Conditional probability: The second ball is black. Find the probability that the first ball was white. Two black balls and three white balls are put in a bag. First, we pull one ball, then the second. The second ball is black. Find the probability that the first ball was white.
I think it could be $\frac{3*2}{3*2+2*1} = 0.75$
as we have $m=3∗2$ (WB - what we need) and $n=3∗2+2∗1$ (WB + BB - all possible options)

Your answer is correct
Another way of getting the same answer is to say that, given the second draw without replacement is white,  there are three black balls and one white ball for the non-second draws, and these are equally likely to be any position making the conditional probability that the first draw is black $\frac34$
