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# Alice, Benjamin, and Carol each try independentl

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Intern
Joined: 19 Oct 2013
Posts: 9
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, 10:18
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Difficulty:

45% (medium)

Question Stats:

72% (01:25) correct 28% (01:32) wrong based on 315 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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Joined: 02 Sep 2009
Posts: 51100
Re: Alice, Benjamin, and Carol each try independentl  [#permalink]

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31 Oct 2013, 10:25
3
2
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.

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Concentration: Finance, Strategy
GMAT Date: 10-31-2013
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Re: Alice, Benjamin, and Carol each try independentl  [#permalink]

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31 Oct 2013, 10:32
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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05 Mar 2015, 10:30
1
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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16 Apr 2017, 19:35
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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Posts: 2279
Re: Alice, Benjamin, and Carol each try independentl  [#permalink]

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16 Apr 2017, 20:16
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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05 Feb 2018, 11:33
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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Alice, Benjamin, and Carol each try independentl  [#permalink]

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01 Jul 2018, 06:29
Bunuel wrote:
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.

Hi Bunuel,
Thanks for the explanation.
I am repeatedly getting confused when to arrange data and when not to while finding out the probability.

Here we just found out the probability of winning and loosing and multiplied it because they were independent events but we didn't arrange them because order of winning does not matter it.
But in some of the questions, say the one below, we are also arranging the probability for one event.

If a fair 6-sided die is rolled three times, what is the probability that exactly one 3 is rolled?

25/216
50/216
25/72
25/36
5/6

So how to determine whether the arrangement is required or not????

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

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04 Jul 2018, 18:07
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

We must individually consider each possible outcome of having two winners and one loser.

If Alice and Benjamin win and Carol loses we have:

1/5 x 3/8 x 5/7 = 1 x 3/8 x 1/7 = 3/56

If Alice and Carol win and Benjamin loses we have:

1/5 x 5/8 x 2/7 = 1 x 1/8 x 2/7 = 2/56

If Benjamin and Carol win and Alice loses we have:

4/5 x 3/8 x 2/7 = 1/5 x 3/2 x 2/7 = 1/5 x 3 x 1/7 = 3/35

Therefore, the probability that two of them will win and one will lose is:

3/56 + 2/56 + 3/35 = 15/280 + 10/280 + 24/280 = 49/280 = 7/40

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

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10 Dec 2018, 07:13
Veritas Prep OFFICIAL EXPLANATION

First, understand that the probability that someone will lose equals 1 minus the winning probability. For example, the probability that Alice will lose is 1 - 1/5, or 4/5. The scenarios in which there are two winners and one loser are: AB win and C loses; AC win and B loses; BC win and A loses. For each of these scenarios, compute the probability by multiplying the winners' probabilities together with the loser's probability, which will be 1-winning probability.

AB win and C loses: 1/5 * 3/8 * 5/7 = 15/280. AC win and B loses: 1/5 * 2/7 * 5/8 = 10/280. BC win and A loses: 3/8 * 2/7 * 4/5 = 24/280. Notice that answers (B), (C), and (D) are these partial answers in reduced form. Don't reduce, yet, because now we must combine these partial answers. Either AB will win or AC will win or BC will win. The question indicates that we should add these probabilities together. 15/280 + 10/280 + 24/280 = 49/280, or 7/40.
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# Alice, Benjamin, and Carol each try independentl

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