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)

The order in which the balls were drawn is really irrelevant. Two balls are drawn. If you're told that one of them is black then the other is equally likely to be any one of the four remaining balls three of which are white.
