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Manager  Joined: 04 Aug 2013
Posts: 94
Location: India
Schools: McCombs '17
GMAT 1: 670 Q47 V35 GPA: 3
WE: Manufacturing and Production (Pharmaceuticals and Biotech)
Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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3
14 00:00

Difficulty:   95% (hard)

Question Stats: 55% (02:42) correct 45% (02:19) wrong based on 117 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
Intern  Joined: 27 Nov 2014
Posts: 43
Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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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
------------------------------------------------------------------------------------------------------------------------------------------------

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
Manager  Joined: 04 Aug 2013
Posts: 94
Location: India
Schools: McCombs '17
GMAT 1: 670 Q47 V35 GPA: 3
WE: Manufacturing and Production (Pharmaceuticals and Biotech)
Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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1
2
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.
Intern  Joined: 08 Jul 2012
Posts: 46
Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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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
_________________
Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time. - Thomas A. Edison
Manager  Joined: 21 Sep 2012
Posts: 208
Location: United States
Concentration: Finance, Economics
Schools: CBS '17
GPA: 4
WE: General Management (Consumer Products)
Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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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
Intern  Joined: 21 Aug 2014
Posts: 3
Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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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
Board of Directors P
Joined: 17 Jul 2014
Posts: 2509
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30 GPA: 3.92
WE: General Management (Transportation)
Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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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
Math Expert V
Joined: 02 Aug 2009
Posts: 7984
Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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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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Target Test Prep Representative D
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8109
Location: United States (CA)
Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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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.

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Non-Human User Joined: 09 Sep 2013
Posts: 13275
Re: Three friends Alan, Roger and Peter attempt to answer a question on an  [#permalink]

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_________________ Re: Three friends Alan, Roger and Peter attempt to answer a question on an   [#permalink] 06 Jun 2019, 10:08
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