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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.

12 years ago

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

2 years ago

P(w/c) can't be less than 1/4..As there are 4 choices to the multiple choice question and even if he copies from someone else's the minimum probability of getting a correct choice is greater than or equal to 1/4.

one year ago

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

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