Chapter 11.7, Problem 8E

1.

To determine

The possible combinations of $3$ callers that can be selected.

Answer to Problem 8E

The possible combination of $3$ callers that can be selected are $1140$ .

Explanation of Solution

Given information:

A radio station takes the name of first $20$ listeners who call in after hearing a certain song. The station will randomly select $3$ of the callers to win tickets to a concert.

Calculation:

To find the number of combinations of $n$ objects taken $r$ at a time, divide the number of permutations of $n$ objects taken $r$ at a time by $r!$ .

  $nCr=\frac{nPr}{r!}$   [Combination formula]

TO-DETERMINE the possible combination of $3$ callers that can be selected from $20$ , find $20C3$

  $\Rightarrow 20C3=\frac{20P3}{3!}$   [Use combination formula]

  $=\frac{20\cdot 19\cdot 18}{3\cdot 2\cdot 1}$   [Write $20P3$ and $3!$ as a product]

  $=\frac{\cancel{\stackrel{10}{20}}\cdot 19\cdot \cancel{\stackrel{6}{18}}}{\cancel{\underset{1}{3}}\cdot \cancel{\underset{1}{2}}\cdot 1}$   [Divide out common factors]

  $=1140$   [Simplify]

Therefore, the number of combinations are $1140$ .

2.

To determine

The possible combinations of $3$ ticket winners to include all the $3$ friends .

Answer to Problem 8E

The possible combinations of $3$ ticket winners to include all the $3$ friends is $1$ .

Explanation of Solution

Given information:

A radio station takes the name of first $20$ listeners who call in after hearing a certain song. The station will randomly select $3$ of the callers to win tickets to a concert. $3$ friends are among the first $20$ callers.

Calculation:

To find the number of combinations of $n$ objects taken $r$ at a time, divide the number of permutations of $n$ objects taken $r$ at a time by $r!$ .

  $nCr=\frac{nPr}{r!}$   [Combination formula]

TO-DETERMINE the possible combination of $3$ ticket winners to include all the $3$ friends, find $3C3$

  $\Rightarrow 3C3=\frac{3P3}{3!}$   [Use combination formula]

  $=\frac{3\cdot 2\cdot 1}{3\cdot 2\cdot 1}$   [Write $3P3$ and $3!$ as a product]

  $=\frac{\cancel{3}\cdot \cancel{2}\cdot 1}{\cancel{3}\cdot \cancel{2}\cdot 1}$   [Divide out common factors]

  $=1$   [Simplify]

Therefore, the number of combinations is $1$ .

3.

To determine

The probability that the $3$ friends win tickets.

Answer to Problem 8E

Probability that the friends will win tickets is $\frac{1}{1140}$ .

Explanation of Solution

Given information:

A radio station takes the name of first $20$ listeners who call in after hearing a certain song. The station will randomly select $3$ of the callers to win tickets to a concert. $3$ friends are among the first $20$ callers.

Calculation:

To find the probability that the $3$ friends win tickets, divide the number of ways the $3$ friends will be selected, to the total number of ways any $3$ callers can be selected.

  $\text{P( friends win tickets)=}\frac{\text{Number of ways in which 3 friends are selected }}{\text{Total number of way in which any 3 people are slected}}$

Number of ways in which $3$ friends are selected= $1$

Number of ways In which any $3$ people ae selected= $1140$

  $\Rightarrow \text{P( friends win tickets)=}\frac{1}{1140}$

Therefore, the probabilitythat the friends will win tickets is $\frac{1}{1140}$ .