A, B, C and D, and we are trying to find the probability that exactly one event occurs. Truthfully, this is a homework problem. I've come across a solution, but I'm really trying to figure out how this works, hopefully at an intuitive level.
We have four events, A, B, C and D, and we are trying to find the probability that exactly one event occurs.
I've seen the "Inclusion exclusion principle", and that would help if I had to find the probability of A or B or C or D. I feel this should be trivial, probability is my weakest math, which is why I'm taking it, but this particular problem is giving me a lot of trouble. Any help would be greatly appreciated.
 A: There are many possible formulas. The appropriate one depends on the information you are given. I assume that you know the basic Inclusion Exclusion Principle for computing $\Pr(A\cup B\cup C\cup D)$. A modification of that idea will take care of your problem. 
When we find 
$$\Pr(A)+\Pr(B)+\Pr(C)+\Pr(D),$$
the situations in which two of $A$, $B$, $C$, and $D$ occur have been taken account of twice. But we want to count them zero times. So we want to subtract 
$$2\left(\Pr(A\cap B)+\Pr(A\cap C+\Pr(A\cap D)+\Pr(B\cap C)+\Pr(B\cap D)+\Pr(C\cap D)\right).$$
But we have been overenthusiastic in our subtractions, for we subtracted too often the probability that three of $A$, $B$, $C$, $D$ occur. Use a picture (Venn Diagram) to decide what we must add back. And there will be also an issue about the probability that all four of $A$, $B$, $C$, and $D$ occur. 
A: Using ${}^c$ to denote complement of an event, "exactly one of $A,B,C,D$" means 
$AB^cC^cD^c \cup A^c B C^c D^c \cup A^c B^c C D^c \cup A^c B^c C^c D$, those four being
mutually exclusive.  So one way to write it is
$$P(AB^cC^cD^c) + P(A^c B C^c D^c) + P(A^c B^c C D^c) + P(A^c B^c C^c D) $$
If you want to express it without any complements, note that e.g. 
$P(A B^c C^c D^c) = P(A) - P(AB \cup AC \cup AD)$, and use inclusion-exclusion on $P(AB \cup AC \cup AD)$ (and similarly for the other three).  
A: Decompose the event into the 4 simpler events $A\cap B^c\cap C^c\cap D^c$, $A^c\cap B\cap C^c\cap D^c$, $A^c\cap B^c\cap C\cap D^c$, and $A^c\cap B^c\cap C^c\cap D$. These are disjoint, so you can simply sum the probabilities of each and you'll be done.
