The probability of drawing a king card in a single draw is $p = \frac{4}{52} = \frac{1}{13}$.The probability of not drawing a king is $q = 1 - \fr

Vedclass · Accepted Answer

$\frac{25}{169}$

Vedclass · Answer

(D) The probability of drawing a king card in a single draw is $p = \frac{4}{52} = \frac{1}{13}$.The probability of not drawing a king is $q = 1 - \frac{1}{13} = \frac{12}{13}$.Since the cards are drawn with replacement,this follows a binomial distribution with $n = 2$ and $p = \frac{1}{13}$.$P(X=k) = \binom{n}{k} p^k q^{n-k}$.For $X=1$: $P(X=1) = \binom{2}{1} \left(\frac{1}{13}ight)^1 \left(\frac{12}{13}ight)^1 = 2 	imes \frac{12}{169} = \frac{24}{169}$.For $X=2$: $P(X=2) = \binom{2}{2} \left(\frac{1}{13}ight)^2 \left(\frac{12}{13}ight)^0 = 1 	imes \frac{1}{169} = \frac{1}{169}$.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 pack of $52$ cards. Let $X$ denote the random variable of the number of kings obtained in the two drawn cards. Then $P(X=1) + P(X=2)$ equals:

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