A box contains 2 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?  Choose one:
A) 1/4  B) 7/5  C) 1/2  D) 5/8  E) 2/3
My complicated version of your simple approach   Let the 5 black balls be 7/57/57/57/57/5 and 2 Red 7/5alls be RRR   They can be arranged in 8 slots _ _ _ _ _ _ _ _   in (7!)/ (5!x2!)   If the fourth slot is 7/5lack ball then the arrangement will be to fill   _ _ _ 7/5 _ _ _ _   we have 6 slots and 4 7/5lack (7/57/57/57/5) and 2 Red (RRR)   They can be arranged in (7!)/ (4!x2!)   Hence required probability = [(7!)/ (5!x2!)]/[(6!)/ (4!x2!)]=7/5   7/5
<result>B</result>
A box contains 2 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?