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# Alice, Benjamin, and Carol each try independentl

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Intern
Joined: 19 Oct 2013
Posts: 10
Location: United States
Concentration: Finance, Technology
GMAT Date: 11-06-2013
GPA: 3.5
WE: Engineering (Investment Banking)
Followers: 0

Kudos [?]: 7 [0], given: 13

Alice, Benjamin, and Carol each try independentl [#permalink]

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31 Oct 2013, 11:18
00:00

Difficulty:

45% (medium)

Question Stats:

78% (02:25) correct 22% (02:21) wrong based on 91 sessions

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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
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 34393
Followers: 6245

Kudos [?]: 79323 [1] , given: 10016

Re: Alice, Benjamin, and Carol each try independentl [#permalink]

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31 Oct 2013, 11:25
1
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Puneethrao wrote:
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

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.

_________________
Intern
Joined: 14 Aug 2013
Posts: 35
Location: United States
Concentration: Finance, Strategy
GMAT Date: 10-31-2013
GPA: 3.2
WE: Consulting (Consumer Electronics)
Followers: 2

Kudos [?]: 60 [0], given: 4

Re: Alice, Benjamin, and Carol each try independentl [#permalink]

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31 Oct 2013, 11:32
Puneethrao wrote:
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

probability that exactly two of the three players will win = probability of (A will win,B will win,C will lose+ A will win , B will lose ,C will win+ A will lose,B will win,C will win)
=>(1/5 * 3/8 * (1- 2/7)) + (1/5 * (1 - 3/8) * 2/7) + ((1 - 1/5) * 3/8 * 2/7)
=>15/280 + 10/280 + 24/280
=> 49/280
=>7/40
Senior Manager
Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 443
Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
Followers: 2

Kudos [?]: 92 [0], given: 169

Alice, Benjamin, and Carol each try independentl [#permalink]

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05 Mar 2015, 11:30
1
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Well. there are three possible outcomes of interest, if 2 of the three have to win and one to lose:

AB...C
AC...B
CB...A

For the wins, we use the given probabilities. For the loss we use the remaining of the given probability. We multiply these together:

1/5 * 3/8 * 5/7 = 15 / 280
1/5 * 2/7 * 5/8 = 10 / 280
3/8 * 2/7 * 4/5 = 24 / 280

We now want to add these individual probabilities, which gives us: 49 / 280 = 7 / 40 ANS E
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11029
Followers: 509

Kudos [?]: 133 [0], given: 0

Re: Alice, Benjamin, and Carol each try independentl [#permalink]

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29 Mar 2016, 08:29
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Alice, Benjamin, and Carol each try independentl   [#permalink] 29 Mar 2016, 08:29
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