/*! 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 67 A shipment of 25 television sets... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 67

A shipment of 25 television sets contains three defective units. In how many ways can a vending company purchase four of these units and receive (a) all good units, (b) two good units, and (c) at least two good units?

Short Answer

Expert verified
Therefore, the number of ways for each scenario are: (a) for all good units, calculate \( \binom{22}{4} \), (b) for two good units, calculate \( \binom{22}{2} \times \binom{3}{2} \), (c) for at least two good units, calculate the sum of \( \binom{22}{2} \times \binom{3}{2} \), \( \binom{22}{3} \times \binom{3}{1} \), and \( \binom{22}{4} \)

Step by step solution

01

Find Total Number of Ways to Select Four Units

There are 25 units in total and the company wants to purchase four units. The total number of ways to select four units out of 25, not considering whether the units are defective or not, can be found using the combination formula: \( \binom{25}{4} = \frac{25!}{4!(25-4)!} \)
02

Find Number of Ways to Select All Good Units

There are 22 good units (25 - 3 defective units). The number of ways to select all good units can be found by substituting 'n' with 22 (total number of good units) and 'r' with 4 (units to purchase) in the combination formula: \( \binom{22}{4} = \frac{22!}{4!(22-4)!} \)
03

Find Number of Ways to Select Two Good Units

Here the company wants to select two good units and two defective units. The number of ways to select two good units out of 22 can be found using the combination formula: \( \binom{22}{2} = \frac{22!}{2!(22-2)!} \). Similarly, the number of ways to select two defective units out of 3 can be found using the same formula: \( \binom{3}{2} = \frac{3!}{2!(3-2)!} \). As we want two good and two defective units, multiply these two results to get the total.
04

Find Number of Ways to Select At Least Two Good Units

In this scenario, there can be 2, 3, or 4 good units. To compute this, find the number of ways for each of these cases and then sum the results. The number of ways to select 2 good and 2 defective units has been calculated in the previous step. For 3 good and 1 defective, use the similar method: \( \binom{22}{3} \times \binom{3}{1} \), and for 4 good units it has been calculated in the step related to 'all good units'. Sum these three sets of results to get the total.

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

You and a friend agree to meet at your favorite fast - food restaurant between \( 5:00 \) and \( 6:00 \) P.M.The one who arrives first will wait \( 15 \) minutes for the other, and then will leave (see figure). What is the probability that the two of you will actually meet,assuming that your arrival times are random within the hour?

In Exercises 53 - 60, the sample spaces are large and you should use the counting principles discussed in Section 9.6. On a game show, you are given five digits to arrange in the proper order to form the price of a car. If you are correct, you win the car. What is the probability of winning, given the following conditions? (a) You guess the position of each digit. (b) You know the first digit and guess the positions of the other digits.

You are dealt five cards from an ordinary deck of \( 52 \) playing cards. In how many ways can you get (a) a full house and (b) a five-card combination containing two jacks and three aces? (A full house consists of three of one kind and two of another. For example, \( A-A-A-5-5 and K-K-K-10-10 \) are full houses.)

What is the relationship between \( _nC_r \) and \( _nC_{n - r} \)?

In Exercises 1 - 7, fill in the blanks. The ________ of an event \( A \) is the collection of all outcomes in the sample space that are not in \( A \).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Statistics

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.