TITLE: From an urn containing a white and b black balls, balls are successively drawn without replacement until only those of the same colour are left.
QUESTION [5 upvotes]: From an urn containing a white and b black balls, balls are successively drawn without replacement until only those of the same colour are left. What is the probability that balls left are white?
This is what I have done.
(https://i.stack.imgur.com/CZH2x.jpg)
But the answer which is given is : 
a/(a+b)
I can't possibly make out how it's happening.

REPLY [5 votes]: Its the number of ways a white ball is the last ball to be chosen divided by the total number of ways of arranging all the balls.
This is:
$$p = \frac{\frac{(a+b-1)!}{(a-1)!b!}}{\frac{(a+b)!}{a!b!}} = \frac{(a+b-1)!}{(a-1)!b!}\cdot \frac{a!b!}{(a+b)!} = \frac{a}{a+b}$$