We need to find the probability that all the red balls are exhausted before the white or black balls are exhausted. So, we can still pick white or black balls, but we cannot pick ALL of the white/black balls.
I tried going case by case but quickly realized there are way too many cases for that. I'm not very good at permutations and combinations so I can't figure out how to get in all the cases.
Few example cases can be: R R R W B W B R, B B B W B R R R R, etc.
Best Answer
Note that each ball has an equal chance of appearing last.
We distinguish two cases, according to the color of the last ball drawn.
Case I: Last ball white (probability $\frac 3{12}=\frac 14$).
Now, consider the cards other than the last (white) one. We want the conditional probability that the reds were exhausted before the blacks, at this point the whites are irrelevant. There are $4$ reds and $5$ blacks in that collection, so the probability that the last of those to be drawn was black is $\frac 59$. Thus the probability of success with a white final ball is $$\frac 14\times \frac 59=\frac 5{36}$$
Case II: Last ball black (probability $\frac 5{12}$).
As before we get that the probability of success with a black final ball is $$\frac 5{12}\times \frac 37=\frac {5}{28}$$
The final result is the sum $$\boxed {\frac 5{36}+\frac {5}{28}=\frac {20}{63}}$$
Sanity Check: if there were $4$ of each color, then symmetry would apply and the answer would be $\frac 13$. As the distribution is close to uniform, we'd expect the answer to be close to $\frac 13$, as it is. Note: I find it hard to have intuition as to whether the result should be greater or less than $\frac 13$. Red is more likely to finish first than Black is, but White is more likely still.