 
 <translate> Question: Antony, Ben, and Charles 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?
Options:
(A) 3/140
(B) 1/28
(C) 7/40
(D) 3/35
(E) 5/40
Stream-of-consciousness: P = P(A wins, B wins, C loses) + P(A wins, B loses, C wins) + P(A loses, B wins, C wins) = 1/5*3/8*5/7 + 1/5*5/8*2/7 + 4/5*3/8*2/7 = 7/40. <ibo> Ajụjụ: Antony, Ben, na Charles na-agbalị n'otu n'otu iji merie n'egwuregwu a na-eme n'ememme. Ọ bụrụ na ohere onye ọ bụla n'ime ha nwere inwe ihe ịga nke ọma bụ otu ụzọ n'ụzọ ise, atọ ụzọ n'ụzọ asatọ, na abụọ ụzọ n'ụzọ asaa, nke ọ bụla n'ime ha, olee otú o si bụrụ na mmadụ abụọ n'ime mmadụ atọ ahụ ga-emeri, ma otu onye ga-emeri? Nhọrọ: (A) 3/140 (B) 1/28 (C) 7/40 (D) 3/35 (E) 5/40 Mgbasa-uche: P = P ((A na-emeri, B na-emeri, C na-efunahụ) + P ((A na-emeri, B na-efunahụ, C na-emeri) + P ((A na-efunahụ, B na-emeri, C na-emeri) = 1/5*3/8*5/7 + 1/5*5/8*2/7 + 4/5*3/8*2/7 = 7/40. 