A box contains 1 red balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black? Choices:
A) 1/4	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 1 Red Balls be R   They can be arranged in 6 slots _ _ _ _ _ _ _ _   in (6!)/ (5!x1!)   If the fourth slot is Black ball then the arrangement will be to fill   _ _ _ B _ _ _ _   we have 5 slots and 4 Black (BBBB) and 1 Red (R)   They can be arranged in (5!)/ (4!x1!)   Hence required probability = [(6!)/ (5!x1!)]/[(5!)/ (4!x1!)]   =6/5   1/2
<result>C</result>
A box contains 1 red balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?