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(A) Let $X$ denote the number of kings in a draw of two cards. $X$ is a random variable which can assume the values $0, 1, 2$.$P(X=0) = \frac{^{48}C_2

Vedclass · Accepted Answer

Mean = $2/13$,Variance = $400/2873$,Standard Deviation = $0.373$

Vedclass · Answer

(A) Let $X$ denote the number of kings in a draw of two cards. $X$ is a random variable which can assume the values $0, 1, 2$.$P(X=0) = \frac{^{48}C_2}{^{52}C_2} = \frac{48 	imes 47}{52 	imes 51} = \frac{188}{221}$$P(X=1) = \frac{^4C_1 	imes ^{48}C_1}{^{52}C_2} = \frac{4 	imes 48}{1326} = \frac{32}{221}$$P(X=2) = \frac{^4C_2}{^{52}C_2} = \frac{6}{1326} = \frac{1}{221}$Probability distribution table:$X$$0$$1$$2$$P(X)$$\frac{188}{221}$$\frac{32}{221}$$\frac{1}{221}$Mean $E(X) = \sum x_i P(x_i) = 0 	imes \frac{188}{221} + 1 	imes \frac{32}{221} + 2 	imes \frac{1}{221} = \frac{34}{221} = \frac{2}{13} \approx 0.1538$$E(X^2) = \sum x_i^2 P(x_i) = 0^2 	imes \frac{188}{221} + 1^2 	imes \frac{32}{221} + 2^2 	imes \frac{1}{221} = \frac{36}{221}$Variance $Var(X) = E(X^2) - [E(X)]^2 = \frac{36}{221} - (\frac{34}{221})^2 = \frac{7956 - 1156}{48841} = \frac{6800}{48841} \approx 0.1392$Standard Deviation $\sigma = \sqrt{Var(X)} = \sqrt{0.1392} \approx 0.373$

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

Two cards are drawn simultaneously (or successively without replacement) from a well-shuffled pack of $52$ cards. Find the mean,variance,and standard deviation of the number of kings.

Similar Questions

The distribution of a random variable $X$ is given below. The value of $k$ is:
$X = x$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(X = x)$ $\frac{1}{10}$ $k$ $\frac{1}{5}$ $2k$ $\frac{3}{10}$ $k$

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