carnival game

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jain2016: Master | Next Rank: 500 Posts; Posts: 313; Joined: Tue Oct 13, 2015 7:01 am; Thanked: 2 times

carnival game

by jain2016 » Mon Mar 14, 2016 9:00 am

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

OAE

Hi Experts ,

Please explain.

Many thanks in advance.

SJ

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Source: Beat The GMAT — Problem Solving | Mon Mar 14, 2016 9:00 am

GMAT/MBA Expert

[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pm; Location: Palo Alto, CA; Thanked: 2867 times; Followed by:511 members; GMAT Score:800

by [email protected] » Mon Mar 14, 2016 9:11 am

Hi jain2016,

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

Contact Rich at [email protected]

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jain2016: Master | Next Rank: 500 Posts; Posts: 313; Joined: Tue Oct 13, 2015 7:01 am; Thanked: 2 times

by jain2016 » Tue Mar 15, 2016 7:01 am

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

Hi Rich ,

Thanks for your reply.

One thing why did you add all ?

Please explain.

Many thanks in advance.

SJ

Quote

GMAT/MBA Expert

[email protected]: Elite Legendary Member; Posts: 10392; Joined: Sun Jun 23, 2013 6:38 pm; Location: Palo Alto, CA; Thanked: 2867 times; Followed by:511 members; GMAT Score:800

by [email protected] » Tue Mar 15, 2016 9:05 am

Hi jain2016,

The prompt asked for the probability that EXACTLY 2 of the people won, while the third person lost. To answer that question, we have to account for all the possible outcomes that 'fit' what the question asks for. I calculated each of the individual outcomes, but I still have to add them all up to get the total probability.

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

Contact Rich at [email protected]

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GMAT/MBA Expert

Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Wed Jul 04, 2018 6:08 pm

jain2016 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

Answer: E

Scott Woodbury-Stewart
Founder and CEO
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