Chapter 8.6, Problem 62E

(a)

To determine

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

Answer to Problem 62E

The probability that a hand will contain only two wild cards is is 0.0756.

Explanation of Solution

Given information:

The deck for a card game is made up of 108 cards. Twenty five given are red, yellow, blue, and green, and eight are wild cards. Given players is randomly dealt a seven card hand.

Calculation:

Total number of ways of selecting 7 cards from 108 is given by,

  $n\left(S\right){=}_{108}{C}_7$

The total number of ways of selecting 2 wild cards from 8 wild cards is ${}_8{C}_2$ . Since, the hand contains exactly two wild cards, the other 5 cards in the hand are selected from 100 colored cards and the number of ways to do so is given by ${}_{100}{C}_5$ . By using the fundamental counting principle, number of ways of getting 2 wild cards in a hand is given by,

  $n\left(E\right){=}_{100}{C}_5{\cdot }_8{C}_2$

The probability that a hand will contain only two wild cards is,

  $\begin{array}{c}P=\frac{n\left(E\right)}{n\left(S\right)}\\ =\frac{{}_{100}{C}_5{\cdot }_8{C}_2}{{}_{108}{C}_7}\\ =0.0756\end{array}$

(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 62E

The probability that a hand will contain only two wild cards is 0.00069.

Explanation of Solution

Given information:

The deck for a card game is made up of 108 cards. Twenty five given are red, yellow, blue, and green, and eight are wild cards. Given players is randomly dealt a seven card hand.

Calculation:

Total number of ways of selecting 7 cards from 108 is given by,

  $n\left(S\right){=}_{108}{C}_7$

The total number of ways of selecting 2 wild cards from 8 wild cards is ${}_8{C}_2$ . The total number of ways of selecting 2 red cards from 25 wild cards is ${}_{25}{C}_2$ . The total number of ways of selecting 3 blue cards from 25 wild cards is ${}_{25}{C}_3$ . By using the fundamental counting principle, number of ways of getting 2 wild cards, 2 red cards and 3 blue cards in a hand is given by.

  $n\left(E\right){=}_8{C}_2{\cdot }_{25}{C}_2{\cdot }_{25}{C}_3$

The probability that a hand will contain only two wild cards is,

  $\begin{array}{c}P=\frac{n\left(E\right)}{n\left(S\right)}\\ =\frac{{}_8{C}_2{\cdot }_{25}{C}_2{\cdot }_{25}{C}_3}{{}_{108}{C}_7}\\ =0.00069\end{array}$