probability space of $100$ homeworks graded by two person 
$100$ homeworks are on the table, with two questions to be graded. Andy is in charge
of grading question one and Patrick is in charge of grading question two. First, Andy
grades some homeworks at random; each homework has probability $0.3$ of being graded.
Next, Patrick randomly grades half of the homeworks. That is, he grades $50$ homeworks.
Assume Andy and Patrick make their choices independently. 
(a) Write down the probability space $(S, P)$ based on the information
given.
(b) Let $N$ be the number of homeworks that Andy graded. What is $\mathbb E(N)$?
(c) Let $M$ be the number of homeworks that Andy graded and Patrick did not grade.
What is $\mathbb E (M)$?

For (a) I guess the $S$ the set of all $100$ homeworks. But I didn't understand how to assign probabibilities on the sigma algebra generated by this set.
For $(b)$ I guess it should be $\mathbb E(N)=100\times0.3=30$. Is it correct?
For $(c)$, the probability of homework graded by Andy and Patrick is $0.3$ and $0.5$ respectively. Then is it $\mathbb E(M)=0.3\times0.5$?
It will be a great help if anyone correct me if I am wrong. Thanks in Advance.
 A: Label the homeworks with $i=1,2,\dots,100$.
a)
The info that each homework has probability $0.3$ to be graded by Andy can be reworded immediately as: Andy graded $30$ homeworks.
Then for $S$ you can take the set:$$S:=\{(A,P)\in\mathcal P([100])^2:|A|=30\text{ and }|P|=50\}$$with equiprobability of all outcomes. Here $A$ stands for the subset of $[100]=\{1,2,\dots,100\}$ contain all labels of homeworks graded by Andy and $P$ corresponds similarly with Patrick.
b)
If we prescribe $A_{i}:S\to\mathbb R$ as $(A,P)\mapsto 1$ if $i\in A$ and $(A,P)\mapsto0$ otherwise then $N$ can be recognized as: $$N:=\sum_{i=1}^{100}A_i$$ and we linearity of expectation and symmetry we find indeed that $\mathbb EN=100\times0.3=30$. So you are correct for question b).
c)
Likewise we can prescribe $P_{i}:S\to\mathbb R$ as $(A,P)\mapsto 1$ if $i\in P$ and $(A,P)\mapsto0$ otherwise and then $M$ can be recognized as $$M:=\sum_{i=1}^{100}A_i(1-P_i)$$
Uptil now you only calculated $\mathbb EA_i(1-P_i)$ but to be found is $\mathbb EM$. Again use linearity of expectation and symmetry.
