In an entrance examination there are multiple choice questions. There are four possible answer to each question of which one is correct. The probability that a student knows the answer to a question is 90%. If the gets the correct answer to the question, then the probability that he was guessing is

$\frac{1}{37}$

Consider the following events:

$E_1 → $ He know the answer, $E_2 → $ He guesses the answer

$A → $ He gets the correct answer.

We have, 

$P(E_1)=\frac{90}{100}=\frac{9}{10}, P(E_2)=\frac{1}{10}, $

$P(A/E_1 = 1. P(A/E-2)=\frac{1}{4}$

∴ Required probability $= P(E_2/A)$

$=\frac{P(E_2)P(A/E_2)}{P(E_1)P(A/E_1)+P(E_2)P(A/E_2)}$

$=\frac{\frac{1}{10}×\frac{1}{4}}{\frac{9}{10}×1+\frac{1}{10}×\frac{1}{4}}=\frac{1}{37}$