Question

Two cards are drawn simultaneously from a pack of 52 cards. Compute the mean and standard deviation of the number of kings.

Answer 1

Let X denote the number of kings in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,

\[P\left( X = 0 \right)\]
\[=P\left( \text{ no king }\right)\]
\[=\frac{{}^{48} C_2}{{}^{52} C_2}\]
\[=\frac{1128}{1326}\]
\[=\frac{188}{221}\]
\[P\left( X = 1 \right)\]
\[=P\left( 1 \text{ king } \right)\]
\[=\frac{{}^4 C_1 \times^{48} C_1}{{}^{52} C_2}\]
\[=\frac{192}{1326}\]
\[=\frac{32}{221}\]
\[P\left( X = 2 \right)\]
\[=P\left( 2 \text{ kings } \right)\]
\[=\frac{{}^4 C_2}{{}^{52} C_2}\]
\[=\frac{6}{1326}\]
\[=\frac{1}{221}\]Thus, the probability distribution of X is given by

x	P(X)
0	\[\frac{188}{221}\]
1	\[\frac{32}{221}\]
2	\[\frac{1}{221}\]

Computation of mean and variance

xi	pi	pixi	pixi2
0	\[\frac{188}{221}\]	0	0
1	\[\frac{32}{221}\]	\[\frac{32}{221}\]	\[\frac{32}{221}\]
2	\[\frac{1}{221}\]	\[\frac{2}{221}\]	\[\frac{4}{221}\]
		`∑`pixi = \[\frac{34}{221}\]	`∑`pixi2= \[\frac{36}{221}\]

\[\text{ Mean}  = \sum p_i x_i = \frac{34}{221}\]
\[\text{Variance} = \sum p_i {x_i}2^{}_{} - \left( \text{ Mean } \right)^2 \]
\[ = \frac{36}{221} - \left( \frac{34}{221} \right)^2 \]
\[ = \frac{7956 - 1156}{48841}\]
\[ = \frac{6800}{48841}\]
\[ = \frac{400}{2873}\]