In a test, an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that makes a guess $$is13$$ . The probability that he copies is $$16$$ and the probability that his answer is correct given that he copied it $$is18$$ . The probability that he knew the answer to the question given that he correctly answered it, is

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In a test, an examinee either guesses or copies or knows the answer to  a multiple choice question with four choices. The probability that makes a  guess $$is13$$     . The probability that he copies is $$16$$     and the probability that his answer is correct given that  he copied it $$is18$$     . The probability that he knew the answer to the question  given that he correctly answered it, is

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Let $$E1$$  be the event that the answer is guessed, $$E2$$  be the event that the answer is copied, $$E3$$  be the event that the examinee knows the answer and E be the event that the examinee answers correctly.Given $$P(E1)=13$$,$$P(E2)=16$$ assume that events, $$E1$$,$$E2$$&$$E3$$ are exhaustive.∴$$P(E1)+P(E2)+P(E3)=1$$∴$$P(E3)=1$$−$$P(E1)$$−$$P(E2)=1$$−$$13$$−$$16=12Now$$, $$P(EE1)=$$ probability of getting correct answer by guessing $$=$$ $$14$$ (since$$ $$4$$ $$alternatives)P(EE2)=$$  probability of answering correctly by copying $$=18And$$ $$P(EE3)=$$  probability of answering correctly by knowing $$=$$ $$1Clearly$$, $$(E3E) is the event he knew the answer to the question that he correctly answered it. Using Baye’s theorem $$P(E3E)$$  $$=$$ $$P(E3)$$.$$P(EE3)P(E1)$$.$$P(EE1)+P(E2)$$.$$P(EE2)+P(E3)$$.$$P(EE3)=12$$×$$113$$×$$14+16$$×$$18+12$$×$$1=2429$$  by Bayes' theorem

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