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?\nOptions:\n(A) 3/140\n(B) 1/280\n(C) 3/280\n(D) 3/35\n(E) 68/280 Stream of consciousness first, then make a decision:
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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).
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