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 \quad 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., $\quad P\left(\frac{E}{E_{1}}\right)=1$
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}} $
$=\frac{\frac{1}{40}}{\frac{9}{10}+\frac{1}{40}}=\frac{\frac{1}{40}}{\frac{36+1}{40}}=\frac{1}{37}$