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Alice, Benjamin, and Carol each try independentl

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Intern
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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

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Alice, Benjamin, and Carol each try independentl [#permalink] New post 31 Oct 2013, 10:18
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

80% (02:51) correct 20% (03:12) wrong based on 30 sessions
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
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Joined: 02 Sep 2009
Posts: 23411
Followers: 3614

Kudos [?]: 28942 [1] , given: 2874

Re: Alice, Benjamin, and Carol each try independentl [#permalink] New post 31 Oct 2013, 10:25
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Expert's post
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.

Answer: E.
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Intern
Intern
User avatar
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: 1

Kudos [?]: 30 [0], given: 4

GMAT ToolKit User
Re: Alice, Benjamin, and Carol each try independentl [#permalink] New post 31 Oct 2013, 10:32
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
Re: Alice, Benjamin, and Carol each try independentl   [#permalink] 31 Oct 2013, 10:32
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Alice, Benjamin, and Carol each try independentl

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