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

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

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

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

This question is slightly tricky due to the phrase "not via cheating." Not via cheating means we're looking for the probability of Roger not correctly answering the question.

P(A, R failing, P) = 1/5 * 1/3 * 1/6 = 5/90
P(A, R failing, P failing) = 1/5 * 1/3 * 1/6 = 1/90
P(A failing, R failing, P) = 4/5 * 1/3 * 5/6 = 20/90

5/90 + 1/90 + 20/90 =26/90 = 13/45

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

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

Is it an official GMAT Question?

No, it's from Veritas Prep. You can check the source of a question among the tags just above the original post.

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