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,

\[P\left( X = 0 \right) = P\left( \text{ no ace }  \right) = \frac{{}^{48} C_4}{{}^{52} C_4}\]
\[P\left( X = 1 \right) = P\left( 1 \text{ ace }  \right) = \frac{{}^4 C_1 \times^{48} C_3}{{}^{52} C_4}\]
\[P\left( X = 2 \right) = P\left( 2 \text{ aces}  \right) = \frac{{}^4 C_2 \times^{48} C_2}{{}^{52} C_4}\]
\[P\left( X = 3 \right) = P\left( 3 \text{ aces } \right) = \frac{{}^4 C_3 \times^{48} C_1}{{}^{52} C_4}\]
\[P\left( X = 4 \right) = P\left( 4 \text{ aces } \right) = \frac{{}^4 C_4}{{}^{52} C_4}\]

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	\[\frac{{}^4 C_4}{{}^{52} C_4}\]