Conditional probability: independence? 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 approach? 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.
 A: The word "independently" here is somewhat hard to interpret.
I expect it is meant to mean that, for any given student, the two events " getting questions $A$, $B$ right" are independent.  However, this does not apply to an unspecified student.
Let's work this out.  You computed the probability that a randomly selected student gets the first question right.  That is $.78$  and, of course, this is also the probability that this student gets the second question right, assuming we don't know how they did on the first one.  But what's the probability that they get both right?   Well, by the same methodology you used, we see that this is $$.8\times .95^2+.2\times .1^2=.724$$
But of course $$.78^2=.6084\neq .724$$
It's unintuitive, somewhat, but you can gain intuition by thinking about an extreme case.  Say a prepared student is never wrong and an unprepared one is never right.  Then, of course, the probability of getting one right is $.8$ since there is a $.8$ chance that the student is prepared.  But the probability of getting two right is also $.8$, not $.8^2$.
For what it's worth, I wouldn't refer to this as "independence" since, as your apparent contradiction indicates, we do in fact get a huge amount of information from the first question and that changes our expectation for the second.  I'd say something like "a prepared student answers each question correctly with a probability of $.95$, independent of all other questions and an unprepared student answers each question correctly with a probability of $.1$, independently of all other questions."
