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: A: BBB___  WWWW
B: BBBB__  _WWW
C: BBBBB_  __WW
D: BBBBBB  ___W

P(w|A) = 4/7
P(w|B) = 3/7
P(w|C) = 2/7
P(w|D) = 1/7

Answer should be P(w|A)*P(A) + P(w|B)*P(B) + P(w|C)*P(C) + P(w|D)*P(D)
P(A): remove 1st black ball (p=6/10); remove second black ball (p=6/10 * 5/9); remove third black ball (p = 6/10 * 5/9 * 4/8)= 120/720
P(B): remove 2 black and one white; also 120/720
P(C): 6/10 * 4/9 * 3/8 = 72/720
P(D): 4/10 * 3/9 * 2/8 = 24/120
Doing the math gives me 0.239. 
A: 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 $\frac{4}{7} * \frac{20}{120} + \frac{3}{7} * \frac{60}{120} + \frac{2}{7} * \frac{36}{120} + \frac{1}{7} * \frac{4}{120} = \frac{2}{5}$
A: You don't actually have to do any calculating.  Since no information was added by removing the balls, the probability can't change. No matter how many balls we remove, the probability that the next ball we choose will always be 2/5.  
A: Think of it this way:
Take the urn with 10 balls as it is.
Draw three, put them to the side. These are the removed ones. Now draw a fourth one. 
It should have a probability of 40% to be white, because each ball has a 40% probability to be white and we do not look at the colors of the balls removed earlier. 
You could have that the three balls that were removed initially all were white. This would then greatly increase your chance of getting a black one with the fourth draw. However, they could also have all been black, thereby increasing your chance of getting a white one with the fourth draw. It can be proven that these will cancel out and the probability stays unchanged, if you do the math correctly.
As a matter of fact, since the three balls are selected randomly, would you intuitively think that there should be a change in expected color of the fourth one after you took out three balls without looking at their colors?
