Alice, Benjamin, and Carol each try independently to win a carnival game. If their individual probabilities for success are 1/5, 3/8, and 2/7, respectively, what is the probability that exactly two of the three players will win but one will lose?A. 3/140B. 1/28C. 3/56D. 3/35E. 7/40

Answer:E. 7/40Step-by-step explanation:The following probabilities are given:P (Alice Wins) = 1/5P (Benj. Wins) = 3/8P (Carol Wins) = 2/7We can deduce the probabilities for losses:P (Alice Loses) = 1 - P (Alice Wins) = 1 - 1/5 =4/5P (Benj. Loses) = 1 - P (Benj. Wins) = 1 - 3/8 = 5/8P (Carol Loses) = 1 - P (Carol Wins) = 1 - 2/7 =5/7The possible outcomes that two players win and one player loses are as follows:Alice Wins, Benj Wins, Carol LosesAlice Loses, Benj Wins, Carol WinsAlice Wins, Benj Loses, Carol WinsWe can compute the probabilities of each of the 3 outcomes above:P(Alice Wins, Benj Wins, Carol Loses) = (1/5) x (3/8) x (5/7) = 3/56Alice Loses, Benj Wins, Carol Wins = (4/5) x (3/8) x (2/7) = 3/35Alice Wins, Benj Loses, Carol Wins = (1/5) x (5/8) x (2/7) =  2/56P ( 2 wins and 1 loss) = 3/56 + 3/35 + 2/56 = 343 / 1960= 7/40