Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Get rich A survey of 4826 randomly selected young adults (aged 19to25) asked, “What do you think are the chances you will have much more than a middle-class income at age 30?” The two-way table summarizes the responses.

Choose a survey respondent at random. Define events G: a good chance, M: male, and N: almost no chance.

a. Find P(G|M). Interpret this value in context.

b. Given that the chosen survey respondent didn’t say “almost no chance,” what’s the probability that this person is female? Write your answer as a probability statement using correct symbols for the events.

Short Answer

Expert verified

a. Required probability is P(GM)0.3083

b. Probability of female respondent didn't say "almost no chance" isPMcNc0.4903

Step by step solution

01

Given Information

It is given that:

02

Determining probability for male young adult having good chance of much more than middle class income at age 30

Using Conditional Probability: P(BA)=P(AB)P(A)=P(AandB)P(A)

G: Good Chance

M: Male

Information about 4826young adults is provided.

From table, 2459/4826young adults are males.

P(M)=Number of favourable outcomesNumberof possible outcomes=24594826

Also, 78/4826young adults have a good chance.

P(GandM)=Number of favourable outcomesNumberof possibleoutcomes=7584826

Using Conditional Probability

P(GM)=P(GandM)P(M)=758482622594826=75824590.3083=30.83%

Hence, 30.83%of young males are of the view that there is good chance for them to have much more than middle aged income at30and probability is0.3083

03

Determining probability of the female respondent didn't say "almost no chance".

As per complement rule,

PAc=P(notA)=1-P(A)

and conditional probability P(BA)=P(AB)P(A)=P(AandB)P(A)

N: Almost no chance

M: Male

From table, 194/4826young adults think that "Almost no chance".

4632young adults do not have such opinion.

PNc=Numberof favourable outcomesNumberof possibleoutcomes=46324826

From table, 96/4826female adults have opinion "Almost no Chance".

As total female young adults are 2367.

2271/4826female young adults dis not have above opinion.

PMcandNc=Numberof favourableoutcomesNumber of possible=22714826

Using Conditional Probability:

PMcNc=PMcandNcPNc=2271422646324826=22714632=75715440.4903=49.03%

Hence, probability for the female respondent didn't say "almost no chance" is0.4903

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 Vaia!

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

Roulette An American roulette wheel has 38 slots with numbers 1through36,0,and 00, as shown in the figure. Of the numbered slots, 18are red, 18are black, and 2—the 0and 00—are green. When the wheel is spun, a metal ball is dropped onto the middle of the wheel. If the wheel is balanced, the ball is equally likely to settle in any of the numbered slots. Imagine spinning a fair wheel once. Define events B: ball lands in a black slot, and E: ball lands in an even-numbered slot. (Treat0and 00as even numbers.)

a. Make a two-way table that displays the sample space in terms of events Band E.

b. Find P(B)andP(E).

c. Describe the event “Band E” in words. Then find the probability of this event.

d. Explain why P(BorE)P(B)+P(E). Then use the general addition rule to compute P(BorE).

Dogs and cats In one large city, 40%of all households own a dog, 32% own a cat, and 18% own both. Suppose we randomly select a household. What’s the probability that the household owns a dog or a cat?

Liar, liar! Sometimes police use a lie detector test to help determine whether a suspect is

telling the truth. A lie detector test isn’t foolproof—sometimes it suggests that a person is

lying when he or she is actually telling the truth (a “false positive”). Other times, the test

says that the suspect is being truthful when he or she is actually lying (a “false negative”).

For one brand of lie detector, the probability of a false positive is 0.08.

a. Explain what this probability means.

b. Which is a more serious error in this case: a false positive or a false negative? Justify

your answer.

Mystery box Ms. Tyson keeps a Mystery Box in her classroom. If a student meets expectations for behavior, she or he is allowed to draw a slip of paper without looking. The slips are all of equal size, are well mixed, and have the name of a prize written on them. One of the “prizes”—extra homework—isn’t very desirable! Here is the probability model for the prizes a student can win:

a. Explain why this is a valid probability model.

b. Find the probability that a student does not win extra homework.

c. What’s the probability that a student wins candy or a homework pass?

Recycling Do most teens recycle? To find out, an AP® Statistics class asked an SRS of 100students at their school whether they regularly recycle. In the sample, 55students said that they recycle. Is this convincing evidence that more than half of the students at the school would say they regularly recycle? The dotplot shows the results of taking 200SRSS of 100students from a population in which the true proportion who recycle is 0.50.

a. Explain why the sample result (55out of 100said "Yes") does not give convincing evidence that more than half of the school's students recycle.

b. Suppose instead that 63students in the class's sample had said "Yes." Explain why this result would give convincing evidence that a majority of the school's students recycle.

See all solutions

Recommended explanations on Math Textbooks

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.

Sign-up for free