Q79 / 160 · TS EAMCET 2018 (05 May Shift 1) · Mathematics
In a test, a student either guesses or copies or knows the answer to a multiple choice question with four choices having one correct answer.
The probability that he guesses the answer is $\frac{1}{3}$ and the probability that he copies it is $\frac{1}{12}$. The probability that his answer is correct given that he copied it is $\frac{1}{6}$. The probability that he knew the answer, given that he has correctly answered it, is
- Correct answerA$\frac{6}{7}$
- B$\frac{15}{49}$
- C$\frac{7}{12}$
- D$\frac{10}{13}$
Solution
$E_1$ be an event that the student guess the answer $E_2$ be the event that the student know that answer $E_3$ be the event that he copies the answer and $A$ be the event that the answer is correct
Given that,
$$
P\left(E_1\right)=\frac{1}{3}, P\left(E_3\right)=\frac{1}{12}
$$
So, $P\left(E_2\right)=1-\frac{1}{3}-\frac{1}{12}$
$$
P\left(E_2\right)=\frac{12-4-1}{12}=\frac{7}{12}
$$
And the probability that the answer is correct if he guess is
$$
\therefore \quad P\left(\frac{A}{E_1}\right)=\frac{1}{4}
$$
Probability that the answer is correct given that he knows the answer is
$$
\therefore
$$
$$
P\left(\frac{A}{E_2}\right)=1
$$
Probability that the answer is correct given that he copies the answer is
$$
\therefore \quad P\left(\frac{A}{E_3}\right)=\frac{1}{6}
$$
by using Bayes theorem
$$
\begin{aligned}
P\left(\frac{E_2}{A}\right) & =\frac{P\left(E_2\right) P\left(\frac{A}{E_2}\right)}{P\left(E_1\right) \cdot P\left(\frac{A}{E_1}\right)+P\left(E_2\right) P\left(\frac{A}{E_2}\right)+P\left(E_3\right) P\left(\frac{A}{E_3}\right)} \\
P\left(\frac{E_2}{A}\right) & =\frac{\frac{7}{12} \cdot 1}{\frac{1}{3} \times \frac{1}{4}+\frac{7}{12} \times 1+\frac{1}{12} \times \frac{1}{6}} \\
& =\frac{\frac{7}{12}}{\frac{1}{12}+\frac{7}{12}+\frac{1}{72}}=\frac{\frac{7}{12}}{\frac{6+42+1}{72}}=\frac{7}{12} \times \frac{72}{49} \\
P\left(\frac{E_2}{A}\right) & =\frac{6}{7}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.