My problem:
Suppose that a random sample of size $n=4$ is taken without replacement from a finite population with $N=10$. Now suppose that $c_{i}=3i+4, i=1,2,...,N$, where $c_i$ is the label on the $i^{th}$ ball in the urn.
Let $Y=X_{1}+3X_{2}$ and $W=X_{1}+X_{3}+2X_{4}$.
Find the covariance of $Y$ and $W$.
My attempt:
So far I have found the variance of $c_{i}$, $$\sigma^{2}= \frac{\sum_{i=1}^{N}\left(x_{i}-\mu\right)^{2}}{N}=\frac{742.5}{10}=74.25$$
After this is where I start to get lost.
I know that for integers $r, s \in \{1,2,...,n\}$ with $r\ne s$, the covariance can be calculated as $$cov\left(X_{r}, X_{s}\right)= -\frac{\sigma^{2}}{N-1}$$
However, I am not sure how to proceed with the multiple variables. Any hints or tips would be greatly appreciated. Thanks.
Remember that $Cov$ is linear in each argument. Letting $\sigma^2=74.25$, you have
$Cov(Y,W)=Cov(X_1+3X_2,X_1+X_3+2X_4)\\=Cov(X_1,X_1)+Cov(X_1,X_3)+2Cov(X_1,X_4)+3Cov(X_2,X_1)+3Cov(X_2,X_3)+6Cov(X_2,X_4)\\=\sigma^2-\frac{1}{9}(1+2+3+3+6)\sigma^2\\=-\frac{2}{3}\sigma^2\\=-49.5$