Probability question : How to find the actual numerical value without the lengthy calculation? 
A class contains $10$ boys and $15$ girls. $8$ students are to be selected
  at random from the class without replacement. Let X be the number of
  boys selected and Y the number of girls selected. Find $E(X −Y )$.

I could chalk out the table:
$E(X-Y)= -8$ when $X=0$ and $Y=8$
$E(X-Y)= -7$ when $X=1$ and $Y=7$
etc etc 
each with probability $\binom{25}{8}\frac{10}{25}^{x}\frac{15}{25}^{y}$
But how do i go about the lengthy calculation and find the actual expected value?
 A: Well you could try $E(X)-E(Y)=E(X-Y)$
A: Following the comment in another answer, observe that $X$ and $Y$ are not independent variables. In fact, $Y=8-X$. So $E[X-Y]=E[X-(8-X)]=E[2X-8]=2E[X]-8$. Now, $X\sim\operatorname{Hypergeometric}(10,10+15,8)$, so $E[X]=8\cdot{10\over25}=\frac{16}5$, therefore $E[X-Y]=-\frac85$.  
If you don’t remember the mean of the hypergeometric distribution, another way to attack this is to observe that the distribution is exchangeable, that is, the probability depends only on the number of boys (and girls) selected, not on the order in which they were selected. Therefore, the probability of choosing a boy is order-independent: the probability that the $i$th selection is a boy is simply ${10\over25}$. If we let $B_i$ be indicator variables for these events and $G_i$ be indicator variables for the events “the $i$th choice is a girl,” then $$E[X]=\sum_{i=1}^8E[B_i]=8\Pr(\text{boy is chosen}) \\ E[Y]=\sum_{i=1}^8E[G_i]=8\Pr(\text{girl is chosen}) \\ E[X-Y]=8\Pr(\text{boy is chosen})-8\Pr(\text{girl is chosen})=8\left({10\over25}-{15\over25}\right)=-\frac85.$$
