/*! 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} Q 62. Get rich A survey of 4826 random... [FREE SOLUTION] | 91影视

91影视

Get rich A survey of 4826 randomly selected young adults (aged 19to25) asked, 鈥淲hat do you think are the chances you will have much more than a middle-class income at age 30?鈥 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鈥檛 say 鈥渁lmost no chance,鈥 what鈥檚 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 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

Tall people and basketball players Select an adult at random. Define events T: person is over 6feet tall, and

B: person is a professional basketball player. Rank the following probabilities from smallest to largest. Justify your answer.

P(T)P(B)P(TB)P(BT)

Matching suits A standard deck of playing cards consists of 52 cards with 13 cards in each of four suits: spades, diamonds, clubs, and hearts. Suppose you shuffle the deck thoroughly and deal 5 cards face-up onto a table.

a. What is the probability of dealing five spades in a row?

b. Find the probability that all 5 cards on the table have the same suit.

Four-sided dice A four-sided die is a pyramid whose four faces are labeled

with the numbers 1,2,3and4(see image). Imagine rolling two fair, four-sided dice and

recording the number that is showing at the base of each pyramid. For instance, you would

record a 4if the die landed as shown in the figure.

a. Give a probability model for this chance process.

b. Define event A as getting a sum of 5. Find P(A).

You read in a book about bridge that the probability that each of the four players is dealt exactly one ace is approximately 0.11. This means that

a. in every 100bridge deals, each player has 1ace exactly 11times.

b. in 1million bridge deals, the number of deals on which each player has 1ace will be exactly 110,000.

c. in a very large number of bridge deals, the percent of deals on which each player has 1ace will be very close to 11%.

d. in a very large number of bridge deals, the average number of aces in a hand will be very close to 0.11.

e. If each player gets an ace in only 2of the first 50deals, then each player should get an ace in more than 11%of the next 50deals.

The dotplot displays the number of made shots in 100simulated sets of 50free throw by someone with probability 0.56of making a free throw.

Which of the following is an appropriate statement about Wilt鈥檚 free-throw shooting

based on this dotplot?

a. If Wilt were still only a 56%shooter, the probability that he would make at least 34of his shots is about0.03.

b. If Wilt were still only a 56%shooter, the probability that he would make at least 34of his shots is about 0.97..

c. If Wilt is now shooting better than 56%, the probability that he would make at least 34of his shots is about 0.03.

d. If Wilt is now shooting better than56%, the probability that he would make at least 34of his shots is about0.97.

e. If Wilt were still only a 56%shooter, the probability that he would make at least 34of his shots is about 0.01.

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.