If two Continuous Random Variables $-1<A,B<1$ are independently positive (negative) with the probability $\frac 12$, what is the probability $A+2B>0?$

Question

If we have two Continuous Random Variables $-1<A<1$ and $-1<B<1$ which are independently positive and negative with probability $\frac 12$, then, based on the given information, can we compute the probability that $A+2B>0 $? Assuming that the random variables $A,B$ are uniformly distributed over the interval.

Answer 1

Once you know the joint probability distribution of $A,B$ you just have to calculate the integral$$P(A+2B>0)=\int_Df_{A,B}(x,y)~dx~dy$$where $D:=\{(x,y)\in\Bbb R^2:x+2y>0\}$. As an example, consider that $A,B$ are independent random variables that are uniformly distributed over $(-1,1)$, i.e. their PDFs are $f_A(x)=f_B(x)=\frac12,-1<x<1$ and $0$ otherwise. Then $f_{A,B}(x,y)=f_A(x)f_B(y)=\frac14,-1<x,y<1$ and $0$ otherwise. The required integral is$$\begin{align*}P(A+2B>0)&=\int_{-1}^1\int_{-x/2}^1\frac{dy~dx}4\\&=\frac14\int_{-1}^1\left(1+\frac x2\right)dx\\&=1/2.\end{align*}$$