Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. Two balls are transferred at random from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred balls were both black.

Let E1, E2, E3 and A be events such that
E1 = Both transferred balls from Bag I to Bag II are red.
E2 = Both transferred balls from Bag I to Bag II are black.
E3 = Out of two transferred balls one is red and other is black.
A = drawing a red ball from Bag II.
Here, P(E2|A) is required.
Now, P(E1) = 3C2/7C2 = 3!/(2!1!) × 2!5!/7! = 1/7
P(E2) = 4C2/7C2 = 4!/(2!2!) × 2!5!/7! = 2/7
P(E3) = 3C1 × 4C1/7C2 = 3! × 4!/7! × 2! × 5!/1! = 4/7
P(A|E1) = 6/11, P(A|E2) = 4/11, P(A|E3) = 5/11
Therefore, P(E2|A) = P(E2).P(A|E2) / [P(E1)P(A|E1) + P(E2)P(A|E2) + P(E3)P(A|E3)]
after solving = (2/7)(4/11) / [(1/7)(6/11) + (2/7)(4/11) + (4/7)(5/11)] = 8/77 / (6/77 + 8/77 + 20/77) = 8/77 × 77/34 = 4/17
Therefore, the probability that the transferred balls were both black = 4/17