Question

Two cards are drawn successively with replacement from a well shuffled deck of $52$ cards. Find the probability distribution of the number of aces.

Answer 1

Total number of cards in a deck $=52$

Number of ace cards $=4$

Let $X$: Number of ace cards

Possibilites of getting $2$ ace cards or $1$ ace card or $0$ ace card

Hence, $X=0,X=1\&X=2$

$P(X=0)=P\left(\text{non-ace and non-ace}\right)$

$P(X=0)=P\left(\text{non-ace}\right)\times P\left(\text{non-ace}\right)$

$P(X=0)=\frac{48}{52}\times \frac{48}{52}=\frac{144}{169}$

$P(X=1)=P\left(\text{ace and non-ace or non-ace and ace}\right)$

$P(X=1)=P\left(\text{ace and non-ace and non-ace}\right)+P\left(\text{non-ace and ace}\right)$

$P(X=1)=P\left(\text{ace}\right).P\left(\text{non-ace}\right)+P\left(\text{non-ace}\right).P\left(\text{ace}\right)$

$P(X=1)=\frac{4}{52}\times \frac{48}{52}+\frac{48}{52}\times \frac{4}{52}=\frac{24}{169}$

$P(X=2)=P\left(\text{ace and ace}\right)$

$P(X=2)=\frac{4}{52}\times \frac{4}{52}=\frac{1}{169}$

Thus, the required probability distribution is

$\begin{array}{cccc}X& 0& 1& 2\\ P\left(X\right)& \frac{144}{169}& \frac{24}{169}& \frac{1}{169}\end{array}$