GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Oct 2019, 12:51

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

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]

### Show Tags

02 Dec 2014, 23:32
3
14
00:00

Difficulty:

95% (hard)

Question Stats:

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

### Show Tags

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
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]

### Show Tags

03 Dec 2014, 01:46
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]

### Show Tags

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
_________________
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]

### Show Tags

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
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, 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
Board of Directors
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]

### Show Tags

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

### Show Tags

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
_________________
Target Test Prep Representative
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]

### Show Tags

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.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

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

### Show Tags

06 Jun 2019, 10:08
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Three friends Alan, Roger and Peter attempt to answer a question on an   [#permalink] 06 Jun 2019, 10:08
Display posts from previous: Sort by