/*! 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 33 A leading manufacturer of kitche... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 33

A leading manufacturer of kitchen appliances advertised its products in two magazines: Good Housekeeping and the Ladies Home Journal. A survey of 500 customers revealed that 140 learned of its products from Good Housekeeping, 130 learned of its products from the Ladies Home Journal, and 80 learned of its products from both magazines. What is the probability that a person selected at random from this group saw the manufacturer's advertisement in a. Both magazines? b. At least one of the two magazines? c. Exactly one magazine?

Short Answer

Expert verified
a. The probability of a random person seeing the advertisement in both magazines is 0.16. b. The probability of a random person seeing the advertisement in at least one magazine is 0.38. c. The probability of a random person seeing the advertisement in exactly one magazine is 0.22.

Step by step solution

01

Define the variables

Let's define the variables as follows: - Total number of people surveyed: \(N\) - Number of people who learned of the products from Good Housekeeping magazine: \(G\) - Number of people who learned of the products from Ladies Home Journal magazine: \(L\) - Number of people who learned of the products from both magazines: \(B\) Given information: - \(N=500\) - \(G=140\) - \(L=130\) - \(B=80\)
02

Calculate the probability of seeing the advertisement in both magazines

We have the total number of people surveyed \(N\) and the number of people who saw the advertisement in both magazines \(B\). To calculate the probability of a random person seeing the advertisement in both magazines, we need to divide the number of people who saw the advertisement in both magazines by the total number of people surveyed. Probability of seeing the advertisement in both magazines = \(\frac{B}{N}\) Let's plug in the given values: Probability of seeing the advertisement in both magazines = \(\frac{80}{500} = 0.16\) a. The probability of a random person seeing the advertisement in both magazines is 0.16.
03

Calculate the probability of seeing the advertisement in at least one of the magazines

To calculate the probability of seeing the advertisement in at least one magazine, we need to first calculate the total number of people who saw the advertisement in either Good Housekeeping or Ladies Home Journal only. To do this, we can use the information about the people who saw the advertisement in both magazines. Number of people who saw the advertisement in Good Housekeeping only: \(G - B\) Number of people who saw the advertisement in Ladies Home Journal only: \(L - B\) Now, we can find the total number of people who saw the advertisement in at least one magazine. Total people = (Good Housekeeping only) + (Ladies Home Journal only) + (Both magazines) Total people = \((G - B) + (L - B) + B\) Plugging in the given values: Total people = \((140 - 80) + (130 - 80) + 80 = 60 + 50 + 80 = 190\) Probability of seeing the advertisement in at least one magazine = \(\frac{Total\:people}{N} = \frac{190}{500} = 0.38\) b. The probability of a random person seeing the advertisement in at least one magazine is 0.38.
04

Calculate the probability of seeing the advertisement in exactly one magazine

Based on the previous step, we already know the total number of people who saw the advertisement in either Good Housekeeping (60 people) or Ladies Home Journal (50 people) only. Probability of seeing the advertisement in exactly one magazine = \(\frac{Total\:people\:in\:Good\:Housekeeping\:or\:Ladies\:Home\:Journal\:only}{N}\) Plugging in the values: Probability of seeing the advertisement in exactly one magazine = \(\frac{60 + 50}{500} = \frac{110}{500} = 0.22\) c. The probability of a random person seeing the advertisement in exactly one magazine is 0.22.

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

There are 12 signs of the Zodiac: Aries, Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpio, Sagittarius, Capricom, Aquarius, and Pisces. Each sign corresponds to a different calendar period of approximately 1 month. Assuming that a person is just as likely to be born under one sign as another, what is the probability that in a group of five people at least two of them a. Have the same sign?

QuaLrr ConTRoL It is estimated that \(0.80 \%\) of a large consignment of eggs in a certain supermarket is broken. a. What is the probability that a customer who randomly selects a dozen of these cggs receives at least one broken egg? b. What is the probability that a customer who selects these eggs at random will have to check three cartons before finding a carton without any broken eggs? (Each carton contains a dozen eggs.)

A survey involving 400 likely Democratic voters and 300 likely Republican voters asked the question: Do you support or oppose legislation that would require trigger locks on guns, to prevent misuse by children? The following results were obtained: $$\begin{array}{lcc} \hline \text { Answer } & \text { Democrats, \% } & \text { Republicans, \% } \\\ \hline \text { Support } & 88 & 71 \\ \hline \text { Oppose } & 7 & 20 \\ \hline \text { Don't know/refused } & 5 & 9 \\ \hline \end{array}$$ If a randomly chosen respondent in the survey answered "support," what is the probability that he or she is a likely Republican voter?

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. Let \(S=\left\\{s_{1}, s_{2}, \ldots, s_{n}\right\\}\) be a uniform sample space for an experiment. If \(n \geq 5\) and \(E=\left\\{s_{1}, s_{2}, s_{5}\right\\}\), then \(P(E)=3 / n\).

A poll was conducted among 250 residents of a certain city regarding tougher gun-control laws. The results of the poll are shown in the table: $$ \begin{array}{lccccc} \hline & \begin{array}{c} \text { Own } \\ \text { Only a } \\ \text { Handgun } \end{array} & \begin{array}{c} \text { Own } \\ \text { Only a } \\ \text { Rifle } \end{array} & \begin{array}{c} \text { Own a } \\ \text { Handgun } \\ \text { and a Rifle } \end{array} & \begin{array}{c} \text { Own } \\ \text { Neither } \end{array} & \text { Total } \\ \hline \text { Favor } & & & & & \\ \text { Tougher Laws } & 0 & 12 & 0 & 138 & 150 \\ \hline \begin{array}{l} \text { Oppose } \\ \text { Tougher Laws } \end{array} & 58 & 5 & 25 & 0 & 88 \\ \hline \text { No } & & & & & \\ \text { Opinion } & 0 & 0 & 0 & 12 & 12 \\ \hline \text { Total } & 58 & 17 & 25 & 150 & 250 \\ \hline \end{array} $$ If one of the participants in this poll is selected at random, what is the probability that he or she a. Favors tougher gun-control laws? b. Owns a handgun? c. Owns a handgun but not a rifle? d. Favors tougher gun-control laws and does not own a handgun?
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Mechanics 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.