Bag A contains 3 white and 2 red balls. Bag B contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from bag A and put into bag B. However if tail appears then 2 balls are drawn at random from bag A and put into bag B. Now one ball is drawn at random from bag B. Given that the drawn ball from B is white, the probability that head appeared on the coin is

Step-by-step solution

P(H) = P(T) = 1 2 If Head: 1 ball drawn from Bag A (3W, 2R), put into Bag B (1W). P(W_B|H) = 3 5 1 + 2 5 1 2 = 3 5 + 1 5 = 4 5 If Tail: 2 balls drawn from Bag A, put into Bag B (1W). P(2W) = ^3C₂ ^5C₂ = 3 10 , P(1W1R) = ^3C₁ ^2C₁ ^5C₂ = 6 10 , P(2R) = ^2C₂ ^5C₂ = 1 10 P(W_B|T) = 3 10 1 + 6 10 2 3 + 1 10 1 3 = 3 10 + 4 10 + 1 30 = 9+12+1 30 = 22 30 = 11 15 By Bayes' theorem: P(H|W_B) = 1 2 4 5 1 2 4 5 + 1 2 11 15 = 4 5 4 5 + 11 15 = 12 15 12+11 15 = 12 23