An urn contains 5 black, 5 red, and 5 white balls. 3 balls are chosen without
replacement at random. We are asked to find the probability of getting 2 black and 1 red ball.

One way to solve it: $\frac{\binom{5}{2}\*\binom{5}{1}}{\binom{15}{3}} \approx 0.109$

Another way is to use conditional probability. We have:

P(2 black $\cap$ 1 red) = P(2 black | 1 red) \* P(1 red) = $\frac{\binom{5}{2}}{\binom{14}{2}} \* \frac{5}{15} \approx 0.036$. Where did I forget to multiply by 3? What did I do wrong? I feel stupid