Question

In a test, a student either guesses, 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 is \( \frac{1}{6} \). The probability that his answer is correct, given that he guessed it, is \( \frac{1}{4} \), and the probability that he copied it and it is correct is \( \frac{1}{8} \). The probability that he knew the answer to the question, given that he answered it correctly, is:

• A. \( \frac{29}{24} \)
• B. \( \frac{22}{29} \)
• C. \( \frac{24}{29} \)
• D. \( \frac{23}{29} \)

Hint:

Bayes' Theorem is perfect for reverse probability — focus on conditional likelihoods and priors.

Answer 1

The Correct Option is C

Let the events be: - \( G \): guessed, \( C \): copied, \( K \): knew - \( P(G) = \frac{1}{3}, P(C) = \frac{1}{6}, P(K) = 1 - \frac{1}{3} - \frac{1}{6} = \frac{1}{2} \) Given: - \( P(Correct|G) = \frac{1}{4} \), \( P(Correct|C) = \frac{1}{8} \), \( P(Correct|K) = 1 \) Use Bayes’ Theorem: \[ P(K | Correct) = \frac{P(K) \cdot P(Correct | K)}{P(G) \cdot P(Correct|G) + P(C) \cdot P(Correct|C) + P(K) \cdot P(Correct|K)} \] Substitute: \[ P(K | Correct) = \frac{\frac{1}{2} \cdot 1}{\frac{1}{3} \cdot \frac{1}{4} + \frac{1}{6} \cdot \frac{1}{8} + \frac{1}{2} \cdot 1} = \frac{\frac{1}{2}}{\frac{1}{12} + \frac{1}{48} + \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{4 + 1 + 24}{48}} = \frac{\frac{1}{2}}{\frac{29}{48}} = \frac{24}{29} \]