Question 2

Let $K$ be the event that the student knows the answer, and $G$ be the event that the student guesses the answer. Let $C$ be the event that the answer is correct. We are given: $P(C|K) = 1$ (as the student always gives the correct answer if they know it). $P(C|G) = \frac{1}{2}$ (given). $P(G|C) = \frac{1}{6}$ (given). Let $P(K) = p$, then $P(G) = 1 - p$. Using Bayes' Theorem: $P(G|C) = \frac{P(G)P(C|G)}{P(G)P(C|G) + P(K)P(C|K)}$ $\frac{1}{6} = \frac{(1-p) \cdot \frac{1}{2}}{(1-p) \cdot \frac{1}{2} + p \cdot 1}$ $\frac{1}{6} = \frac{1-p}{1-p+2p} = \frac{1-p}{1+p}$ Solving for $p$: $1+p = 6 - 6p \implies 7p = 5 \implies p = \frac{5}{7}$ Thus, the probability that the student knows the answer is $\frac{5}{7}$.