Alice, Benjamin, and Carol each try independently to win a

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AAPL: Moderator; Posts: 2599; Joined: Sun Oct 29, 2017 2:08 pm; Followed by:2 members

Alice, Benjamin, and Carol each try independently to win a

by AAPL » Mon Apr 15, 2019 5:07 am

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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

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Source: Beat The GMAT — Problem Solving | Mon Apr 15, 2019 5:07 am

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Mon Apr 15, 2019 6:11 am

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 - Creator of GMATPrepNow.com

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Tue Apr 16, 2019 4:54 pm

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

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