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Three friends Alan, Roger and Peter attempt to answer a question on an

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Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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New post 02 Dec 2014, 23:32
3
9
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

54% (05:12) correct 46% (02:15) wrong based on 149 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
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Posts: 54
Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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New post 03 Dec 2014, 00:28
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
------------------------------------------------------------------------------------------------------------------------------------------------

Prob.(alan) = 1/5
Prob.(roger) without cheating = 2/3-1 = 1/3
Prob. (peter) = 5/6

Total Probability = 1/5*1/3*/5/6 = 1/18

Hence A!

Kindly clarify what am i missing why ans is E not A ?

Regards
SG
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Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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New post 03 Dec 2014, 01:46
1
1
There are three cases in which the correct answer will be supplied without Roger doing it by cheating:

1) Alan and Peter are both right, Roger is wrong.

Probability is 1/5 * 5/6 * (1 – 2/3)

=1/6 * 1/3

= 1/18

2) Alan is right, Peter and Roger are both wrong

1/5 * (1 – 5/6) * (1 – 2/3)

= 1/5 * 1/6 * 1/3

= 1/90

3) Peter is right, Alan and Roger are both wrong

5/6 * (1 – 1/5) * (1 – 2/3)

= 5/6 * 4/5 * 1/3

= 2/9

Then sum the three probabilities that all lead to "correct answer, no cheating" and you'll find that 2/9 + 1/90 + 1/18 = 13/45.
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Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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New post 03 Dec 2014, 02:25
There can be 3 cases where we can get correct answer but without cheating:
(A's ans. is correct and R and P's ans. are wrong) or (P's ans. is correct and R and A's ans. are wrong) or (A & P's ans. are correct and R's ans. is wrong)
this translates to:
(1/5*(1-2/3)*(1-5/6))+(5/6*(1-1/5)*(1-2/3))+(1/5*5/6*(1-2/3))
= (1/5*1/3*1/6) + (5/6*4/5*1/3) + (1/5*5/6*1/3)
= (1/90)+(2/9)+(1/18)
= 13/45
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Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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New post 03 Dec 2014, 02:31
What is the probability that the question is answered correctly, but not via cheating? In other words we need to find probability of
(Alan OR Peter being right) AND Roger being wrong

= [ 1/5 + 5/6 - (1/5*5/6) ] * 1/3
= [ 1/5 + 5/6 - 1/6 ] * 1/3
= [ 1/5 + 2/3 ] *1/3
=13/15 * 1/3
= 13/45

Ans = E
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Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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New post 11 Feb 2015, 05:34
1
1
Probability that the question is answered correctly, but not via cheating = 1 - [P(all incorrect) + P(cheating)]

1 - [(4/5*1/6*1/3) + 2/3] =
1 - (2/45 + 2/3) =
1 - (2/45 + 30/45) =
1 - 32/45 =
13/45
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Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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New post 28 Mar 2016, 19:31
anceer 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


3 cases:
A+ R- P+

or
A+ R- P-

or
A- R- P+

the result will be the sum of all 3 cases
1st case: 1/5*1/3*5/6 = 1/18
2nd case: 1/5*1/3*1/6 = 1/90
3rd case: 4/5*1/3*5/6 = 2/9

now:
1/18+2/9+1/90
LCM of 18, 9, and 90 is 90.
1st fraction multiply by 5/5
2nd fraction by 10/10
3rd leave as is:
5/90+20/90+1/90 = 26/90
simplify:
13/45

E
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Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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New post 28 Mar 2016, 19:52
mvictor wrote:
anceer 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


3 cases:
A+ R- P+

or
A+ R- P-

or
A- R- P+

the result will be the sum of all 3 cases
1st case: 1/5*1/3*5/6 = 1/18
2nd case: 1/5*1/3*1/6 = 1/90
3rd case: 4/5*1/3*5/6 = 2/9

now:
1/18+2/9+1/90
LCM of 18, 9, and 90 is 90.
1st fraction multiply by 5/5
2nd fraction by 10/10
3rd leave as is:
5/90+20/90+1/90 = 26/90
simplify:
13/45

E



Hi,

another way..


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 \(\frac{26}{30}*(1-\frac{2}{3})\)
=>\(\frac{26}{30}*\frac{1}{3}=\frac{13}{45}\)
E
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Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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New post 20 Jan 2018, 07:44
anceer 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) = ⅕ x ⅓ x ⅚ = 5/90

P(A, not R, not P) = ⅕ x ⅓ x ⅙ = 1/90

P(not A, not R, P) = ⅘ x ⅓ x ⅚ = 20/90

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

Answer: E
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Re: Three friends Alan, Roger and Peter attempt to answer a question on an &nbs [#permalink] 20 Jan 2018, 07:44
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