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Chapter 11: Problem 5

A group consists of four men and five women. Three people are selected to attend a conference. a. In how many ways can three people be selected from this group of nine? b. In how many ways can three women be selected from the five women? c. Find the probability that the selected group will consist of all women.

Short Answer

Expert verified
(a) The total number of ways to select three people out of nine is 84 ways. (b) The total number of ways to select three women out of five is 10 ways. (c) The probability that all individuals selected are women is \( \frac{5}{42} \).

Step by step solution

01

Calculate total combinations of selecting three of nine people

Use the combination formula with \( n = 9 \) (the total number of people) and \( r = 3 \) (the number of people to be selected). Therefore, \[ ^{9}C_{3} = \frac{9!}{3!(9-3)!} = 84 \].
02

Calculate combinations of selecting three women out of five

Now, use the combination formula with \( n = 5 \) (the total number of women) and \( r = 3 \) (the number of women to be selected). Hence, \[ ^{5}C_{3} = \frac{5!}{3!(5-3)!} = 10 \].
03

Find the probability of selecting all women

To find the probability, divide the number of ways to select 3 women from the total (found in step 2) by the total number of ways to select any 3 people from the total group (found in step 1). Therefore, \[ P (\text{all women}) = \frac{^{5}C_{3}}{^{9}C_{3}} = \frac{10}{84} = \frac{5}{42} \].

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