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

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Question

(C) Let us define the following events:$E_1$: Student guesses the answer.$E_2$: Student copies the answer.$E_3$: Student knows the answer.$E$: The ans

Vedclass · Accepted Answer

$\frac{24}{29}$

Vedclass · Answer

(C) Let us define the following events:$E_1$: Student guesses the answer.$E_2$: Student copies the answer.$E_3$: Student knows the answer.$E$: The answer is correct.Given probabilities:$P(E_1) = \frac{1}{3}$,$P(E_2) = \frac{1}{6}$.Since the events are exhaustive,$P(E_3) = 1 - (P(E_1) + P(E_2)) = 1 - (\frac{1}{3} + \frac{1}{6}) = 1 - \frac{1}{2} = \frac{1}{2}$.Conditional probabilities:$P(E|E_1) = \frac{1}{4}$ (since there are $4$ choices).$P(E|E_2) = \frac{1}{8}$ (given).$P(E|E_3) = 1$ (since he knows the answer).Using Bayes' Theorem,the probability that he knew the answer given that he answered correctly is:$P(E_3|E) = \frac{P(E|E_3)P(E_3)}{P(E|E_1)P(E_1) + P(E|E_2)P(E_2) + P(E|E_3)P(E_3)}$$P(E_3|E) = \frac{1 	imes \frac{1}{2}}{(\frac{1}{4} 	imes \frac{1}{3}) + (\frac{1}{8} 	imes \frac{1}{6}) + (1 	imes \frac{1}{2})}$$P(E_3|E) = \frac{\frac{1}{2}}{\frac{1}{12} + \frac{1}{48} + \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{4+1+24}{48}} = \frac{\frac{1}{2}}{\frac{29}{48}} = \frac{1}{2} 	imes \frac{48}{29} = \frac{24}{29}$.

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