Question: Urn A contains 6 green and 4 red marbles, and urn B contains 2 green and 7 red marbles. A marble is to be selected at random from A and placed in B. One marble is then selected from B. Given that the marble selected from B is green, what is the probability that the marble selected from A will also be green?

Solution:

Probability of marble selected from B is green, if the marble selected from A and placed in B is green = 6/10 x (2+1)/(9+1) = 6/10 x 3/10 = 18/100. Probability of marble selected from B is green; if the marble selected from A and placed in B is red = 4/10 x 2/9+1 = 4/10 x 2/10 = 8/100.

The joint probability of green marble selected from B = 18/100 + 8/100 = 26/100. The probability, given that the marble selected from B is green, the marble selected from A will also be green.

=18/100 / 26/100 = 18/26.