Question

In an entrance test, there are multiple choice questions. There are four possible options 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. $\frac{1}{37}$
• B. $\frac{36}{37}$
• C. $\frac{1}{4}$
• D. $\frac{1}{49}$

Answer 1

Answer: (A)

$\frac{1}{37}$

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
Thus $P\left(A_1\right)=\frac{9}{10},P\left(A_2\right)=1-\frac{9}{10}=\frac{1}{10},P\left(\frac{E}{A_1}\right)=1,P\left(\frac{E}{A_2}\right)=\frac{1}{4}$
$\therefore$ required probability $=P\left(\frac{A_2}{E}\right)=\frac{P\left(A_2\right)P\left(\frac{E}{A_2}\right)}{P\left(A_1\right)P\left(\frac{E}{A_1}\right)+P\left(A_2\right)P\left(\frac{E}{A_2}\right)}$
$=\frac{\frac{1}{10}.\frac{1}{4}}{\frac{9}{10}.1+\frac{1}{10}.\frac{1}{4}}=\frac{1}{36+1}=\frac{1}{37}$