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# Three friends Alan, Roger and Peter attempt to answer

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Intern
Joined: 20 May 2019
Posts: 31
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Kudos [?]: 2 [1] , given: 0

Three friends Alan, Roger and Peter attempt to answer [#permalink]  08 Jun 2019, 05:56
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Question Stats:

0% (00:00) correct 100% (00:01) wrong based on 1 sessions
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
[Reveal] Spoiler: OA
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Joined: 18 Apr 2015
Posts: 8128
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Kudos [?]: 1709 [0], given: 7477

Re: Three friends Alan, Roger and Peter attempt to answer [#permalink]  08 Jun 2019, 09:25
Expert's post
Prob that it will be answered correctly by Alan and Peter= $$\frac{1}{5}+ \frac{5}{6}-\frac{1}{5}*\frac{5}{6}=\frac{26}{30}$$

But we have to ensure that this does not include when even Roger answers correctly.. so $$\frac{26}{30}*(1-\frac{2}{3})$$

$$\frac{26}{30}*\frac{1}{3}=\frac{13}{45}$$
_________________
Re: Three friends Alan, Roger and Peter attempt to answer   [#permalink] 08 Jun 2019, 09:25
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