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Question

(A) Let $E_1$ be the event that the student guesses the answer and $E_2$ be the event that the student knows the answer. Let $A$ be the event that the

Vedclass · Accepted Answer

$\frac{1}{37}$

Vedclass · Answer

(A) Let $E_1$ be the event that the student guesses the answer and $E_2$ be the event that the student knows the answer. Let $A$ be the event that the student answers correctly.Given that the student knows $90 \%$ of the answers,$P(E_2) = \frac{9}{10}$.Consequently,the probability that the student guesses is $P(E_1) = 1 - P(E_2) = 1 - \frac{9}{10} = \frac{1}{10}$.If the student knows the answer,the probability of answering correctly is $P(A \mid E_2) = 1$.If the student guesses,since there are $4$ options and only one is correct,the probability of answering correctly is $P(A \mid E_1) = \frac{1}{4}$.We need to find the probability that the student was guessing given that they answered correctly,which is $P(E_1 \mid A)$.Using Bayes' Theorem:$P(E_1 \mid A) = \frac{P(E_1) \cdot P(A \mid E_1)}{P(E_1) \cdot P(A \mid E_1) + P(E_2) \cdot P(A \mid E_2)}$$P(E_1 \mid A) = \frac{\frac{1}{10} \cdot \frac{1}{4}}{(\frac{1}{10} \cdot \frac{1}{4}) + (\frac{9}{10} \cdot 1)}$$P(E_1 \mid A) = \frac{\frac{1}{40}}{\frac{1}{40} + \frac{9}{10}} = \frac{\frac{1}{40}}{\frac{1+36}{40}} = \frac{1}{37}$.

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