Bag A holds 4 red and 6 white balls; Bag B holds 7 red and 3 white balls. One of the two bags is chosen at random and a single ball is drawn from it. The ball drawn turns out to be red. What is the probability that it came from Bag A?

P(Bag A and red) = (1/2)(4/10) = 4/20 and P(Bag B and red) = (1/2)(7/10) = 7/20, so a red ball turns up with probability 11/20. The question runs the other way — given that the ball IS red, how likely is Bag A — and that is (4/20) ÷ (11/20) = 4/11. The option 2/5 is P(red given Bag A), the same two events read in the reverse order; 11/20 is the chance of a red ball at all; and 1/2 is the chance of picking Bag A before the colour is seen, which the red ball has now made less likely.