Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is draw from Bag II . The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is

Step-by-step solution

Given, In bag 1 → 3 R ,   4 B   &   3 W And in bag 2 → 2 R ,   5 B   &   2 W Let A : Drawn ball from bag II is black And let B : Red ball transferred Now by total probability theorem we have P A = 3 9 × 5 10 + 4 9 × 6 10 + 3 9 × 5 10 = 54 90 And P A ∩ B = 3 9 × 5 10 So, by bayes theorem we have, P B A = P A ∩ B P A = 3 9 × 5 10 3 9 × 5 10 + 4 9 × 6 10 + 3 9 × 5 10 = 15 15 + 24 + 15 = 15 54 = 5 18