Let E1​,E2​,E3​ and A be the events defined asE1​= the examinee guesses the answerE2​= the examinee copies the answerE3​= the examinee knows the answerand A= the examinee answer correctlyWe have,P(E1​)=31​,P(E2​)=61​Since, E1​,E2​,E3​ are mutually exclusive and exhaustive events.∴P(E1​)+P(E2​)+P(E3​)=1⇒P(E3​)=1−31​−61​=21​If E1​ has already occured, then the examinee guesses. Since, there are four choices out of which only one is correct, therefore the probability that he answer correctly given that he has made a guess is 1/4.i.e. P(A/E1​)=41​It is given that, P(A/E2​)=81​and P(A/E3​)= probability that he answer correctly given that he know the answer =1By Bayes theorem, we have∴​P(E3​/A)=[P(E1​)⋅P(A/E1​)+P(E2​)⋅P(A/E2​)+P(E3​)⋅P(A/E3​)​]P(E3​)⋅P(A/E3​)​P(E3​/A)=(31​×41​)+(61​×81​)+(21​×1)21​×1​=2924​​