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Chapter 1: Problem 42

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.

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Most popular questions from this chapter

Urn 1 contains two white balls and one black ball, while urn 2 contains one white ball and five black balls. One ball is drawn at random from urn 1 and placed in an urn 2\. A ball is then drawn from urn 2. It happens to be white. What is the probability that the transferred ball was white?

An individual uses the following gambling system at Las Vegas. He bets \(\$ 1\) that the roulette wheel will come up red. If he wins, he quits. If he loses then he makes the same bet a second time only this time he bets \(\$ 2\); and then regardless of the outcome, quits. Assuming that he has a probability of \(\frac{1}{2}\) of winning each bet, what is the probability that he goes home a winner? Why is this system not used by everyone?

For a fixed event \(B\), show that the collection \(P(A \mid B)\), defined for all events \(A\), satisfies the three conditions for a probability. Conclude from this that $$ P(A \mid B)=P(A \mid B C) P(C \mid B)+P\left(A \mid B C^{c}\right) P\left(C^{C} \mid B\right) $$ Then directly verify the preceding equation.

(a) A gambler has in his pocket a fair coin and a two-headed coin. He selects one of the coins at random, and when he flips it, it shows heads. What is the probability that it is a fair coin? (b) Suppose that he flips the same coin a second time and again it shows heads. Now, what is the probability that it is a fair coin? (c) Suppose that he flips the same coin a third time and it shows tails. Now, what is the probability that it is a fair coin?

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?
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