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Question

(C) Let $X$ be the random variable representing the number of kings drawn.Since the cards are drawn with replacement,the trials are independent.Total

Vedclass · Accepted Answer

$X$$0$$1$$2$$P(X)$$\frac{144}{169}$$\frac{24}{169}$$\frac{1}{169}$

Vedclass · Answer

(C) Let $X$ be the random variable representing the number of kings drawn.Since the cards are drawn with replacement,the trials are independent.Total number of cards = $52$.Number of kings = $4$.Probability of drawing a king in a single trial,$p = \frac{4}{52} = \frac{1}{13}$.Probability of not drawing a king in a single trial,$q = 1 - p = 1 - \frac{1}{13} = \frac{12}{13}$.Since there are $n = 2$ trials,$X$ follows a binomial distribution $B(n, p) = B(2, \frac{1}{13})$.The probability distribution is given by $P(X = k) = {}^nC_k p^k q^{n-k}$.For $X = 0$: $P(X = 0) = {}^2C_0 (\frac{1}{13})^0 (\frac{12}{13})^2 = 1 \cdot 1 \cdot \frac{144}{169} = \frac{144}{169}$.For $X = 1$: $P(X = 1) = {}^2C_1 (\frac{1}{13})^1 (\frac{12}{13})^1 = 2 \cdot \frac{1}{13} \cdot \frac{12}{13} = \frac{24}{169}$.For $X = 2$: $P(X = 2) = {}^2C_2 (\frac{1}{13})^2 (\frac{12}{13})^0 = 1 \cdot \frac{1}{169} \cdot 1 = \frac{1}{169}$.Thus,the probability distribution is:$X=0, P(X)=\frac{144}{169}$$X=1, P(X)=\frac{24}{169}$$X=2, P(X)=\frac{1}{169}$Therefore,option $(C)$ is correct.

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

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