Is the result of an exam a binomial? Here is a simple statistics question I was given. I'm not really sure I understand it. 

X = the number of aquired points in an exam (multiple choice and a right answer is one point). Is X binomial distributed?

The professor's answer was:

Yes, because there is only right or wrong answers.

My answer:

No, because each question has a different "success-probability" p. As I did understand a binomial distribution is just a series of Bernoulli-experiments, which each have a simple outcome (success or failure) with a given success-probability p (and all are "identical" regarding p).  E.g., Flipping a (fair) coin 100 times, this is 100 Bernoulli-experiments and all have p=0.5 . But here the questions have
  different kinds of p right?

 A: The answer to this problem depends on the framing of the question and when information is gained.  Overall, I tend to agree with the professor but think the explanation of his/her answer is poor and the professor's question should include more information up front.  
If you consider an infinite number of potential exam questions, and you draw one at random for question 1, draw one at random for question 2, etc. Then going into the exam:


*

*Each question has two outcomes (right or wrong)

*There are a fixed number of trials (questions)

*Each trial could be considered independent (going into question two, your probability $p$ of getting it right is the same as when going into question one)


Under this framework, the assumptions of a binomial experiment are met.
Alas, ill-proposed statistical problems are very common in practice, not just on exams.  I wouldn't hesitate to defend your rationale to your professor.
A: I would agree with your answer. Usually this kind of data would nowadays be modeled with some kind of Item Response Theory model. For example, if you used the Rasch model, then the binary answer $X_{ni}$ would be modeled as
$$
\Pr \{X_{ni}=1\} =\frac{e^{{\beta_n} - {\delta_i}}}{1 + e^{{\beta_n} - {\delta_i}}}
$$
where $\beta_n$ can be thought as $n$-th persons ability and $\delta_i$ as $i$-th question difficulty. So the model enables you to catch the fact that different persons vary in abilities and questions vary in difficulty, and this is the simplest of the IRT models.
Your professors answer assumes that all questions have same probability of "success" and are independent, since binomial is a distribution of a sum of $n$ i.i.d. Bernoulli trials. It ignores the two kinds of dependencies described above.
As noticed in the comments, if you looked at the distribution of answers of a particular person (so you don't have to care about between-person variability), or answers of different people on the same item (so there is no between-item variability), then the distribution would be Poisson-binomial, i.e. the distribution of the sum of $n$ non-i.i.d. Bernoulli trials. The distribution could be approximated with binomial, or Poisson, but that's all. Otherwise you're making the i.i.d. assumption.
Even under "null" assumption about guessing, this assumes that there is no guessing patterns, so people do not differ in how they guess and items do not differ in how they are guessed--so the guessing is purely random.
A: If there are n questions, and I can answer any one question correctly with probability p, and there is enough time to attempt answering all questions, and I did 100 of these tests, then my scores would be normal distributed with a mean of np. 
But it's not me repeating the test 100 times, it's 100 different candidates doing one test, each with his own probability p. The distribution of these p's will be the overriding factor. You might have a test where p = 0.9 if you studied the subject well, p = 0.1 if you didn't, with very few people between 0.1 and 0.9. The distribution of points will have very strong maxima at 0.1n and 0.9 n and will be nowhere near normal distribution. 
On the other hand, there are tests where everybody can answer any question, but take different amounts of time, so some will answer all n questions, and others will answer fewer because they run out of time. If we can assume that the speed of the candidates is normal distributed, then the points will be close to normal distributed. 
But many tests will contain some very hard and some very easy questions, intentionally so that we can distinguish between the best candidates (who will answer all questions up to some degree of difficulty) and the worst candidates (who will only be able to answer very simple questions). This would change the distribution of points quite strongly.
A: By definition, a binomial distribution is a set of $n$ independent and identically distributed Bernoulli trials. In the case of a multiple choice exam, each of the $n$ questions would be one of the Bernoulli trials.
The issue here arises because we can't reasonably assume that the $n$ questions:


*

*Are identically distributed. As you said, the probability a student knows the answer to question $1$ is almost certainly not going to be the same as the probability they know the answer question $2$, and so on. 

*Are independent. Many exams ask questions that are built upon the answers to the previous question(s). Who's to say for sure that that wouldn't happen on the exam in this question? There are other factors that could make answers to exam questions not independent of one another, but I think this one is the most intuitively obvious. 


I have seen questions in Statistics classes that model exam questions as binomials, but they are framed something along the lines of:

What probability distribution would model the number of questions answered correctly on a multiple choice exam where every question has four choices, and the student taking the exam is guessing every answer at random? 

In this scenario, of course it would be represented as a binomial distribution with $p= \frac{1}{4}$
.
