Question

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

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

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

Use the conditional probability formula

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

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

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

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.