Question

Two cards drawn simultaneously (or successively without replacement) from a well shuffled pack of $52$ cards. Find the mean and variance of the numbers of kings.

Answer 1

Let $X$ represent the number of kings drawn.
Then $X=\{0,1,2\}$
Numbers of kings in a deck of $52$ cards $=4$

$X$	$0$	$1$	$2$
$P\left(X\right)$	$\frac{{}^{48}{C}_2{\times }^4{C}_0}{{}^{52}{C}_2}=\frac{188}{221}$	$\frac{{}^{48}{C}_1{\times }^4{C}_1}{{}^{52}{C}_2}=\frac{32}{221}$	$\frac{{}^{48}{C}_0{\times }^4{C}_2}{{}^{52}{C}_2}=\frac{1}{221}$
$XP\left(X\right)$	$0$	$\frac{32}{221}$	$\frac{2}{221}$
$X^{2}P\left(X\right)$	$0$	$\frac{32}{221}$	$\frac{4}{221}$

Mean, $E\left(X\right)=\sum XP\left(X\right)=0+\frac{32}{221}$+$\frac{2}{221}=\frac{2}{13}$
Variance $=\sum X^{2}P\left(X\right)-{(\sum XP(X\left)\right)}^2=0+\frac{32}{221}+\frac{4}{221}-\frac{4}{169}=\frac{400}{2873}$