Bag M contains 5 white balls and 2 red balls. Bag N contains 3 white balls and 4 red balls. a) A ball is randomly selected from Bag M and Placed in Bag N. A ball is then randomly selected from Bag N. What is the probability that the ball selected form Bag N is white? b) If a white ball is selected from Bag N, what is the probability that a red ball was transferred from Bag M to Bag N? a) I got this answer which is right: 13/28. I got this by taking the probablity that a white ball is chosen from bag m and then that a white ball is chosen from bag n (which now has one more white ball) and then multiplying that number by the probability of choosing a red ball from bag m, but then choosing a white ball which is the same, just one more red ball to choose from. That sounds horribly complicated. Overall: (5/7)(4/8) + (2/7)(3/8) = 13/28 b) This is where I get confused... I know it's a conditional probability problem, so GIVEN THAT the white ball is selected from bag n (my previous answer), aka (13/28) must be on the bottom... but how do I get what's on the top? A red ball transferred from Bag M to N... so I thought it must be (2/7)(5/8)... PLEASEHELP!