In the new pattern of IIT-JEE paper there are multiple choice questions. There are 4 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

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$\frac{1}{37}$

$\frac{4}{37}$

$\frac{1}{4}$

$\frac{37}{40}$

answer is A.

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Detailed Solution

$A_{1}$ : He knows the answer
$A_{2}$ : He guess the answer
E : He gets the correct answer
$P (A_{1}) = \frac{9}{10}, P (A_{2}) = 1 - \frac{9}{10} = \frac{1}{10}, P (\frac{E}{A_{1}}) = 1, P (\frac{E}{A_{2}}) = \frac{1}{4}$
∴ required probability=P( he was guessing /he gets the correct answer)

$P (\frac{A_{2}}{E}) = \frac{P (A_{2}) P (\frac{E}{A_{2}})}{P (A_{1}) P (\frac{E}{A_{1}}) + P (A_{2}) P (\frac{E}{A_{2}})}$
$= \frac{\frac{1}{10} \cdot \frac{1}{4}}{\frac{9}{10} \cdot 1 + \frac{1}{10} \cdot \frac{1}{4}} = \frac{1}{36 + 1} = \frac{1}{37}$