 <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. <yor> Ìbéèrè: Ọ̀kọ̀ọ̀kan Antony, Ben, àti Charles ló ń gbìyànjú láti borí nínú eré ayẹyẹ kan. Tó bá jẹ́ pé ìdánilójú pé wọ́n á ṣàṣeyọrí jẹ́ 1/5, 3/8, àti 2/7, kí ni ìdánilójú pé méjì nínú àwọn ọ̀dọ́kùnrin mẹ́ta náà ló máa ṣẹ́gun, àmọ́ ọ̀kan nínú wọn ló máa ṣẹ́gun? Àwọn àyè: (A) 3/140 (B) 1/28 (C) 7/40 (D) 3/35 (E) 5/40 Ìṣàn ìmọ̀lára: P = P ((A gba, B gba, C pàdánù) + P ((A gba, B pàdánù, C gba) + P ((A pàdánù, B gba, C gba) = 1/5*3/8*5/7 + 1/5*5/8*2/7 + 4/5*3/8*2/7 = 7/40. 