/*! 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 42 There are three coins in a box. ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

There are three coins in a box. One is a two-headed coin, another is a fair coin, and the third is a biased coin that comes up heads 75 percent of the time. When one of the three coins is selected at random and flipped, it shows heads. What is the probability that it was the two-headed coin?

Short Answer

Expert verified
The probability that it was the two-headed coin, given that the coin flip shows heads, is \(\frac{4}{11}\) or approximately 0.36.

Step by step solution

01

Identify the probabilities

Let's identify the probabilities we know: - The probability of selecting each coin: P(A) = P(B) = P(C) = 1/3 - The probability of getting heads from the two-headed coin (Coin A): P(Heads|A) = 1 - The probability of getting heads from the fair coin (Coin B): P(Heads|B) = 1/2 - The probability of getting heads from the biased coin (Coin C): P(Heads|C) = 3/4
02

Use the conditional probability formula

The formula for finding the conditional probability is: P(A|Heads) = P(Heads and A) / P(Heads)
03

Find the probability of Heads

To find the probability of getting heads, we can use the law of total probability: P(Heads) = P(Heads|A) * P(A) + P(Heads|B) * P(B) + P(Heads|C) * P(C) Plugging in the values we have: P(Heads) = (1) * (1/3) + (1/2) * (1/3) + (3/4) * (1/3) = 1/3 + 1/6 + 1/4 = 11/12
04

Find the probability of Heads and A

Since Coin A is a two-headed coin, the probability of selecting it and flipping heads is the same as the probability of selecting Coin A: P(Heads and A) = P(A) = 1/3
05

Calculate the conditional probability

Now we can plug the values we found into the conditional probability formula: P(A|Heads) = (1/3) / (11/12) Multiply the numerator and denominator by 12 to get rid of the fraction: P(A|Heads) = (4) / (11) So, the probability that the coin selected is the two-headed coin (Coin A) given that the coin flip shows heads is 4/11 or approximately 0.36.

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 91Ó°ÊÓ!

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

Sixty percent of the families in a certain community own their own car, thirty percent own their own home, and twenty percent own both their own car and their own home. If a family is randomly chosen, what is the probability that this family owns a car or a house but not both?

An urn contains \(b\) black balls and \(r\) red balls. One of the balls is drawn at random, but when it is put back in the urn \(c\) additional balls of the same color are put in with it. Now suppose that we draw another ball. Show that the probability that the first ball drawn was black given that the second ball drawn was red is \(b /(b+r+c)\).

Three prisoners are informed by their jailer that one of them has been chosen at random to be executed, and the other two are to be freed. Prisoner \(A\) asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information, since he already knows that at least one will go free. The jailer refuses to answer this question, pointing out that if \(A\) knew which of his fellows were to be set free, then his own probability of being executed would rise from \(\frac{1}{3}\) to \(\frac{1}{2}\), since he would then be one of two prisoners. What do you think of the jailer's reasoning?

The dice game craps is played as follows. The player throws two dice, and if the sum is seven or eleven, then she wins. If the sum is two, three, or twelve, then she loses. If the sum is anything else, then she continues throwing until she either throws that number again (in which case she wins) or she throws a seven (in which case she loses). Calculate the probability that the player wins.

Suppose that 5 percent of men and \(0.25\) percent of women are color-blind. A color-blind person is chosen at random. What is the probability of this person being male? Assume that there are an equal number of males and females.

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.