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Alice, Benjamin, and Carol each try independentl

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

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New post 31 Oct 2013, 11:18
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A
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Question Stats:

72% (01:23) correct 28% (01:24) wrong based on 235 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
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Re: Alice, Benjamin, and Carol each try independentl [#permalink]

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New post 31 Oct 2013, 11:25
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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]

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New post 31 Oct 2013, 11:32
1
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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Alice, Benjamin, and Carol each try independentl [#permalink]

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New post 05 Mar 2015, 11:30
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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
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Re: Alice, Benjamin, and Carol each try independentl [#permalink]

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New post 16 Apr 2017, 20:35
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



In order to solve this question we should set up the equation

Prob (Exactly #success Exactly # failures)= P(A success x B success x C failure) + P(A success B failure C success) + P(A failure x B success x C success)
Prob(Exactly # 2 success and Exactly 1 Failure)= P( 1/5 x 3/8 x 5/7) + P(1/5 x 5/8 x 2/7) + P( 4/5 x 3/8 x 2/7) = 49/280

Thus
7/40
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Re: Alice, Benjamin, and Carol each try independentl [#permalink]

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New post 16 Apr 2017, 21:16
1
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


To solve this question we need to consider three different cases -

    • A and B wins and C loses
    • The probability of the above case can be written as -
      o \(P(A_wB_wC_L) = \frac{1}{5} * \frac{3}{8} * (1-\frac{2}{7}) = \frac{3}{56}\)
    • B and C wins and A loses
    • The probability of the above case can be written as -
      o \(P(A_LB_wC_w) = (1-\frac{1}{5}) * \frac{3}{8} * \frac{2}{7} = \frac{3}{35}\)
    • C and A wins and B loses
    • The probability of the above case can be written as -
      o \(P(A_wB_LC_w) = \frac{1}{5} * (1-\frac{3}{8}) * \frac{2}{7} = \frac{1}{28}\)
    • The overall probability \(= P(A_wB_wC_L) + P(A_LB_wC_w) + P(A_wB_LC_w)\)
      \(= \frac{3}{56} + \frac{3}{35} + \frac{1}{28}\)
      \(= \frac{3}{56} + \frac{3}{35} + \frac{1}{28}\)
      \(= \frac{(15+24+10)}{280}\)
      \(=\frac{7}{40}\)


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

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New post 05 Feb 2018, 12:33
Hi All,

Since we have the probabilities of each person winning a game, we can 'map out' the 3 situations in which 2 of them win and 1 of them loses:

A wins, B wins, C loses = (1/5)(3/8)(5/7) = 15/240
A wins, B loses, C wins = (1/5)(5/8)(2/7) = 10/240
A loses, B wins, C wins = (4/5)(3/8)(2/7) = 24/240

Total probability = 15/280 + 10/280 + 24/280 = 49/280 = 7/40

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Re: Alice, Benjamin, and Carol each try independentl   [#permalink] 05 Feb 2018, 12:33
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