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.