Three friends Alan, Roger and Peter

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pm; Followed by:23 members

Three friends Alan, Roger and Peter

by BTGmoderatorDC » Thu Nov 23, 2017 3:34 am

Three friends Alan, Roger and Peter attempt to answer a question on an exam. Alan randomly guesses the answer, giving him a 1/5 probability of guessing correctly. Roger cheats by looking at the paper of the student in front of him, giving him a 2/3 probability of answering correctly. And Peter dutifully performs the calculations, then marks the answer, giving him a 5/6 probability of a correct answer. What is the probability that the question is answered correctly, but not via cheating?

A 1/18
B 1/9
C 23/90
D 5/18
E 13/45

Can some experts help me find the solution in this?

OA E

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Source: Beat The GMAT — Problem Solving | Thu Nov 23, 2017 3:34 am

DavidG@VeritasPrep: Legendary Member; Posts: 2663; Joined: Wed Jan 14, 2015 8:25 am; Location: Boston, MA; Thanked: 1153 times; Followed by:128 members; GMAT Score:770

by DavidG@VeritasPrep » Thu Nov 23, 2017 6:14 am

lheiannie07 wrote:Three friends Alan, Roger and Peter attempt to answer a question on an exam. Alan randomly guesses the answer, giving him a 1/5 probability of guessing correctly. Roger cheats by looking at the paper of the student in front of him, giving him a 2/3 probability of answering correctly. And Peter dutifully performs the calculations, then marks the answer, giving him a 5/6 probability of a correct answer. What is the probability that the question is answered correctly, but not via cheating?

A 1/18
B 1/9
C 23/90
D 5/18
E 13/45

Can some experts help me find the solution in this?

OA E

This question could use some cleaning up. It seems to be asking what the probability is that at least one of the friends answers the question correctly, so long as that friend is not Roger, the cheater. (It wouldn't make much sense to assume that the friends are working together and answering the problem as a cohesive unit.)

If so, this could happen one of three ways:

1) Alan and Peter both answer correctly, and Roger is wrong.
2) Alan answers correctly, and Peter and Roger are wrong
3) Peter answers correctly, and Alan and Roger are wrong

Scenario 1: P(Alan correct) = 1/5; P(Peter correct)= 5/6; P(Roger wrong) = 1/3 --> 1/5 *5/6 * 1/3 = 1/18
Scenario 2: P(Alan correct) = 1/5; P(Peter wrong)= 1/6; P(Roger wrong) = 1/3 --> 1/5 *1/6 * 1/3 = 1/90
Scenario 3: P(Peter correct) = 5/6; P(Alan wrong)= 4/5; P(Roger wrong) = 1/3 --> 5/6 * 4/5 * 1/3 = 2/9

Add the scenarios together: 1/18 + 1/90 + 2/9 = 5/90 + 1/90 + 20/90 = 26/90 = 13/45. The answer is E

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pm; Followed by:23 members

by BTGmoderatorDC » Wed Jan 17, 2018 9:40 pm

DavidG@VeritasPrep wrote:
lheiannie07 wrote:Three friends Alan, Roger and Peter attempt to answer a question on an exam. Alan randomly guesses the answer, giving him a 1/5 probability of guessing correctly. Roger cheats by looking at the paper of the student in front of him, giving him a 2/3 probability of answering correctly. And Peter dutifully performs the calculations, then marks the answer, giving him a 5/6 probability of a correct answer. What is the probability that the question is answered correctly, but not via cheating?

A 1/18
B 1/9
C 23/90
D 5/18
E 13/45

Can some experts help me find the solution in this?

OA E
This question could use some cleaning up. It seems to be asking what the probability is that at least one of the friends answers the question correctly, so long as that friend is not Roger, the cheater. (It wouldn't make much sense to assume that the friends are working together and answering the problem as a cohesive unit.)

If so, this could happen one of three ways:

1) Alan and Peter both answer correctly, and Roger is wrong.
2) Alan answers correctly, and Peter and Roger are wrong
3) Peter answers correctly, and Alan and Roger are wrong

Scenario 1: P(Alan correct) = 1/5; P(Peter correct)= 5/6; P(Roger wrong) = 1/3 --> 1/5 *5/6 * 1/3 = 1/18
Scenario 2: P(Alan correct) = 1/5; P(Peter wrong)= 1/6; P(Roger wrong) = 1/3 --> 1/5 *1/6 * 1/3 = 1/90
Scenario 3: P(Peter correct) = 5/6; P(Alan wrong)= 4/5; P(Roger wrong) = 1/3 --> 5/6 * 4/5 * 1/3 = 2/9

Add the scenarios together: 1/18 + 1/90 + 2/9 = 5/90 + 1/90 + 20/90 = 26/90 = 13/45. The answer is E

Thanks a lot!

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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

Hi

by Scott@TargetTestPrep » Sat Jan 20, 2018 6:23 am

lheiannie07 wrote:Three friends Alan, Roger and Peter attempt to answer a question on an exam. Alan randomly guesses the answer, giving him a 1/5 probability of guessing correctly. Roger cheats by looking at the paper of the student in front of him, giving him a 2/3 probability of answering correctly. And Peter dutifully performs the calculations, then marks the answer, giving him a 5/6 probability of a correct answer. What is the probability that the question is answered correctly, but not via cheating?

A 1/18
B 1/9
C 23/90
D 5/18
E 13/45

Let A, R, P = the probability that Alan, Roger and Peter correctly answer the question, respectively. Then the probability the question is answered correctly but not by cheating is the probability that Roger (the cheater) answers it incorrectly but either Alan or Peter or both answer it correctly.

P(A, not R, P) = 1/5 x 1/3 x 5/6 = 5/90
P(A, not R, not P) = 1/5 x 1/3 x 1/6 = 1/90
P(not A, not R, P) = 4/5 x 1/3 x 5/6= 20/90

Thus the probability is 5/90 + 1/90 + 20/90 = 26/90 = 13/45.

Answer: E

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