Consider the following problem:

A prepared student answers correctly to each oral question independently from one another with probability $ 0.95$ . If, however, he is not prepared, he answers correctly with probability $ 0.1$ . $ 80$ % of the students taking the exam are prepared. Donald answers wrong to the first question. What is the probability he is prepared? Donald asks his teacher if he can have a second question. Determine the probability that he will answer correctly.

So, as far as the first question is concerned, we easily compute:$ $ P(P|W_1)=\frac{P(P)\cdot P(W_1|P)}{P(P)\cdot P(W_1|P)+P(P^c)\cdot P(W_1|P^c)}=\frac{0.8\cdot 0.05}{0.8\cdot 0.05+0.2\cdot 0.9}$ $ according to Bayes’s theorem. I’m actually concerned with the second question. I approached it making the following thought: since we are given that each student answers each question independently of one another, it shouldn’t matter whether Donald answered wrong or right to the first question in oder to answer correctly to the second question (I’m assuming from the text that $ P(R_2|W_1)=P(R_2)$ ). Thus I computed:$ $ P(R_2)=P(P)\cdot P(R_2|P)+P(P^c)\cdot P(R_2|P^c)=0.8\cdot 0.95+0.2\cdot 0.1$ $ which is $ 0.78$ . According to my teacher, right answer is:$ $ P(R_2|W_1)=\frac{P(W_1R_2|P)\cdot P(P)+P(W_1R_2|P^c)\cdot P(P^c)}{P(W_1)}$ $ $ P(W_1)$ is computed as before in the previous equation. Whereas the numerator is:$ $ P(W_1|P)\cdot P(R_2|P)\cdot P(P)+P(W_1|P^c)\cdot P(R_2|P^c)\cdot P(P^c)$ $ this gives $ 0.25$ as overall result, which is way lower than mine. He’s assuming that each $ R_i$ , $ W_i$ is idependent from each other knowing that either event $ P$ or $ P^c$ occured. Thus he’s assuming, if I’m not mistaken, conditional independence. So, what is the correct way to solve this problem? Which is the best appraoch? What will you assume reading that text? I would probably stick with my solution, since it’s weird that one answers independently of each question only if he’s either prepared or not.