/*! 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 9 A retail establishment accepts e... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 9

A retail establishment accepts either the American Express or the VISA credit card. A total of 24 percent of its customers carry an American Express card, 61 percent carry a VISA card, and 11 percent carry both cards. What percentage of its customers carry a credit card that the establishment will accept?

Short Answer

Expert verified
74% of the customers carry a credit card that the retail establishment will accept, using the principle of inclusion and exclusion: |A 鈭� V| = |A| + |V| - |A 鈭� V|, with |A| = 24, |V| = 61, and |A 鈭� V| = 11.

Step by step solution

01

Understanding the principle of inclusion and exclusion

The principle of inclusion and exclusion is a counting method used in combinatorics to find the number of elements in the union of multiple sets. In this exercise, we have two sets: the set of customers carrying an American Express card (A) and the set of customers carrying a VISA card (V). We want to find the percentage of customers who carry either an American Express or a VISA card (which is the union of these two sets). The principle of inclusion and exclusion states that: |A 鈭� V| = |A| + |V| - |A 鈭� V| where |A 鈭� V| is the number of elements in the union of sets A and V, |A| is the number of elements in set A, |V| is the number of elements in set V, and |A 鈭� V| is the number of elements in the intersection of sets A and V (i.e., customers carrying both cards).
02

Assigning percentages to each set

We are given the following percentages: - 24% of customers carry an American Express card, so |A| = 24. - 61% of customers carry a VISA card, so |V| = 61. - 11% of customers carry both cards, so |A 鈭� V| = 11.
03

Apply the principle of inclusion and exclusion

Using the principle of inclusion and exclusion, we can find the percentage of customers who carry at least one of the accepted credit cards: |A 鈭� V| = |A| + |V| - |A 鈭� V| Substitute the given values: |A 鈭� V| = 24 + 61 - 11
04

Calculate the percentage of customers carrying an accepted card

Now, calculate the sum: |A 鈭� V| = 13 + 61 = 74 Therefore, 74% of the customers carry a credit card that the retail establishment will accept.

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

The chess clubs of two schools consist of, respectively, 8 and 9 players. Four members from each club are randomly chosen to participate in a contest between the two schools. The chosen players from one team are then randomly paired with those from the other team, and each pairing plays a game of chess. Suppose that Rebecca and her sister Elise are on the chess clubs at different schools. What is the probability that (a) Rebecca and Elise will be paired? (b) Rebecca and Elise will be chosen to represent their schools but will not play cach other? (c) either Rebecca or Elise will be chosen to represent her school?

If 8 rooks (castles) are randomly placed on a chessboard, compute the probability that none of the rooks can capture any of the others. That is, compute the probability that no row or file contains more than one rook.

A town contains 4 people who repair televisions. If 4 sets break down, what is the probability that exactly \(i\) of the repairers are called? Solve the problem for \(i=1,2,3,4 .\) What assumptions are you making?

An instructor gives her class a set of 10 problems with the information that the final exam will consist of a random selection of 5 of them. If a student has figured out how to do 7 of the problems, what is the probability that he or she will answer correctly (a) all 5 problems? (b) at least 4 of the problems?

A system is comprised of 5 components, each of which is either working or failed. Consider an experiment that consists of observing the status of each component, and let the outcome of the experiment be given by the vector \(\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right)\) where \(x_{i}\) is equal to 1 if component \(i\) is working and is equal to 0 if component \(i\) is failed. (a) How many outcomes are in the sample space of this experiment? (b) Suppose that the system will work if components 1 and 2 are both working, or if components 3 and 4 are both working, or if components \(1,3,\) and 5 are all working. Let \(W\) be the event that the system will work. Specify all the outcomes in \(W\) (c) Let \(A\) be the event that components 4 and 5 are both failed. How many outcomes are contained in the event \(A ?\) (d) Write out all the outcomes in the event \(A W\).
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

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.