Chapter 9.7, Problem 46E

To determine

To find: the probability that the hand drawn is a full house.

Answer to Problem 46E

  $0.00144$

Explanation of Solution

Given:

A full house consists of three of one kind and two of another.

Let E be the event that the hand drawn is a full house.

Total choices for the number that shows on the first three cards is = $C{}_1$

Total number of choices of suits for those three cards is = $C{}_3$

Total choices for the number that shows on the remaining two cards is = $C{}_1$

Total number of choices of suits for those two cards is = $C{}_2$

So,

  $\begin{array}{c}n\left(E\right)=C{}_1\times C{}_3\times C{}_1\times C{}_2\\ =3744\end{array}$

Total number of cards is 52. So, total number of ways to choose 5 cards out of 52 cards is:

  $\begin{array}{c}n\left(S\right)=C{}_5\\ =\frac{\text{52!}}{\text{5!}\left(52-5\right)\text{!}}\\ =\frac{\text{52!}}{\text{5!47!}}\\ =2,598,960\end{array}$

Thus, the probability of that the hand drawn is a full house is:

  $\begin{array}{c}P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}\\ =\frac{3,744}{2,598,960}\\ =0.00144\end{array}$