Example of where expectation of product of r.v's does not split as product of individual expectations

Question

Can someone give an example of two r.v's $X,Y$ such that $E(XY)\neq E(X)E(Y)$? I know that this means $Cov(X,Y)\neq 0$, so I've been trying r.v's where we have $Y=aX+b$ for some constants $a,b$, but I keep getting $E(XY)=E(X)E(Y)$ for these examples.

Answer 1

Consider: $(X, Y) = (1, 0)$ with $1/2$ probability, and $(0, 1)$ with $1/2$ probability. Then $E[X] = 1/2$, $E[Y] = 1/2$, $E[XY] = 0$.

Note that in this case, $Y=1-X$.