Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 04 Aug 2013
Posts: 104
Location: India
GPA: 3
WE: Manufacturing and Production (Pharmaceuticals and Biotech)

Three friends Alan, Roger and Peter attempt to answer a question on an [#permalink]
Show Tags
02 Dec 2014, 22:32
2
This post received KUDOS
10
This post was BOOKMARKED
Question Stats:
54% (05:18) correct 46% (02:18) wrong based on 143 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.



Manager
Joined: 27 Nov 2014
Posts: 66

Re: Three friends Alan, Roger and Peter attempt to answer a question on an [#permalink]
Show Tags
02 Dec 2014, 23:28
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 = 2/31 = 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: 104
Location: India
GPA: 3
WE: Manufacturing and Production (Pharmaceuticals and Biotech)

Re: Three friends Alan, Roger and Peter attempt to answer a question on an [#permalink]
Show Tags
03 Dec 2014, 00:46
1
This post received KUDOS
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.



Manager
Joined: 08 Jul 2012
Posts: 50

Re: Three friends Alan, Roger and Peter attempt to answer a question on an [#permalink]
Show Tags
03 Dec 2014, 01:25
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 *(12/3) *(15/6)) +(5/6 *(11/5) *(12/3)) +(1/5 *5/6 *(12/3)) = (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: 218
Location: United States
Concentration: Finance, Economics
GPA: 4
WE: General Management (Consumer Products)

Re: Three friends Alan, Roger and Peter attempt to answer a question on an [#permalink]
Show Tags
03 Dec 2014, 01:31
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



Intern
Joined: 21 Aug 2014
Posts: 3

Re: Three friends Alan, Roger and Peter attempt to answer a question on an [#permalink]
Show Tags
11 Feb 2015, 04:34
1
This post received KUDOS
1
This post was BOOKMARKED
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
Joined: 17 Jul 2014
Posts: 2736
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: Three friends Alan, Roger and Peter attempt to answer a question on an [#permalink]
Show Tags
28 Mar 2016, 18:31
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



Math Expert
Joined: 02 Aug 2009
Posts: 5662

Re: Three friends Alan, Roger and Peter attempt to answer a question on an [#permalink]
Show Tags
28 Mar 2016, 18:52
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
_________________
Absolute modulus :http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
BANGALORE/



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 2192
Location: United States (CA)

Re: Three friends Alan, Roger and Peter attempt to answer a question on an [#permalink]
Show Tags
20 Jan 2018, 06:44
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
_________________
Scott WoodburyStewart
Founder and CEO
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions




Re: Three friends Alan, Roger and Peter attempt to answer a question on an
[#permalink]
20 Jan 2018, 06:44






