Q108 / 160 · AP EAMCET 2012 · Mathematics

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
    $\frac{37}{40}$
  • Correct answer
    B
    $\frac{1}{37}$
  • C
    $\frac{36}{37}$
  • D
    $\frac{1}{9}$

Solution

Let $E_1$ : The event that the students knows the answer and $E_2$ : The event that the student guesses the answer $\therefore \quad P\left(E_1\right)=\frac{9}{10} \quad$ and $\quad P\left(E_2\right)=1-\frac{9}{10}=\frac{1}{10}$ 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 $\frac{1}{4}$. $P\left(\frac{E}{E_2}\right)=\frac{1}{4}$ By using Baye's theorem, $\begin{aligned} P\left(\frac{E_2}{E}\right) & =\frac{P\left(\frac{E}{E_2}\right) P\left(E_2\right)}{P\left(\frac{E}{E_1}\right) P\left(E_1\right)+P\left(\frac{E}{E_2}\right) P\left(E_2\right)} \\ & =\frac{\frac{1}{4} \times \frac{1}{10}}{1 \times \frac{9}{10}+\frac{1}{4} \times \frac{1}{10}} \\ & =\frac{\frac{1}{40}}{\frac{9}{10}+\frac{1}{40}}=\frac{\frac{1}{40}}{\frac{36+1}{40}}=\frac{1}{37}\end{aligned}$

Hence, the correct answer is option: 'B'.

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