Chapter 9.7, Problem 44E

a.

To determine

To find: the probability that a hand will contain exactly two wild cards.

Answer to Problem 44E

  $0.076$

Explanation of Solution

Given:

The deck for a card game contains 108 cards. Twenty-five each are red, yellow, blue, and green, and eight are wild cards. Each player is randomly dealt a seven-card hand.

Let E be the event that you exactly two wild cards.

Since, there are 8 wild cards. So, total number of ways to choose 2 wild cards is:

  $C{}_2=\frac{8\text{!}}{2\text{!}\left(8-2\right)\text{!}}$

Total possible number of ways to choose 5 cards out of 100 cards is:

  $C{}_5=\frac{\text{100!}}{\text{5!}\left(100-5\right)\text{!}}$

So, the probability of event E is:

  $n\left(E\right)=C{}_2\times C{}_5$

Total possible number of ways to choose 7 cards out of 108 cards is:

  $\begin{array}{c}n\left(S\right)=C{}_7\\ =\frac{\text{108!}}{\text{7!}\left(108-7\right)\text{!}}\end{array}$

Thus, the probability of that a hand will contain exactly two wild cards is:

  $\begin{array}{c}P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}\\ =\frac{\frac{8\text{!}}{2\text{!}6\text{!}}\times \frac{\text{100!}}{\text{5!95!}}}{\frac{\text{108!}}{\text{7!}101\text{!}}}\\ =0.076\end{array}$

b.

To determine

To find: the probability that a hand will contain two wild cards, two red cards, and three blue cards.

Answer to Problem 44E

  $0.00069$

Explanation of Solution

Given:

The deck for a card game contains 108 cards. Twenty-five each are red, yellow, blue, and green, and eight are wild cards. Each player is randomly dealt a seven-card hand.

Let E be the event that you two wild cards, two red cards, and three blue cards.

Since, there are 8 wild cards. So, total number of ways to choose 2 wild cards is:

  $C{}_2=\frac{8\text{!}}{2\text{!}\left(8-2\right)\text{!}}$

Since, there are 25 red cards. So, total number of ways to choose 2 red cards is:

  $C{}_2=\frac{\text{25!}}{2\text{!}\left(25-2\right)\text{!}}$

Since, there are 25 blue cards. So, total number of ways to choose 3 blue cards is:

  $C{}_3=\frac{\text{25!}}{\text{3!}\left(25-3\right)\text{!}}$

So, the probability of event E is:

  $n\left(E\right)=C{}_2\times C{}_2\times C{}_3$

Total possible number of ways to choose 7 cards out of 108 cards is:

  $\begin{array}{c}n\left(S\right)=C{}_7\\ =\frac{\text{108!}}{\text{7!}\left(108-7\right)\text{!}}\end{array}$

Thus, the probability that a hand will contain two wild cards, two red cards, and three blue cards is:

  $\begin{array}{c}P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}\\ =\frac{C{}_2\times C{}_2\times C{}_3}{C{}_7}\\ =\frac{\frac{8\text{!}}{2\text{!}6\text{!}}\times \frac{\text{25!}}{\text{2!23!}}\times \frac{\text{25!}}{\text{3!22!}}}{\frac{\text{108!}}{\text{7!}101\text{!}}}\\ =0.00069\end{array}$