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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Alice, Benjamin, and Carol each try independently to win a tagged by: AAPL ##### This topic has 2 expert replies and 0 member replies ### Top Member ## Alice, Benjamin, and Carol each try independently to win a ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult Veritas Prep 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. 3/140 B. 1/28 C. 3/56 D. 3/35 E. 7/40 OA E ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12975 messages Followed by: 1249 members Upvotes: 5254 GMAT Score: 770 AAPL wrote: Veritas Prep 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. 3/140 B. 1/28 C. 3/56 D. 3/35 E. 7/40 OA E P(exactly 2 win) = P(A wins and B wins and C loses OR B wins and C wins and A loses OR A wins and C wins and B loses) = P(A wins and B wins and C loses) + P(B wins and C wins and A loses) + P(A wins and C wins and B loses) Let's calculate each probability P(A wins and B wins and C loses) = P(A wins) x P(B wins) x P(C loses) = 1/5 x 3/8 x 5/7 = 15/280 P(B wins and C wins and A loses) = P(B wins) x P(C wins) x P(A loses) = 3/8 x 2/7 x 4/5 = 24/280 P(A wins and C wins and B loses) = P(A wins) x P(C wins) x P(B loses) = 1/5 x 2/7 x 5/8 = 10/280 So, P(exactly 2 win) = 15/280 + 24/280 + 10/280 = 49/280 = 7/40 Answer: E Cheers, Brent _________________ Brent Hanneson â€“ Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2791 messages Followed by: 18 members Upvotes: 43 AAPL wrote: Veritas Prep 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. 3/140 B. 1/28 C. 3/56 D. 3/35 E. 7/40 OA E We must individually consider each possible outcome of having two winners and one loser. If Alice and Benjamin win and Carol loses, we have: 1/5 x 3/8 x 5/7 = 1 x 3/8 x 1/7 = 3/56 If Alice and Carol win and Benjamin loses, we have: 1/5 x 5/8 x 2/7 = 1 x 1/8 x 2/7 = 2/56 If Benjamin and Carol win and Alice loses, we have: 4/5 x 3/8 x 2/7 = 1/5 x 3/2 x 2/7 = 1/5 x 3 x 1/7 = 3/35 Therefore, the probability that two of them will win and one will lose is: 3/56 + 2/56 + 3/35 = 15/280 + 10/280 + 24/280 = 49/280 = 7/40 Answer: E _________________ Scott Woodbury-Stewart Founder and CEO scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews • Free Practice Test & Review How would you score if you took the GMAT Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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