A box contains 3 red balls and 3 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?
Choose the correct choice from the following choices:
A) 1/4 B) 1/2 C) 1/2 D) 2 E) 2/3
My complicated version of your simple approach   Let the 3 black balls be BBB and 3 Red Balls be RRR   They can be arranged in 8 slots _ _ _ _ _ _ _ _   in (6!)/ (3!x3!)   If the fourth slot is Black ball then the arrangement will be to fill   _ _ _ B _ _ _ _   we have 5 slots and 2 Black (BB) and 3 Red (RRR)   They can be arranged in (5!)/ (2!x3!)   Hence required probability = [(6!)/ (3!x3!)]/[(5!)/ (2!x3!)]   =2   2
<result>D</result>
A box contains 3 red balls and 3 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?