Antony, Ben, and Charles each try independently to win a carnival game. If their individual probabilities for success are 2/5, 3/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose?
Options:
(A) 3/140
(B) 1/280
(C) 3/280
(D) 3/35
(E) 68/280 Stream of consciousness first, then make a decision:
P = P(A wins, B wins, C loses) + P(A wins, B loses, C wins) + P(A loses, B wins, C wins) = 2/5*3/8*5/7 + 2/5*5/8*2/7 + 3/5*3/8*2/7 = 68/280.
Thus, the answer is (E).