In a test an examinee either guesses or copies or knows that answer to a multiple choice question which has 4 choices. The probability that he makes a guess is 1/3 and the probability that he copies is 1/6. The probability that his answer is correct, given the copied it, is 1/8. Find the probability that he knew the answer to the question, given that he answered it correctly.

P(g) = probability of guessing = 1/3

        P(c) = probability of copying = 1/6

        P(k) = probability of knowing = 1 - 1/3 - 1/6 = 1/2

        (Since the three-event g, c and k are mutually exclusive and exhaustive)

        P(w) = probability that answer is correct

       P(k/w)=(P(w/k).P(k))/(P(w/c)P(c)+P(w/k)P(k)+P(w/g)P(g))        (using Baye's theorem)

         = (1×1/2)/((1/8,1/6)+(1×1/2)+(1/4×1/3) )=24/29     

Approved

Dear Student

Let define the following events:
E1​:Examine guesses the answer to the question.
E2​:Examine copies the answer to the question.
E3​:Examine know the answer to the question.
E:answer is correct
HereP(E1​) = 1/3, P(E2​) = 1/6​
∴P(E3​) = 1−(1/3​+1/6​) = 1/2​
Since the question is a multiple choice question with four choice so the probability that the answer is correct
when is guessed isP(E/E1​) = 1/4​
Also the probability that his answer is correct, given that he copied it is81​
i.eP(E/E2​) = 1/8
Moreover his answer is correct, given he knew the answer is a sure event, soP(E/E3​​) = 1
Hence by Baye's theorem the required probability is


I hope this answer will help you.
Thanks & Regards
Yash Chourasiya