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Alice, Benjamin, and Carol each try independentl

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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:

  45% (medium)

Question Stats:

76% (02:50) correct 24% (02:42) wrong based on 50 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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Re: Alice, Benjamin, and Carol each try independentl [#permalink] New post 31 Oct 2013, 10:25
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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.

Answer: E.
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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
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Kudos [?]: 47 [0], given: 105

Alice, Benjamin, and Carol each try independentl [#permalink] New post 05 Mar 2015, 10:30
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
Alice, Benjamin, and Carol each try independentl   [#permalink] 05 Mar 2015, 10:30
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