Marbles in the Bag

$\dfrac{2}{3}$

We know that we do not have Bag B (two black marbles) so there are three possibilities

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.

P(A or B)=P(A and B)/P(B)

A = 2nd white

B= 1st white

P(A and B)=1/3 only one bag allows this

P(B)=(1/3) (1/2)+(1/3) 1+(1/3)*0=1/6+1/3=1/2

P(A or B)=P(A and B)/P(B)=(1/3)/(1/2)

P(A or B)=2/3

2+3=5

You choose one of the bags at random, randomly remove a marble and then remove the second marble.

Imagine repeating this experiment 6n times, and letting n tend to infinity. It is easy to see that, in the limit

2n times you will choose the black bag and withdraw (black,black)

2n times you will choose the mixed bag, and so

n times you will withdraw (black,white) and

n times you will withdraw (white,black)

2n times you will choose the white bag and so withdraw (white,white)

Given that the first ball withdrawn is white, one of the last two options must have occurred, and we can see that (white,white) is twice as likely as (white,black) and so

$P(white,white)=\dfrac{2}{3}$

$P(white,black)=\dfrac{1}{3}$

and the answer is $2+3=\boxed{5}$