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Problem 12 - Entrance Test

A bag contains 4 red and 4 black balls. Another bag contains 2 red and 6 black balls. One of the two bags is selected at random. If a ball drawn from the selected bag is found to be red, find the probability that the ball is drawn from the first bag.

Correct: A

Let E1 be the event that the first bag is selected, and E2 be the event that the second bag is selected. Let A be the event that a red ball is drawn. We are given P(E1) = P(E2) = 1/2. P(A|E1) = probability of drawing a red ball from the first bag = 4/8 = 1/2. P(A|E2) = probability of drawing a red ball from the second bag = 2/8 = 1/4. We want to find P(E1|A). Using Bayes' theorem, P(E1|A) = [P(A|E1) * P(E1)] / [P(A|E1) * P(E1) + P(A|E2) * P(E2)] = [(1/2) * (1/2)] / [(1/2) * (1/2) + (1/4) * (1/2)] = (1/4) / (1/4 + 1/8) = (1/4) / (3/8) = (1/4) * (8/3) = 2/3.