Bag A holds 3 red and 2 white balls; Bag B holds 2 red and 4 white balls. One ball is taken at random from Bag A and dropped into Bag B without being looked at. A ball is then drawn at random from Bag B. What is the probability that it is red?

Bag B is not the same bag at the second draw, so split the work by what was moved. The transferred ball is red with probability 3/5, leaving Bag B with 3 red among 7, so a red follows with probability 3/7. The transferred ball is white with probability 2/5, leaving 2 red among 7, so a red follows with probability 2/7. Altogether (3/5)(3/7) + (2/5)(2/7) = 9/35 + 4/35 = 13/35. The value 3/7 assumes the transferred ball was red, which is only one of the two branches and the more likely one but not a certainty; 1/3 = 2/6 is Bag B's red fraction before the transfer, which ignores the extra ball entirely. The value 5/11 pools the two bags into one and asks for 5 red out of 11 — but the balls of Bag A never had an equal chance of being drawn, since only one of them ever reached Bag B.