In a test, a student either guesses, copies, | vedclass

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Question

(A) Let $E_1$ be the event that the student guesses the answer, $E_2$ be the event that the student knows the answer, and $E_3$ be the event that he c

Vedclass · Accepted Answer

$\frac{6}{7}$

Vedclass · Answer

(A) Let $E_1$ be the event that the student guesses the answer, $E_2$ be the event that the student knows the answer, and $E_3$ be the event that he copies the answer. Let $A$ be the event that the answer is correct.Given that,$P(E_1) = \frac{1}{3}, P(E_3) = \frac{1}{12}$.Since the events are exhaustive, $P(E_2) = 1 - P(E_1) - P(E_3) = 1 - \frac{1}{3} - \frac{1}{12} = \frac{12-4-1}{12} = \frac{7}{12}$.The probability that the answer is correct if he guesses is $P(A|E_1) = \frac{1}{4}$ (since there are $4$ choices).The probability that the answer is correct if he knows the answer is $P(A|E_2) = 1$.The probability that the answer is correct if he copies is $P(A|E_3) = \frac{1}{6}$.Using Bayes' theorem, the probability that he knew the answer given that he answered correctly is:$P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$$P(E_2|A) = \frac{\frac{7}{12} 	imes 1}{(\frac{1}{3} 	imes \frac{1}{4}) + (\frac{7}{12} 	imes 1) + (\frac{1}{12} 	imes \frac{1}{6})}$$P(E_2|A) = \frac{\frac{7}{12}}{\frac{1}{12} + \frac{7}{12} + \frac{1}{72}} = \frac{\frac{7}{12}}{\frac{6 + 42 + 1}{72}} = \frac{7}{12} 	imes \frac{72}{49} = \frac{6}{7}$.

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