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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.
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
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
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.
Dear StudentLet 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 correctHereP(E1) = 1/3, P(E2) = 1/6∴P(E3) = 1−(1/3+1/6) = 1/2Since the question is a multiple choice question with four choice so the probability that the answer is correctwhen is guessed isP(E/E1) = 1/4Also the probability that his answer is correct, given that he copied it is81i.eP(E/E2) = 1/8Moreover his answer is correct, given he knew the answer is a sure event, soP(E/E3) = 1Hence by Baye's theorem the required probability isI hope this answer will help you.Thanks & RegardsYash Chourasiya
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