Probability of Knowing Answer in a True False Quiz Using Bayes Theorem

To find the probability that the student knows the answer to a randomly chosen question, let us define the following events:

• $K$: The student knows the answer to the question.
• $G$: The student guesses the answer to the question.
• $C$: The student answers the question correctly.

Let the probability that the student knows the answer be $P(K) = p$. Since the student either knows the answer or guesses it, these events are mutually exclusive and exhaustive: $P(G) = 1 - P(K) = 1 - p$

From the given information:

1. When the student knows the answer, the answer is always correct: $P(C \mid K) = 1$
2. The probability of answering correctly given that the student guessed: $P(C \mid G) = \frac{1}{2}$
3. The probability that the student guessed the answer given that the answer is correct: $P(G \mid C) = \frac{1}{6}$

Using the Law of Total Probability, the total probability that an answer is correct is: $P(C) = P(C \mid K)P(K) + P(C \mid G)P(G)$ $P(C) = 1 \cdot p + \frac{1}{2}(1 - p) = \frac{1 + p}{2}$

Now, by Bayes' Theorem: $P(G \mid C) = \frac{P(C \mid G)P(G)}{P(C)}$

Substituting the known values: $\frac{1}{6} = \frac{\frac{1}{2}(1 - p)}{\frac{1 + p}{2}}$ $\frac{1}{6} = \frac{1 - p}{1 + p}$

Cross-multiplying to solve for $p$: $1 + p = 6(1 - p)$ $1 + p = 6 - 6p$ $7p = 5$ $p = \frac{5}{7}$

Thus, the probability that the student knows the answer to a randomly chosen question is $\frac{5}{7}$.

Correct Option: (C)