Quote:
Originally Posted by eshmrt
Can you elaborate on the hint for this problem? It seems to me that in your alternative problem you aren't using Bayes' Theorem. Wouldn't using Bays' dictate that you would say P[AB] stands for the probability that the second ball is black assuming that the first ball is black, which in your problem would be 100% and P[B] stands for the probability that the first ball is black, which would be 50% because either the bag with two white balls or the bag with two black balls could be chosen. Therefore P[AB]P[B] = 1/2.
I understand your result intuitively. I just don't understand how Bayes' Theorem is being applied to arrive at that result as direction by the hint in the textbook.

Your calculation is correct, but what you are calculating is the joint probability. What the problem asks for is the conditional probability. You got both right.
In the original problem, the conditional probability is not as straightforward to compute as in the simplified case here. One way to use Bayes' theorem in the original problem is to determine the joint P(A,B) for all combinations of A and B, and from that you can calculate all other probabilities including the conditional probability.