3 bags of 2 marbles

Problem:

You have three bags, each containing two marbles. Bag A contains two white marbles, Bag B contains two black marbles, and Bag C contains one white marble and one black marble. You pick a random bag and take out one marble. It is a white marble. What is the probability that the remaining marble from the same bag is also white?

Solution:

Reasoning:

A = {W,W}
B = {B, B}
C = {B, W}


We have already selected a bag, which has a white marble, so randomly at first we have selected bag A or C. Now we have to take the marble from the same selected bag.

Let A, B, C denotes the event that bag A, B, C is picked. Let W_k denotes the event that the kth marble picked is white. By the law of total probability and Bayes' theorem,
P(W_2|W_1)
= P(W_2|A,W_1)*P(A|W_1)+P(W_2|C,W_1)*P(C|W_1)
= P(W_2|A,W_1)*P(W_1|A)*P(A)/P(W_1)+0
= (1)(1)*(1/3)/P(W_1)
= (1/3)/P(W_1).
By the law of total probability again,
P(W_1)
= P(W_1|A)*P(A)+P(W_1|B)*P(B)+P(W_1|C)*P(C)
= (1)(1/3)+0+(1/2)(1/3) = 1/2.
Therefore
P(W_2|W_1)
= (1/3)/(1/2)
= 2/3.

In simpler terms : 
Now there are 3 probabilities :

You chose Bag A, first white marble. The other marble will be white
You chose Bag A, second white marble. The other marble will be white
You chose Bag C, the white marble. The other marble will be black

So 2 out of 3 possibilities are white.

Why not 1/2? You are selecting marbles, not bags.Because in bags A and C, we have 3 marbles left (2W and 1 B) and probability of selecting 1W ball is 2/3.

Answer:

                  2/3 (not 1/2)