Understanding $E[x^2]$ in a variance question I know variance is equal to $V[x]=E[x^2] - (E[x])^2 $, but how do you expand $E[x^2]$ for some x if your given the necessary information...
What I mean for example is if you suppose $x = ys+(1-y)r$ then would $E[x^2] = E[(ys+(1-y)r)^2] = E[(ys+(1-y)r) * (ys+(1-y)r)]$?
I'm trying to solve a problem dealing with variance, but feel like I maybe just don't understand how to properly use $E[x^2]$. 

That's right. If $x=ys+(1-y)r$ then $E(x^2)=E((ys+(1-y)r)^2)$. You can expand this and use properties like: if $x$ is a random variable and $a,b\in \mathbb R$ then $E(ax+b)=aE(x)+b$.
