Question

Four cards are drawn simultaneously from a well shuffled pack of 52 playing cards. Find the probability distribution of the number of aces.

Answer 1

Let X denote the number of aces in a sample of 4 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take values 0, 1, 2, 3 and 4.
Now,
$$\begin{gathered} P\left(X=0\right)=P\left(\mathrm{no}\mathrm{ace}\right)=\frac{{}^{48}{C}_4}{{}^{52}{C}_4} \\ P\left(X=1\right)=P\left(1\mathrm{ace}\right)=\frac{{}^4{C}_1{\times }^{48}{C}_3}{{}^{52}{C}_4} \\ P\left(X=2\right)=P\left(2\mathrm{aces}\right)=\frac{{}^4{C}_2{\times }^{48}{C}_2}{{}^{52}{C}_4} \\ P\left(X=3\right)=P\left(3\mathrm{aces}\right)=\frac{{}^4{C}_3{\times }^{48}{C}_1}{{}^{52}{C}_4} \\ P\left(X=4\right)=P\left(4\mathrm{aces}\right)=\frac{{}^4{C}_4}{{}^{52}{C}_4} \end{gathered}$$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{{}^{48}{C}_4}{{}^{52}{C}_4}$
1	$\frac{{}^4{C}_1{\times }^{48}{C}_3}{{}^{52}{C}_4}$
2	$\frac{{}^4{C}_2{\times }^{48}{C}_2}{{}^{52}{C}_4}$
3	$\frac{{}^4{C}_3{\times }^{48}{C}_1}{{}^{52}{C}_4}$
4	$\frac{{}^4{C}_4}{{}^{52}{C}_4}$