What is the conditional probability distribution for who wrote this particular exam? 
A small course has 3 students, $B$, $C$ and $D$. Based on the previous
  testscores of these students we know that $B$ usually scores $80$%
  correct on the exam questions, $C$ scores $60$% and $D$ scores $40$%.
  The anonymous test that is now corrected has $4$ correct answers out
  of $8$. What is the conditional probability distribution for who took
  this particular exam?

This just doesn't want to work. I know Baye's theorem is of use here and possibly the Law of total probability. 
Let $B=1, \ C=2$ and $D=3.$ Now also let $A_i, \ i=1,2,3$ be the event that person $A_i$ took the test and $X= \ $number of correctly answered question on the test. We are thus looking for 
$$P(A_i|X=4)=\frac{P(X=4|A_i)P(A_i)}{P(X=4)}.$$
This did not make anything simpler. The only thing I now in the RHS is that $P(A_i)=1/3.$
 A: Based on the information provided, it is not possible to give an answer. You are given B, C and D's mean (or, depending on interpretation of "usual", median or mode), but that is not directly relevant. You need to know not what A$_i$'s "usual" score is, but what the percentage of the exams they get 50% is. The only approach I can see is assuming that each student's scores have a binomial distribution with p = .8, .6, and .4, respectively. You could then calculate the P(X = 4) for each of those distributions. (Note that by symmetry, C and D should have the same probability of being the person who took the exam).
You also need a prior distribution, and while "all priors are 1/3" is a reasonable assumption, it is not explicitly stated.
Also note that in American English, to write an exam means to come up with the questions on the exam (and thus, the person who writes the exam is generally the instructor), not to take the exam. 
A: $P(X=4|A) = {8\choose 4} 0.8^4(1-0.8)^4\\
P(X=4|B) = {8\choose 4} 0.6^4(1-0.6)^4$
etc.
$P(B|X = 4) = \frac {P(X=4|B)}{P(X=4|A) + P(X=4|B) + P(X=4|C)}$
