Question

Two cards are drawn with replacement from a well shuffled deck of $52$ cards. Find the mean and variance for the number of aces.

Answer 1

Let $X$ denote the number of aces drawn in drawing two cards with replacement.
$\therefore X$ can take the values $0,1$ and $2$.
Let $P\left(S\right)=P$ (getting aces) $=\frac{4}{52}\Rightarrow P\left(F\right)$
$=P$ (not getting aces) $=\frac{48}{52}$
$P(X=0)=P\left(noace\right)=P\left(FF\right)=P\left(F\right)$
$P\left(F\right)=\frac{48}{52}\cdot \frac{48}{52}=\frac{144}{169}$
$P(X=1)=P$ (one ace) $=P\left(SForFS\right)=P\left(S\right).P\left(F\right)+P\left(F\right)P\left(S\right)$
$=2\left(PS\right).P\left(F\right)=2\times \frac{4}{52}\times \frac{48}{52}=\frac{24}{169}$
$P(X-2)=P$ (two aces) $=P\left(SS\right)=P\left(S\right)$.
$P\left(S\right)=\frac{4}{52}\cdot \frac{4}{52}=\frac{1}{169}$

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

Mean $=E\left[X\right]=\sum _{-\infty }^{\infty }{x}_i{p}_i=\left(0\right)\left(\frac{144}{169}\right)+\left(1\right)\left(\frac{24}{169}\right)+\left(2\right)\left(\frac{1}{169}\right)$
$=\frac{24+2}{169}=\frac{26}{169}=\frac{2}{13}$
$E\left(X^2\right)=\sum _{x=0}^2{x}_i^2{p}_i=\left(0^2\right)\left(\frac{144}{169}\right)+\left(1^2\right)\left(\frac{24}{169}\right)+\left(2^2\right)\left(\frac{1}{169}\right)$
$=\frac{24+4}{169}=\frac{28}{169}$
Variance $\left(X\right)=E\left[X^2\right]-\left(E\right(X){)}^2$
$=\frac{28}{169}-{\left(\frac{2}{13}\right)}^2=\frac{28}{169}-\frac{4}{169}=\frac{24}{169}$