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 = $\text{P} \left(\frac{\left(\text{E}\right)_{3}}{\text{E}}\right)$
$= \frac{\text{P} \left(\frac{\text{E}}{\left(\text{E}\right)_{3}}\right) \cdot \text{P} \left(\left(\text{E}\right)_{3}\right)}{\text{P} \left(\frac{\text{E}}{\left(\text{E}\right)_{1}}\right) \cdot \text{P} \left(\left(\text{E}\right)_{1}\right) + \text{P} \left(\frac{\text{E}}{\left(\text{E}\right)_{2}}\right) \cdot \text{P} \left(\left(\text{E}\right)_{2}\right) + \text{P} \left(\frac{\text{E}}{\left(\text{E}\right)_{3}}\right) \cdot \text{P} \left(\left(\text{E}\right)_{3}\right)}$
$= \frac{1 \times \frac{1}{2}}{\left(\frac{1}{4} \times \frac{1}{3}\right) + \left(\frac{1}{8} \times \frac{1}{6}\right) + \left(1 \times \frac{1}{2}\right)} = \frac{2 4}{2 9}$