Question

Urn 1 has five white and seven black balls. Urn 2 has three white and twelve black balls. We flip a fair coin. If the outcome is headed, then a ball from urn 1 is selected, while if the outcome is tails, then a ball from urn 2 is selected. Suppose that a white ball is selected. What is the probability that the coin landed tails?

Short answer

The probability that the coin landed tails given that a white ball was chosen is \(\frac{4}{9}\).

Step-by-step solution

Identify the Probability of Each Event Separately

We need to find the probabilities for each event occurring individually in order to find the joint probabilities later on. P(Tails) = 1/2 (since the coin is fair) P(Heads) = 1/2 (since the coin is fair) P(White ball | Urn 1) = 5/12 (there are 5 white balls out of a total of 12 balls in Urn 1) P(White ball | Urn 2) = 3/15 (there are 3 white balls out of a total of 15 balls in Urn 2)

Find the Joint Probabilities

Now we need to find the probability of both events happening together (intersection of events). P(Tails ∩ White ball) = P(Tails) * P(White ball | Urn 2) = (1/2) * (3/15) = 1/10 P(Heads ∩ White ball) = P(Heads) * P(White ball | Urn 1) = (1/2) * (5/12) = 5/24

Calculate the Probability of Choosing a White Ball

We will sum the joint probabilities of choosing a white ball from both urns. P(White ball) = P(Tails ∩ White ball) + P(Heads ∩ White ball) = (1/10) + (5/24) = 9/40

Use the Conditional Probability Formula

Now we will use the conditional probability formula to find the probability that the coin landed tails given that a white ball was selected. P(Tails | White ball) = P(Tails ∩ White ball) / P(White ball) = (1/10) / (9/40) = (1/10) * (40/9) = 4/9 So the probability that the coin landed tails given that a white ball was chosen is 4/9.