Computing Expected Value I am completely lost on what the question is saying. I am trying to think of a Probability Distribution to find the expected value, but I cant understand what is going on. I know how to compute the expected value but how do I figure out the probability distribution. Here is the question:
A college offers 10 courses and each student takes only one per semester. Three of the courses have enrollment 100, four courses have enrollment 25 and the rest three courses have enrollment 10. 
X is a random variable that tells the enrollment of an randomly chosen class. 
Y is a random variable that tells, for a randomly chosen student, what’s the enrollment of his/her class.
a) find E(X) and E(Y). 
b) What do you think of these two expectations? If you are a student, which of the above expectations is more important to you?
My main trouble is with understanding what X and Y are I think. I am so lost on how to think about it.
 A: I think the main struggle is in differentiating between selecting an individual at random compared to a group at random.  The possible values for both X and Y are the same, corresponding to the enrollment of a class (100,25,10), but the probabilities for those values are different: since X is the enrollment of a randomly chosen class, pr(X=100)=3/10, pr(X=25)=4/10, pr(X=10)=3/10.  Y, on the other hand, deals with picking a student at random, which means you're more likely to pick a student in a class with a higher enrollment.  For example, you have 430 total students, 300 of which are in classes of enrollment 100, so pr(Y=100)=300/430. Hopefully that helps!
A: When we choose one of the classes at random, there is a probability of $\frac{3}{10}$ that this class has enrollment $100$, since there are $3$ out of the $10$ courses that have this number of students. So $$\mathbb{P}(X=100)=\frac{3}{10}$$
Similary
$$\mathbb{P}(X=25)=\frac{4}{10}\;\;\;\;\;\mathbb{P}(X=10)=\frac{3}{10}$$
And for the students, there are $300$ out of $3*100+4*25+3*10=430$ students that are in a class with $100$ members, so:
$$\mathbb{P}(Y=100)=\frac{300}{430}$$
And similarly
$$\mathbb{P}(Y=25)=\frac{100}{430}\;\;\;\;\; \mathbb{P}(Y=10)=\frac{30}{430}$$
Now you know the distributions of $X$ and $Y$, and I think you can take on from here with what you already know.
