/*! 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 .77. Free downloads? Illegal music do... [FREE SOLUTION] | 91影视

91影视

Free downloads? Illegal music downloading is a big problem: 29% of Internet users download music files, and 67% of downloaders say they don鈥檛 care if the music is copyrighted. 18 Find the probability that a randomly selected Internet user downloads music and doesn鈥檛 care if it鈥檚 copyrighted.

Short Answer

Expert verified

Probability for the randomly selected internet user downloads music and doesn't care about copyrighted music is 0.1943.

Step by step solution

01

Step 1:Given information

Music files are downloaded by 29% of Internet users.

67 %of downloaders think it doesn't matter if the music is protected by copyright.

02

Step 2:Calculaton

According to general multiplication rule,

P(AandB)=P(AB)=P(A)P(BA)=P(B)P(AB)

Let

D: Internet users download music files

C: Downloaders don't care about copyrighted music

Now,

The corresponding probabilities:

Probability for the internet users download music files,

P(D)=0.29

Probability for the downloaders don't care about copyrighted music,

P(CD)=0.67

Apply the general multiplication rule:

P(CandD)=P(CD)=P(D)P(CD)

=0.290.67

=0.1943

Thus

Probability for the randomly selected internet user downloads music and doesn't care about copyrighted music is=0.1943

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

Which of the following is a correct way to perform the simulation?

a. Let integers from 1to34represent making a free throw and 35to50represent missing a free throw. Generate 50random integers from1to50. Count the number

of made free throws. Repeat this process many times.

b. Let integers from 1to34represent making a free throw and 35to50represent missing a free throw. Generate 50 random integers from 1 to 50 with no repeats

allowed. Count the number of made free throws. Repeat this process many times.

c. Let integers from1to56represent making a free throw and 57to100represent missing a free throw. Generate 50 random integers from1to100.Count the number of made free throws. Repeat this process many times.

d. Let integers from localid="1653986588937" 1to56represent making a free throw and localid="1653986593808" 57to100represent missing a free throw. Generate 50 random integers from localid="1653986598680" 1to100with no repeats allowed. Count the number of made free throws. Repeat this process many times.

e. None of the above is correct.

AARP, and Medicare (4.1) To find out what proportion of Americans support proposed

Medicare legislation to help pay medical costs, the AARP conducted a survey of their

members (people over age 50 who pay membership dues). One of the questions was:

鈥淓ven if this plan won鈥檛 affect you personally either way, do you think it should be passed

so that people with low incomes or people with high drug costs can be helped?鈥 Of the

respondents, 75% answered 鈥淵es.鈥

a. Describe how undercoverage might lead to bias in this study. Explain the likely

direction of the bias.

b. Describe how the wording of the question might lead to bias in this study. Explain the

likely direction of the bias.

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

telling the truth. A lie detector test isn鈥檛 foolproof鈥攕ometimes it suggests that a person is

lying when he or she is actually telling the truth (a 鈥渇alse positive鈥). Other times, the test

says that the suspect is being truthful when he or she is actually lying (a 鈥渇alse negative鈥).

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.

Smartphone addiction? A media report claims that 50%of U.S. teens with smartphones feel addicted to their devices. A skeptical researcher believes that this figure is too high. She decides to test the claim by taking a random sample of 100U.S. teens who have smartphones. Only 40of the teens in the sample feel addicted to their devices. Does this result give convincing evidence that the media report鈥檚 50%claim is too high? To find out, we want to perform a simulation to estimate the probability of getting 40or fewer teens who feel addicted to their devices in a random sample of size 100from a very large population of teens with smartphones in which 50% feel addicted to their devices.

Let 1= feels addicted and 2= doesn鈥檛 feel addicted. Use a random number generator to produce 100random integers from 1to 2. Record the number of 1鈥檚 in the simulated random sample. Repeat this process many, many times. Find the percent of trials on which the number of 1鈥檚 was40 or less.

Household size In government data, a household consists of all occupants of a dwelling unit. Choose an American household at random and count the number of people it contains. Here is the assignment of probabilities for the outcome. The probability of finding 3people in a household is the same as the probability of finding 4people.

a. What probability should replace 鈥?鈥 in the table? Why?

b. Find the probability that the chosen household contains more than 2people.

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.