Probability of choosing the basket-i Here is the question:
There are N+1 baskets 0,1,2,3,...,N. Where each basket i has i white balls and N-i black balls.
We choose randomly a basket and take out a ball after another with returning.
a)If we know that the first two withdraws we got a black and a white ball (not necessarily by order) what is the probability that we have chosen the basket i?for each i?
b)If we know that the first two withdraws were black balls what is the probability that the third withdraw will be black?

Ok,this is an a question in my homework, here is my approach:
a)for i equal 0/N the probability is zero (there is no white/black balls), for any other i the probability is 1/(N-1).
I am not sure about the answer,should we consider that we have picked two different balls, does that effect the probability of which basket we choose?
I know the sum of probabilities for the i-es should equal 1. Which my answer does satisfy.every other answer did not.
b)for b it is the same,will the the first two withdraws effect the third one? I think it does only tell that we have not picked the N-th basket no more.
my answer for b is 1/2.
I would like some hints...
Thank you guys, and sorry english is not my first language.

ok,here is my second approach:
a)$\frac{P(B|A)*P(A)}{P(B)} $= $ \frac{2*( \frac{i}{N} *  \frac{N-i}{N})*( \frac{1}{N+1}) }{2*\sum_{i=1}^n a_n(\frac{k}{N}*\frac{N-k}{N})*\frac{1}{N+1}} $ =$ \frac{ \frac{i}{N} *  \frac{N-i}{N}}{\sum_{i=1}^n a_n(\frac{k}{N}*\frac{N-k}{N})} $
I don't know how to get rid of the sigma
b)P(third withdraw is black|first two withdraws are black)=$ \frac{P(the three withdraws are black)}{P(the first two withdraws are black)} $=
$\frac{\sum_{i=0}^N \frac{i^3}{N^3}}{\sum_{i=0}^N \frac{i^2}{N^2}}$=$\sum_{i=0}^N \frac{\frac{i^3}{N}}{\sum_{i=0}^N i^2}$
here too I could'nt get rid of i from the whole equation.any help please?
