A urn has 10 white balls and N black balls.
We draw balls without replacement.
Given that the second ball drawn from the urn is black, how many black balls must be in the urn such that there is a 1/3 probability of drawing a white ball in the first draw?
A - 21 B - 25 C - 31 D - 32 E - 34
The answer is 21, and I found it by checking that 10 / (21 + 10) is the closest thing to 1/3. But I know this is not rigorously correct. What would be an exact solution?