/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 A parent-teacher committee consi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 14

A parent-teacher committee consisting of four people is to be selected from fifteen parents and five teachers. Find the probability of selecting two parents and two teachers.

Short Answer

Expert verified
The probability is approximately 1.083 (or 108.3%), indicating that there's something wrong with our calculations. In step 2, a mistake has been made while counting the combinations. The correct calculation would yield \( C(15, 2) \times C(5, 2) = 45 \times 10 = 450 \), not 5250. Therefore, the desired probability is \( \frac{450}{4845} \approx 0.093 \) or 9.3%.

Step by step solution

01

Calculate Total Combinations

We need to find the total number of ways to select a committee of 4 people from total 20 people (15 parents and 5 teachers). The combination formula is used here: \( C(n, k) = \frac{n!}{k!(n-k)!} \). Substituting n=20 and k=4 gives \( C(20, 4) = \frac{20!}{4!(20-4)!} \). After calculation, the total combinations are 4845.
02

Calculate Desired Combinations

Now we need to find the number of ways to select exactly 2 parents from 15 and 2 teachers from 5. This will be the product of combination of selecting 2 parents from 15 and combination of selecting 2 teachers from 5. So we calculate \( C(15, 2) \times C(5, 2) = \frac{15!}{2!(15-2)!} \times \frac{5!}{2!(5-2)!} \), which gives 5250.
03

Calculate Probability

Finally, the probability that we seek can be found by dividing the number of desired combinations by total combinations. So, the desired probability is \( \frac{5250}{4845} \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Performing at a concert are eight rock bands and eight jazz groups. How many ways can the program be arranged if the first, third, and eighth performers are jazz groups?

The probability that a region prone to flooding will flood in any single year is \(\frac{1}{10}\). a. What is the probability of a flood two years in a row? b. What is the probability of flooding in three consecutive years? c. What is the probability of no flooding for ten consecutive years? d. What is the probability of flooding at least once in the next ten years?

In how many different ways can a police department arrange eight suspects in a police lineup if each lineup contains all eight people?

A game is played using one die. If the die is rolled and shows 1 , the player wins \(\$ 5\). If the die shows any number other than 1 , the player wins nothing. If there is a charge of \(\$ 1\) to play the game, what is the game's expected value? What does this value mean?

In Exercises 21-26, a fair coin is tossed two times in succession. The set of equally likely outcomes is \(\\{H H, H T, T H, T T\\}\). Find the probability of getting two heads.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.