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?	Pick:
A) 1 / 4
B) 7 / 5
C) 1 / 2
D) 5 / 8
E) 2 / 3

Rationale: "my complicated version of your simple approach let the 5 black balls be bbbbb and 2 red balls be rrr they can be arranged in 8 slots _ _ _ _ _ _ _ _ in ( 7 ! ) / ( 5 ! x 2 ! ) if the fourth slot is black ball then the arrangement will be to fill _ _ _ b _ _ _ _ we have 6 slots and 4 black ( bbbb ) and 2 red ( rrr ) they can be arranged in ( 7 ! ) / ( 4 ! x 2 ! ) hence required probability = [ ( 7 ! ) / ( 5 ! x 2 ! ) ] / [ ( 6 ! ) / ( 4 ! x 2 ! ) ] = 7 / 5 b"
Answer: B