A bag A contains 4 red and 6 black marbles .

Question

A bag A contains 4 red and 6 black marbles . Another bag B contains 6 red and 4 black marbles . A marble is drawn from bag A and put into bag B ( without seeing its color ) . Now from bag B , a marble is drawn . What is the probability that the marble drawn from Bag B is red in color ?

Answer 1

P(red marble drawn from B) = P(red marble drawn from B|red marble drawn from A)(P(red marble drawn from A) + P(red marble drawn from B|black marble drawn from A)P(black marble drawn from A)

Now, let's figure each of these probabilities on the right side.

P(red marble drawn from B|red marble drawn from A) - if a red marble is drawn from A, bag B now has 7 red and 4 black marbles. Then, the probability of drawing a red marble from bag B will be 7/11

P(red marble drawn from A) - as 4 of 10 marbles in A are red, the probability of drawing a red marble is 4/10 = 2/5

P(red marble drawn from B|black marble drawn from A) - if a black marble is drawn from A, bag B now has 6 red and 5 black marbles. Then, the probability of drawing a red marble from bag B will be 6/11

P(black marble drawn from A) - as 6 of 10 marbles in A are black, the probability of drawing a black marble is 6/10 = 3/5

Then, P(red marble drawn from B) = 7/11 * 2/5 + 6/11 * 3/5 = (14 + 18)/55 = 32/55

Note how this is slightly less than 3/5 = 6/10, which is what it would have been if there had not been a bag A involved.