Q:

Conditional Probability Question (Marbles)

I am having an issue with a conditional probability question. The question is:
Suppose you have three bags containing only red marbles and blue marbles. Bag one has two red marbles and four blue marbles, bag two has eight red marbles and four blue marbles, and bag three has one red marble and three blue marbles. I pick one marble (at random) from each bag.
(a) What is the probability that I picked exactly two red marbles? 
(b) Suppose I picked two red marbles but forgot which bag they came from. What is the probability that I picked a red marble from bag one?
For part (a), I have the following possibilities - where r = red marble and b = blue marble:
RRB with a probability of 1/6
BRR with a probability of 1/9
RBR with a probability of 1/36 
P(exactly 2 red marbles) = (1/6)+(1/9)+(1/36) = 11/36 
For part (b), I understand that this is where conditional probability is involved. I defined E to be the event where a red marble is picked from bag one and E' to be the event where two apples have been picked.
P(E|E')=P(EnE')/P(E') = ____/(11/36). 
I guess that I am having trouble with the numerator, P(EnE'). I have to find the probability of picking a red marble from bag one given that two red marbles where picked. I am torn between the following choices:
(i) P(EnE') = (1+6)+(1/9); which was found in part (a) of the problem
(ii) P(EnE') = (2/3); I figured this would be a possibility since there were three possible outcomes (RRB, BRR, RBR) where two of the three choices have a red marble chosen from bag one.
Any help would be greatly appreciated!

A:

For part a, correct.
For part b, $\frac{P(RRB)+P(RBR)}{P(RRB)+P(RBR)+P(BRR)}\ \ \ \ \ \ \ \ =\frac{\frac{1}{6}+\frac{1}{36}}{\frac{1}{6}+\frac{1}{36}+\frac{1}{9}}=\frac{7}{11}$
Hint: $P(A\ and\ B)=P(A|B)P(B)$ with $A$ as red ball from bag #1 and $B$ as $2$ red balls.

