91Ó°ÊÓ

First, we calculate the probability of selecting a white ball from each urn. For urn 1, there are 5 white balls and a total of 5 + 7 = 12 balls. The probability of selecting a white ball from urn 1 is: \[P(W_1) = \frac{5}{12}\] For urn 2, there are 3 white balls and a total of 3 + 12 = 15 balls. The probability of selecting a white ball from urn 2 is: \[P(W_2) = \frac{3}{15}\]

Since the coin is fair, the probability of it landing tails is: \[P(Tails) = \frac{1}{2}\] Similarly, the probability of the coin landing heads is: \[P(Heads) = \frac{1}{2}\]

Now that we have the probabilities of selecting a white ball from each urn and the probability of the coin landing tails, we can use Bayes' theorem to find the probability of the coin landing tails given that a white ball has been selected: \[P(Tails | W) = \frac{P(W | Tails) \times P(Tails)}{P(W)}\] We know that \(P(W | Tails)\) is equal to the probability of selecting a white ball from urn 2, which is \(\frac{3}{15}\), and \(P(Tails)\) is \(\frac{1}{2}\). The denominator, \(P(W)\), can be calculated using the law of total probability: \[P(W) = P(W | Heads) \times P(Heads) + P(W | Tails) \times P(Tails)\] Plugging in the values, we get: \[P(W) = \frac{5}{12} \times \frac{1}{2} + \frac{3}{15} \times \frac{1}{2}\] Now, we have all the necessary values to plug into Bayes' theorem: \[P(Tails | W) = \frac{\frac{3}{15} \times \frac{1}{2}}{\frac{5}{12} \times \frac{1}{2} + \frac{3}{15} \times \frac{1}{2}}\]

Simplify the equation to get the probability of the coin landing tails given that a white ball has been selected: \[P(Tails | W) = \frac{\frac{3}{30}}{\frac{10}{60} + \frac{6}{60}} = \frac{\frac{3}{30}}{\frac{16}{60}}\] \[P(Tails | W) = \frac{3}{30} \times \frac{60}{16} = \frac{3}{8}\] So, the probability that the coin landed tails given that a white ball has been selected is \(\frac{3}{8}\).