Question

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. $\frac{1}{9}$

Answer 1

The correct option is B

$\frac{1}{37}$

Explanation for the correct answer.

Finding the probability:

Let, $A$ be the event that the student knows the answer. So, $P\left(A\right)=\frac{9}{10}$.

Let, $B$ be the event that the student does not know the answer. So, $P\left(B\right)=1-P\left(A\right)=\frac{1}{10}$.

Let, $E$ be the event that the student gets the correct answer.

So, the probability that the student knows the answer and he gets it correct is $P(E/A)=1$ and the probability the student does not know the answer but he gets it correct is $P\left(\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$B$}\right.\right)=\frac{1}{4}$ (as one out of the four options is the correct answer).

Now, in this question we have to find the probability that the student gets the correct answer even when he does not know the answer, that is we have to find $P\left(\raisebox{1ex}{$B$}\!\left/ \!\raisebox{-1ex}{$E$}\right.\right)$.

By applying Bayes' Theorem,

$\begin{array}{rcl}P\left(\raisebox{1ex}{$B$}\!\left/ \!\raisebox{-1ex}{$E$}\right.\right)& =& \frac{P\left(B\right)\times P\left({\displaystyle \raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$B$}\right.}\right)}{P\left(A\right)\times P\left({\displaystyle \raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$A$}\right.}\right)+P\left(B\right)\times P\left({\displaystyle \raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$B$}\right.}\right)}\\ & =& \frac{{\displaystyle \frac{1}{10}}\times {\displaystyle \frac{1}{4}}}{\left({\displaystyle \frac{9}{10}}\times 1\right)+\left({\displaystyle \frac{1}{10}}\times {\displaystyle \frac{1}{4}}\right)}\\ & =& \frac{{\displaystyle \frac{1}{40}}}{{\displaystyle \frac{36+1}{40}}}\\ & =& \frac{1}{37}\end{array}$

Hence, option B is correct.