Alice, Benjamin, and Carol each try independentl

Question

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

A. 3/140
B. 1/28
C. 3/56
D. 3/35
E. 7/40

Bunuel · Accepted Answer

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.

BrentGMATPrepNow · Answer

P(exactly 2 win) = P(A wins and B wins and C loses   OR   B wins and C wins and A loses   OR   A wins and C wins and B loses)
= P(A wins and B wins and C loses)   +   P(B wins and C wins and A loses)   +   P(A wins and C wins and B loses)

Let's calculate each probability

P(A wins and B wins and C loses) = P(A wins) x P(B wins) x P(C loses)
= 1/5 x 3/8 x 5/7
= 15/280

P(B wins and C wins and A loses) = P(B wins) x P(C wins) x P(A loses)
= 3/8 x 2/7 x 4/5
= 24/280

P(A wins and C wins and B loses) = P(A wins) x P(C wins) x P(B loses)
= 1/5 x 2/7 x 5/8
= 10/280

So, P(exactly 2 win) = 15/280   +   24/280   +   10/280
= 49/280
= 7/40

Answer: E

Cheers,
Brent

kusena · Answer

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

pacifist85 · Answer

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

Nunuboy1994 · Answer

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

EgmatQuantExpert · Answer

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 e-GMAT Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts

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EMPOWERgmatRichC · Answer

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

Final Answer:  E

GMAT assassins aren't born, they're made, 
Rich

suramya26 · Answer

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

ScottTargetTestPrep · Answer

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

Answer: E

itisSheldon · Answer

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.

hagaip · Answer

Hi, I have a question about a different approach to solving this.
Why not calculate it this way P(2 wins and 1 lose) = 1 - P(3 wins)-P(3 losses)?
I tried doing that and im getting wrong answer and I don't understand why this approach is not good, I would appreciate any insight here.
Thanks

EMPOWERgmatRichC · Answer

Hi hagaip,

The approach that you are considering is not particularly efficient because you have to calculate a lot more than what you are considering. The full equation for that idea would be:
 
1 - P(all 3 win) - P(all 3 lose) - P(of the 3 different ways that just 1 person wins).

That would end up being 5 different calculations that you would THEN have to subtract from "1." It's far easier to calculate the three different options that give us what we 'want' (2 win and 1 loses) and then add the three calculations together.
 
GMAT assassins aren't born, they're made,
Rich

Basshead · Answer

P(A win, B win, C lose) = 1/5 * 3/8 * 5/7 = 24/280
P(A win, B lose, C win) = 1/5 * 5/8 * 2/7 = 10/280
P(A lose, B win, C win) = 4/5 * 3/8 * 2/7 = 15/280

= 49/280 = 7/40

kittle · Answer

Hi is the order not important in this question is A wins, C loses and B wins same as A wins, Bwins and C loses?

EMPOWERgmatRichC · Answer

Hi kittle,

The prompt doesn't state anything about the 'order' of the outcomes mattering (not in the specific question that is asked nor in the details about the order in which the 3 individuals play the game), so the order of the events does not matter. I listed the 3 possible 'acceptable' outcomes in alphabetical order for organizational purposes. 
 
GMAT assassins aren't born, they're made,
Rich

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