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Three friends Alan, Roger and Peter attempt to answer
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08 Jun 2019, 05:56
08 Jun 2019, 05:56
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Question Stats:
50% (01:25) correct
50% (02:24) wrong based on 6 sessions
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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
A 1/18
B 1/9
C 23/90
D 5/18
E 13/45
Re: Three friends Alan, Roger and Peter attempt to answer
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08 Jun 2019, 09:25
08 Jun 2019, 09:25
1
Expert Reply
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 \(\frac{26}{30}*(1-\frac{2}{3})\)
\(\frac{26}{30}*\frac{1}{3}=\frac{13}{45}\)
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
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02 Feb 2021, 06:33
02 Feb 2021, 06:33
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