Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 1/8, 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) 27/280\n(D) 3/35\n(E) 7/56 Think first, then make a decision. Some random thoughts:
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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) = 1/5*1/8*5/7 + 1/5*7/8*2/7 + 4/5*1/8*2/7 = 27/280. So the answer is (C).
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