Probability: Find Dispersion of X + Y $X = \operatorname{Bi}(3,\frac14), Y=\operatorname{Bi}(4,\frac12), \operatorname{Cov}(X,Y) = -\frac34.$
Dispersion of $X+Y =?$
$D(X) = npq = \frac9{16}. D(Y) = npq = 1$
$D(X + Y) = D(X) + D(Y) = \frac{25}{16}$
Something isn't right. I didn't use the given covariance anywhere. What did I do wrong?

Assuming that by $D(X)$, you mean the variance, $Var(X)$, we have:
$Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y)$. So you need to add $-3/2$ to the value that you already have.
Note that $Var(X+Y)$ is equal to $Var(X)+Var(Y)$ if and only if $X$ and $Y$ are independent, i.e. when $Cov(X,Y)=0$.
