Bag I contains n red balls and 4 black balls. Bag II contains 3 red balls and 5 black balls. One ball is transferred at random from Bag I to Bag II, and then a ball is drawn from Bag II. If the ball drawn is found to be red, and the probability that the transferred ball was black is 1 3 , then the value of n is

Step-by-step solution

Let E₁ be the event that a red ball is transferred, and E₂ be the event that a black ball is transferred. The probabilities of these events are: P(E₁) = n n+4 P(E₂) = 4 n+4 Let A be the event that the ball drawn from Bag II is red. If a red ball is transferred, Bag II contains 4 red and 5 black balls. Thus, P(A|E₁) = 4 9 . If a black ball is transferred, Bag II contains 3 red and 6 black balls. Thus, P(A|E₂) = 3 9 . By Bayes' theorem, the probability that the transferred ball was black given that the drawn ball is