Q79 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics
In a certain recruitment test with multiple choice questions, there are four options to answer each question, out of which only one is correct. An intelligent student knows $90 \%$ correct answers while a weak student known only $20 \%$ correct answers. If an intelligent student gets the correct answer for a question, then the probability that he was guessing it, is
- Correct answerA$\frac{1}{37}$
- B$\frac{1}{10}$
- C$\frac{9}{37}$
- D$\frac{1}{2}$
Solution
Let the events $E_1=$ he guesses the answer $E_2=$ he knows the answer $A=$ he answers correctly Then,
$$
\begin{aligned}
P\left(E_2\right) & =\frac{9}{10}, & P\left(E_1\right) & =\frac{1}{10} \\
P\left(A \mid E_2\right) & =1, & P\left(A \mid E_1\right) & =\frac{1}{4}
\end{aligned}
$$
So,
$$
\begin{aligned}
P\left(E_1 \mid A\right) & =\frac{P\left(E_1\right) \cdot P\left(A \mid E_1\right)}{P\left(E_1\right) \cdot P\left(A \mid E_1\right)+P\left(E_2\right) \cdot P\left(A \mid E_2\right)} \\
& =\frac{\frac{1}{10} \times \frac{1}{4}}{\left(\frac{1}{10} \times \frac{1}{4}\right)+\left(\frac{9}{10} \times 1\right)}=\frac{1}{1+36}=\frac{1}{37}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.