Conditional probabilitiy in balls in urn

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?

• You have good instincts. That is an okay way to answer a multi-choice exam question. Still, as you say, it is not rigorous and it is best to understand the "why it works". $10/(21+10)$ is close to $1/3$, but $10/((21-1)+10)$ is equal to $1/3$, and you can't get much closer than that. So ask yourself why would you need to subtract one of the black balls from the number of black balls in the urn? – Graham Kemp Apr 11 '16 at 23:10
• can you see the answer I posted below? is it correct? – Rodrigo Stv Apr 12 '16 at 1:28