Question

In a test, an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is $\frac{1}{3}$ and the probability that he copies the answer is $\frac{1}{6}$ . The probability that his answer is correct given that he copies it is $\frac{1}{8}$ . The probability that his answer is correct, given that he guesses it is $\frac{1}{4}$ . The probability that he knew the answer to the question given that he correctly answered, is

• A. $\frac{24}{31}$
• B. $\frac{17}{24}$
• C. $\frac{24}{29}$
• D. $\frac{29}{31}$

Answer 1

Answer: (C)

Let $E_1,E_2$ and $E_3$ are the events that the examinee guesses, copies and knows the answer and $E$ is the event that he answers correctly.
Then, $\text{P}\left({\left(\text{E}\right)}_1\right)=\frac{1}{3}\text{,}\text{P}\left({\left(\text{E}\right)}_2\right)=\frac{1}{6}$
and $\text{P}\left({\left(\text{E}\right)}_3\right)=1-\left(\frac{1}{3}+\frac{1}{6}\right)=\frac{1}{2}$
∴ Required probability = $\text{P}\left(\frac{{\left(\text{E}\right)}_3}{\text{E}}\right)$
$=\frac{\text{P}\left(\frac{\text{E}}{{\left(\text{E}\right)}_3}\right)\cdot \text{P}\left({\left(\text{E}\right)}_3\right)}{\text{P}\left(\frac{\text{E}}{{\left(\text{E}\right)}_1}\right)\cdot \text{P}\left({\left(\text{E}\right)}_1\right)+\text{P}\left(\frac{\text{E}}{{\left(\text{E}\right)}_2}\right)\cdot \text{P}\left({\left(\text{E}\right)}_2\right)+\text{P}\left(\frac{\text{E}}{{\left(\text{E}\right)}_3}\right)\cdot \text{P}\left({\left(\text{E}\right)}_3\right)}$
$=\frac{1\times \frac{1}{2}}{\left(\frac{1}{4}\times \frac{1}{3}\right)+\left(\frac{1}{8}\times \frac{1}{6}\right)+\left(1\times \frac{1}{2}\right)}=\frac{24}{29}$