urn with two colors of balls; probability of selecting specific color 10 balls in an urn, 6 Black and 4 White. Three are removed, color not noted. What is probability that a white ball will be chosen next?
The answer is 2/5, so my reasoning below must be faulty.
After the initial three balls are removed, there will be 4 possible configurations:
You calculated $P(A)$, $P(B)$, $P(C)$ and $P(D)$ incorrectly. A can happen in $\binom{6}{3} = 20$ ways, B in $\binom{6}{2} * \binom{4}{1} = 60$ ways, C can happen in $\binom{6}{1} * \binom{4}{2} = 36$ ways, D can happen in $4$ ways. To check, there are $20+60+36+4=120$ total ways of removing $3$ balls at random, which is $\binom{10}{3}$.
The answer is then $\frac{4}{7} * \frac{20}{120} + \frac{3}{7} * \frac{60}{120} + \frac{2}{7} * \frac{36}{120} + \frac{1}{7} * \frac{4}{120} = \frac{2}{5}$
