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In an entrance test there are multiple choice questions. There are four possible answers to each question of which one is correct.The probability that a student knows the answer to a question is 90%. If he gets the correct answer to a question, then the probability that he was guessing is (a) \(\frac{37}{40}\) (b) \(\frac{1}{37}\) (c) \(\frac{36}{37}\) (d) \(\frac{1}{9}\) Select the correct answer from above options

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Answer: (b) \(\frac{1}{37}\) We define the given events as: A1: Student knows the answer A2: Student does not know the answer E: He gets the correct answer. P(A1) =\(\frac{9}{10}\) , P(A2) = 1-\(\frac{9}{10}\) = \(\frac{1}{10}\) \(\therefore\) P(E/A1) = P(Student gets the correct answer when he knows the answer) = 1 P(E/A2) = P(Student gets the correct answer when he does not know the correct answer) = 1/4 \(\therefore\) Required probability \(P(A_2/E) = \frac{P(A_2).P(E/A_2)}{P(A_1).P(E/A_1)+P(A_2).P(E/A_2)}\) = \(\frac{\frac{1}{10}.\frac{1}{4}}{\frac{9}{10}.1+\frac{1}{10}.\frac{1}{4}}\) = \(\frac{\frac{1}{40}}{\frac{37}{40}}\) = \(\frac{1}{37}\)

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