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There are w white balls and b black balls in a bowl. Randomly select a ball from the bowl and then return it to the bowl along with n additional balls of the same color. Another single ball is randomly selected from the bowl(now containing w+b+n balls) and it is black. Show that the conditional probability that the first ball selected was white is w/(w+b+n)

The attempt at a solution

This is my last question of my assignment and I can't figure out even how to get the first step. The condition we know here is an event happened afterward, so I am even confused with that. I try to list the sample points which are A(1st-w, 2nd-w), B(1st-w, 2nd-b), C(1st-b, 2nd-b), D(1st-b, 2nd-w), and the possible sample points should be B or D. Then I try the conditional probability for B:

Sample point B :

Being the first selecting:

P(w)=w/(w+b) P(b)=b/(w+b)

By sample point B, it supports that the first selected ball is white, second is black, so using the equation:

P(blw)=P(bnw)/P(w)=P(bnw)/[w/(w+b)]=b/(w+n+b)

so, P(bnw)=[w/(w+b)]*[b/(w+n+b)]

Here it already looks strange because B is just one of the sample points, but I still continue:

P(wlb)=P(wnb)/P(b)={[w/(w+b)]*[b/(w+n+b)]}/[b/(w+b)]=w/(w+b+n)

I got the answer, but I think it is not a correct process because I was just putting something into an equation with no reason. However, when I tried other ways, it even went worse. So is there anyone can give me some ideas just like how I should start to prove and that will be great. Thanks a lot.

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# Finding conditional probability

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