A group of four women and six men must select a five-person committee. Find the probability of getting a committee of three women and two men. Find the probability of getting a committee of one woman and four men.
A group of four women and six men must select a five-person committee. Find the probability of getting a committee of three women and two men. Find the probability of getting a committee of one woman and four men.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Question
![**Problem Statement:**
A group of four women and six men must select a five-person committee.
1. Find the probability of getting a committee of three women and two men.
2. Find the probability of getting a committee of one woman and four men.
**Solution:**
To solve these problems, we will use the concept of combinations.
Let's denote:
- \( C(n, r) \) as the combination function, which calculates the number of ways to choose \( r \) items from \( n \) items.
**Problem 1: Probability of a committee of three women and two men**
- Total ways to choose 5 people from a group of 10 (4 women + 6 men):
\[
C(10, 5) = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252
\]
- Ways to choose 3 women from 4:
\[
C(4, 3) = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4
\]
- Ways to choose 2 men from 6:
\[
C(6, 2) = \frac{6 \times 5}{2 \times 1} = 15
\]
- Total ways to choose this specific committee:
\[
C(4, 3) \times C(6, 2) = 4 \times 15 = 60
\]
- Probability:
\[
\frac{60}{252} = \frac{5}{21} \approx 0.238
\]
**Problem 2: Probability of a committee of one woman and four men**
- Ways to choose 1 woman from 4:
\[
C(4, 1) = 4
\]
- Ways to choose 4 men from 6:
\[
C(6, 4) = \frac{6 \times 5 \times 4 \times 3}{4 \times 3 \times 2 \times 1} = 15
\]
- Total ways to choose this specific committee:
\[
C(4, 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feb60c3aa-6e6b-4864-9c9b-e91dc503105c%2Fdd18140f-760d-4293-b457-c8b90f5953a2%2Folywzk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A group of four women and six men must select a five-person committee.
1. Find the probability of getting a committee of three women and two men.
2. Find the probability of getting a committee of one woman and four men.
**Solution:**
To solve these problems, we will use the concept of combinations.
Let's denote:
- \( C(n, r) \) as the combination function, which calculates the number of ways to choose \( r \) items from \( n \) items.
**Problem 1: Probability of a committee of three women and two men**
- Total ways to choose 5 people from a group of 10 (4 women + 6 men):
\[
C(10, 5) = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252
\]
- Ways to choose 3 women from 4:
\[
C(4, 3) = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4
\]
- Ways to choose 2 men from 6:
\[
C(6, 2) = \frac{6 \times 5}{2 \times 1} = 15
\]
- Total ways to choose this specific committee:
\[
C(4, 3) \times C(6, 2) = 4 \times 15 = 60
\]
- Probability:
\[
\frac{60}{252} = \frac{5}{21} \approx 0.238
\]
**Problem 2: Probability of a committee of one woman and four men**
- Ways to choose 1 woman from 4:
\[
C(4, 1) = 4
\]
- Ways to choose 4 men from 6:
\[
C(6, 4) = \frac{6 \times 5 \times 4 \times 3}{4 \times 3 \times 2 \times 1} = 15
\]
- Total ways to choose this specific committee:
\[
C(4, 1
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