A box contains 3 red balls and 4 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?	Pick.
A) 7/4  B) 1/2  C) 1/2  D) 5/8  E) 2/3
My complicated version of your simple approach   Let the 5 black balls be BBBBB and 3 Red Balls be RRR   They can be arranged in 8 slots _ _ _ _ _ _ _ _   in (7!)/ (4!x3!)   If the fourth slot is Black ball then the arrangement will be to fill   _ _ _ B _ _ _ _   we have 7 slots and 3 Black (BBB) and 3 Red (RRR)   They can be arranged in (6!)/ (3!x3!)   Hence required probability = [(7!)/ (4!x3!)]/[(6!)/ (3!x3!)]   7/4ns = 7/4   7/4
<result>A</result>
A box contains 3 red balls and 4 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?