Trivia

Subject:
Response To:

Larry Barrett
I also have trouble with setting up the Bayesian solution, but this is how I approached this problem.
Let A be the event of drawing two white balls from bag C, and let B be the event that the first ball drawn is a white ball.
So we want to compute P(A|B).
Bayes theorem says that P(A|B) = P(A)*P(B|A)/P(B)

Bag C could contain two white balls or it could contain one white and one black. Both are equally likely. It might help to think of bag C being either bag C1 with two white balls or bag C2 with one white ball, one black ball.
So P(A) is the probability of selecting two white balls, which would happen if you selected bag C1 with probability = 1/2.
P(B|A) is the probability of selecting one white ball, given that you already selected two white balls, so P(B|A) = 1.
P(B) is the probability of selected one white ball. This is the same as 1 - the probability of selecting one black ball. And the probability of selecting the black ball is is the probability of first selecting bag C2, which is 1/2, and then selecting the black ball from bag C2, which is also 1/2.
So P(B) = 1 - (1/2)*(1/2) = 3/4.

Therefore P(A|B) = ((1/2)*1)/3/4 = 4/6 = 2/3.