Quote:
Originally Posted by leduran
Hi, think of this problem in the following way: if the first ball withdrawn was black , then for the second ball to be also black it must come from the bag containing the two black balls. So this problem is equivalent to finding the probability that the bag with the two black balls was choosen given that the first withdrawn ball was black.
Call Bag1 the event of choosing the bag with the 2 black balls, and B the event that the first ball is black, then:
P(bag1B) = P(Bbag1)*P(bag1)/P(B) = P(Bbag1)*P(bag1)/P(Bbag1)*P(bag1)+P(Bbag2)*P(bag2)
= (1)(0.5)/(1)(0.5)+(0.5)(0.5)=2/3
Hope it helps

leduran's reply doesn't have enough parentheses, which has to be corrected