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Three friends Alan, Roger and Peter attempt to answer

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Three friends Alan, Roger and Peter attempt to answer [#permalink] New post 08 Jun 2019, 05:56
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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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Re: Three friends Alan, Roger and Peter attempt to answer [#permalink] New post 08 Jun 2019, 09:25
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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 \(\frac{26}{30}*(1-\frac{2}{3})\)

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

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