Q. 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 be 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) $\frac{1}{9}$
Explanation
(b)
We define the following events:
$A_{1}:He$ knows the answer;
$A_{2}:$He does not know the answer;
$E:$He gets the correct answer
Then, $P\left( A_{1} \right) = \frac{9}{10},P\left( A_{2} \right) = 1 = \frac{9}{10} = \frac{1}{10}$,
$P\left( E \middle| A_{1} \right) = 1,P\left( E \middle| A_{2} \right) = \frac{1}{4}$
$\therefore$ Required probability $= P(A_{2}|E)$
$= \frac{P\left( A_{2} \right)P(E/A_{2})}{P\left( A_{1} \right)P\left( E \middle| A_{1} \right) + P\left( A_{2} \right)P(E|A_{2})} = \frac{\frac{1}{10}.\frac{1}{4}}{\frac{9}{10}.1 + \frac{1}{10}.\frac{1}{4}} = \frac{1}{37}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.