Computing probability of one random variable less than another -- how to approach?

Question

I have 2 random variables X~U[-2,2], Y~U[0,1]. Lets define two new random variables $A=(X+Y)^2$, $B=X^2-Y^2$ .

What is the exact numerical value of the probability of $Pr(A<B)$?

My approach to solve this problem is to make a new random variable $Z=A-B$ and then to compute $F_Z(z)=Pr(Z<z)$ and finally solve for $F_z(0)=Pr(Z<0)=(A-B<0)$.

This is the only approach I can think of to solve this. Does this seem like an efficient approach?

Answer 1

Yes it is. We have $$\Pr(Z<0){=\Pr(A<B)\\=\Pr(2XY+Y^2<-Y^2)\\=\Pr(X<-Y)\\={3\over 8}}$$