Q:

Find the probability that Alice selects the second blue ball.

An urn contains 2 blue and 3 red balls.  Alice and David withdraw
  balls from the urn consecutively without replacement until the second
  blue ball is selected.  Alice draws first, then David, and so on. Find
  the probability that Alice selects the second blue ball.

I've tried to solve by finding the possible ways Alice selects the second blue ball, i.e. the third or fifth ball is blue. So we have: brb, rbb, rrrbb, rrbrb, rbrrb, brrrb. 
Not sure what to do from here. I was thinking about finding the probability for each outcome but would there be a better way to do this?
Thanks in advance!

A:

There are only $10$ possible orderings and these are equi-probable.  They are $$brrrb,brrbr,brbrr,bbrrr,rbrrb,rbrbr, rbbrr,rrbrb,rrrbbr,rrrbb$$ and it is easy to see that of these only numbers $1,3,5,7,8,10$ pass your test so the answer is $$\frac 6{10}=\boxed {\frac 35}$$

