Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin 3 times and notes the number of heads. If she gets 1, 2, 3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1, 2, 3 or 4 with the die?

Let E1 be the event that the outcome on the die is 5 or 6. Let E2 be the event that the outcome on the die is 1, 2, 3 or 4. Therefore, P(E1) = 2/6 = 1/3 P(E2) = 4/6 = 2/3 Let A be the event of getting exactly one head. Then, P(A|E1) = 3/8 P(A|E2) = 1/2 Using Bayes' theorem, P(E2|A) = P(E2) * P(A|E2) / [P(E1) * P(A|E1) + P(E2) * P(A|E2)] P(E2|A) = (2/3) * (1/2) / [(1/3) * (3/8) + (2/3) * (1/2)] = (1/3) / [(1/8) + (1/3)] = (1/3) / [(3/24) + (8/24)] = (1/3) / (11/24) = (1/3) * (24/11) = 8/11.