Probability problem involving hyper-geometric distribution So, I have the following problem: 
"$n$ balls are chosen randomly and without replacement from an urn containing $N$ white balls and $M$ black balls. Give the probability mass function of the random variable $X$ which counts the number of white balls chosen. Show that the expectation of $X$ is $\frac{Nn}{(M+N)}$. 
Hint: Do not use the hypergeometric distribution. Instead, write $X = X_1 + X_2 + ... + X_N$ where $X_i$ equals $1$ if the $i$th white ball was chosen."
If I could simply use the hypergeometric distribution, finding the probability mass function would be easy, and it would be the following: $$ P(X=k) = \frac{{N\choose k} {M\choose n-k}}{{N+M\choose n}} $$ 
Now, the teacher doesn't want us to use this. I guess that his technique makes computations easier when it comes to finding the expected value (because we would have: $E(X) = E(X_1) + E(X_2) + ... + E(X_N)$ (because of the linearity of the expected value).
Now my problem is that I do not know how to find $P(X_i = 0) \text{ and }  P(X_i = 1)$
PS: I get it that to compute the expected value, trying to find out $p(X_i = 0) $ is not very useful, as $E(X_i) = 0 \cdot P(X_i = 0) + 1 \cdot P(X_i=1)$, however, I still want to understand how I could compute it. 
 A: Okay... So I managed to do it on my own. For those of you who are interested here's the solution: 
$$P(X_i = 0) = \frac{N+M-1}{N+M} \cdot \frac{N+M-2}{N+M-1} \cdot \ \ ... \ \ \cdot \frac{N+M-n}{N+M-(n-1)} $$ which, when simplified gives: $$P(X_i=0) = \frac{N+M-n}{N+M} $$
The way to think about this is the following.  $X_i=0$ denotes the fact of not picking the $i$th white ball. So, in the $n$ draws, we're allowed to pick any other ball. Therefore, during the first draw, we can pick $N+M-1$ balls (as we don't want to pick the $i$th white ball), and the probability to do so is $p = \frac{N+M-1}{N+M}$. 
When we do the second draw we are allowed to pick $N+M-2$ balls (as we can't pick the $i$th white ball AND since there's only $N-M-1$ balls in the urn after the first draw),and the probability to do so is $p = \frac{N+M-2}{N+M-1}$
Applying the same logic to the $n$ draws we come to the conclusion that $P(X_i=0) = \frac{N+M-n}{N+M} $
Now, we have $$P(X_i =1) = 1-P(X_i=0) \\ = \frac{n}{M+N}$$
Finally, to get the expected value of $X$, first we notice that $E(X_i) = P(X_i=1)$ (see post for more details), then from the linearity of the expected value, we have that $$E(X) = E(X_1)+E(X_2)+...+E(X_N)\\= \sum_{i=1}^{N} E(X_i)\\=\frac{N\cdot n}{N+M}$$ which is the correct solution. 
Side note: A more formal way of finding $P(X_i=0)$ is the following:
First we need to recall that for two events $E$ and $F$, $P(E\cap F) = P(F)\cdot P(E|F) $
Now, let $D_j$ denote the the fact of not picking the $i$th ball at the $j$th draw. Thus we have that $$P(X_i=0) = P(D_1 \cap D_2 \  \cap \  ... \  \cap  \ D_{n}) \\ = P(D_1)\cdot  P(D_2|D_1) \cdot P(D_3|D_1 \cap \  D_2) \cdot \ ... \ \cdot P(D_n|D_1 \cap D_2 \  \cap \  ... \  \cap  \ D_{n-1})$$ which is exactly what we computed.
