Chapter 8.6, Problem 58E

(a).

To determine

To find:

The probability that a hand will contain exactly two wild cards.

Answer to Problem 58E

The required probability is $0.076$ .

Explanation of Solution

Given:

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

Concept Used:

Apply the formula of the probability:

  $\begin{array}{c}P\left(A\right)=\frac{NumberofFavourableOutcome}{TotalNumberofFavourableOutcome}\\ =\frac{n\left(A\right)}{n\left(S\right)}\end{array}$

Calculation:

Consider the total number cards to be $108$ .

Number of ways of possible outcomes for seven-card hand:

  $C{}_7=\frac{108!}{7!101!}$

So, $n\left(S\right)=\frac{108!}{7!101!}$ .

Assume event $A$ , exactly two wild cards.

The total number of ways to choose two wild cards out of eight is $C{}_2$ .

The total number of ways to choose five other cards out of $100$ cards is $C{}_5$ .

So, the total number of ways to choose exactly two wild cards is:

  $\begin{array}{c}n\left(A\right)=C{}_2\cdot C{}_5\\ =\frac{8!}{2!6!}\cdot \frac{100!}{5!95!}\\ =\frac{8\cdot 7\cdot 6!}{2\cdot 1\cdot 6!}\cdot \frac{100\cdot 99\cdot 98\cdot 97\cdot 96\cdot 95!}{5\cdot 4\cdot 3\cdot 2\cdot 1\cdot 95!}\\ =\frac{8\cdot 7}{2}\cdot \frac{100\cdot 99\cdot 98\cdot 97\cdot 96}{5\cdot 4\cdot 3\cdot 2\cdot 1}\\ =2108050560\end{array}$

Thus, the required probability is:

  $\begin{array}{c}P\left(A\right)=\frac{2108050560}{\frac{108!}{7!101!}}\\ =\frac{2108050560}{27883218168}\\ =0.07560\\ \approx 0.076\end{array}$

Conclusion:

Hence, the required probability is $0.076$ .

(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 58E

The required probability is $0.00693$ .

Explanation of Solution

Given:

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

Concept Used:

Apply the formula of the probability:

  $\begin{array}{c}P\left(A\right)=\frac{NumberofFavourableOutcome}{TotalNumberofFavourableOutcome}\\ =\frac{n\left(A\right)}{n\left(S\right)}\end{array}$

Calculation:

Consider the total number cards to be $108$ .

Number of ways of possible outcomes for seven-card hand:

  $C{}_7=\frac{108!}{7!101!}$

  $\begin{array}{c}n\left(S\right)=\frac{108!}{7!101!}\\ =27883218168\end{array}$

Assume event $A$ , two wild cards, two red cards and three blue cards.

The total number of ways to choose two wild cards out of eight is $C{}_2$ .

The total number of ways to choose $2$ red cards from the $25$ is $C{}_2$ .

The total number of ways to choose $3$ blue cards from the $25$ is $C{}_3$ .

So, the total number of ways to choose exactly two wild cards is:

  $\begin{array}{c}n\left(A\right)=C{}_2\cdot C{}_2\cdot C{}_3\\ =\frac{8!}{2!6!}\cdot \frac{25!}{2!23!}\cdot \frac{25!}{3!22!}\\ =\frac{8\cdot 7}{2}\cdot \frac{25\cdot 24}{2}\cdot \frac{25\cdot 24\cdot 23}{3\cdot 2}\\ =28\cdot 25\cdot 12\cdot 25\cdot 4\cdot 23\\ =19320000\end{array}$

Thus, the required probability is:

  $\begin{array}{c}P\left(A\right)=\frac{19320000}{27883218168}\\ =0.006928\\ \approx 0.00693\end{array}$

Conclusion:

Hence, the required probability is $0.00693$ .