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 know the answer to a question is $9/10$. 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

Answer: (B)

Let $E_1:$ The event that the students knows the answer and
$E_2:$ The event that the student guesses the answer
$\therefore P\left(E_1\right)=\frac{9}{10}$ and $P\left(E_2\right)=1-\frac{9}{10}=\frac{1}{10}$
Let $E:$ The answer is correct.
The probability that the student answered correctly, given that he knows the answer i.e., $P\left(\frac{E}{E_1}\right)=1$
The probability that the students answered correctly, given that he guessed is $\frac{1}{4}$.
i.e., $P\left(\frac{E}{E_2}\right)=\frac{1}{4}$
By using Baye's theorem,
$P\left(\frac{E_2}{E}\right)=\frac{P\left(\frac{E}{E_2}\right)P\left(E_2\right)}{P\left(\frac{E}{E_1}\right)P\left(E_1\right)+P\left(\frac{E}{E_2}\right)P\left(E_2\right)}$
$=\frac{\frac{1}{4}\times \frac{1}{10}}{1\times \frac{9}{10}+\frac{1}{4}\times \frac{1}{10}}$
$=\frac{\frac{1}{40}}{\frac{9}{10}+\frac{1}{40}}=\frac{\frac{1}{40}}{\frac{36+1}{40}}=\frac{1}{37}$