Probablity: 2 Urns I'm trying to find the solution for this problem:

There are 2 urns: urn 1 has 2 red balls and 1 blue ball and urn 2 has
  1 red ball and 2 blue balls. You're supposed to randomly select one
  urn and put your hand inside it to retrieve a ball. 
What is the probability you chose urn2, if a blue ball was retrieved?

Though I have been unable to solve it, I have tried reasoning it out like this:

There are 6 total possibilities since there are 6 balls. 
  There are 2 blue  balls in urn 2.
  There are 2 urns.
  Therefore: (2 blue balls / 6 total balls) * 2 urns = 2/3

I would appreciate any help. Thanks.
 A: Your reasoning is pretty good. Here's a more systematic way to see it.
The expression that you want to calculate is
$$\frac{\textrm{Probability of picking the 2nd urn and picking a blue ball}}{\textrm{Probability of picking a blue ball}} $$
Since there are equal numbers of balls in each urn, and each urn is equally likely to be picked, you can do this by counting. There are 3 ways that you can pick a blue ball. Of those 3 ways, 2 of them involve picking a blue ball from the 2nd urn. Since all possibilities are equally likely, the probability of the ball being from the 2nd urn, given that you chose a blue ball, is
$$\frac{\textrm{Number of ways to pick a blue ball from the 2nd urn}}{\textrm{Number of ways to pick a blue ball}} = \frac{2}{3}$$
Alternatively, you can apply Bayes' theorem. Letting $B$ be the event 'You pick a blue ball' and $2$ be the event 'you pick from the 2nd urn', then you have
$$P(B) = \frac{1}{2}\cdot \left( \frac{1}{3} + \frac{2}{3} \right) = \frac{1}{2}$$
$$P(2) = \frac{1}{2}$$
$$P(B|2) = \frac{2}{3}$$
and hence
$$P(2|B) = \frac{P(2)}{P(B)}\cdot P(B|2) = \frac{1/2}{1/2}\cdot\frac{2}{3} = \frac{2}{3}$$
A: 
This is a very basic and a previous question from IIT(1976). It uses Bayes theorem. Please go through the diagram to understand the problem.
