Chapter 12.4, Problem 3CYU

To determine

To calculate: the probability that 5 Cards contain 3 of one suit and two of another suit.

Answer to Problem 3CYU

The required probability is =0.1030

Explanation of Solution

Given information:

Justice draws 5 cards from a standard deck of 52 cards

Calculation:

Since there are total 52 cards so number of possible ways of selecting 5 cards out of 52 cards is,

  $\begin{array}{l}C(52,5)=\frac{52!}{5!47!}\\ =2598960\end{array}$

There are 4 suits and each suit has 13 cards. Number of ways one suit out of 4 and then 3 cards from the selected suit is,

  $\begin{array}{l}C(4,1)C(13,3)=4\cdot \frac{13!}{3!10!}\\ =1144\end{array}$

Number of ways one suit out of 3 and then 2 cards from the selected suit is,

  $\begin{array}{l}C(3,1)C(13,2)=3\cdot \frac{13!}{2!!1!}\\ =234\end{array}$

So number of ways of selecting 5 cards, 3 from one suit and 2 from other is,

  $1144\cdot 234=267696$

So, the probability that those 5 cards contain 3 of one suit and two of another suit is,

  $\begin{array}{cc}P(3\text{ of one suit and }2\text{ from other})=& \frac{267696}{2598960}\\ & =0.1030\end{array}$

Hence, required probability is =0.1030