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# Alice, Benjamin, and Carol each try independentl

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Intern
Joined: 19 Oct 2013
Posts: 10

Kudos [?]: 13 [1], given: 13

Location: United States
Concentration: Finance, Technology
GMAT Date: 11-06-2013
GPA: 3.5
WE: Engineering (Investment Banking)
Alice, Benjamin, and Carol each try independentl [#permalink]

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31 Oct 2013, 11:18
1
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5
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Difficulty:

45% (medium)

Question Stats:

74% (01:30) correct 26% (01:23) wrong based on 167 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

Kudos [?]: 13 [1], given: 13

Math Expert
Joined: 02 Sep 2009
Posts: 41871

Kudos [?]: 128507 [1], given: 12179

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

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31 Oct 2013, 11:25
1
KUDOS
Expert's post
3
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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.

_________________

Kudos [?]: 128507 [1], given: 12179

Intern
Joined: 14 Aug 2013
Posts: 35

Kudos [?]: 92 [0], given: 4

Location: United States
Concentration: Finance, Strategy
GMAT Date: 10-31-2013
GPA: 3.2
WE: Consulting (Consumer Electronics)
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

Kudos [?]: 92 [0], given: 4

Senior Manager
Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 437

Kudos [?]: 139 [0], given: 169

Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
Alice, Benjamin, and Carol each try independentl [#permalink]

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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

Kudos [?]: 139 [0], given: 169

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16763

Kudos [?]: 273 [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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Kudos [?]: 273 [0], given: 0

Director
Joined: 12 Nov 2016
Posts: 784

Kudos [?]: 35 [0], given: 164

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

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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

Kudos [?]: 35 [0], given: 164

e-GMAT Representative
Joined: 04 Jan 2015
Posts: 746

Kudos [?]: 2069 [1], given: 123

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

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16 Apr 2017, 21:16
1
KUDOS
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

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
Quant Expert
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Kudos [?]: 2069 [1], given: 123

Re: Alice, Benjamin, and Carol each try independentl   [#permalink] 16 Apr 2017, 21:16
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