Let the joint density of two random variables $X$ and $Y$ is given by
$f(x,y)=1,$ where $ 0\le x,y \le 1$
Find, $P(X >2Y)$
What I tried:-
$P(X >2Y)=1- P(X <2Y)=1-[\int_{0}^{1} \int_{0}^{2y} f(x,y) dx dy]=1-[\int_{0}^{1} \int_{0}^{2y} 1 dx dy]$
I am getting the answer $0$
Am I correct?
If you draw the square $[0,1]\times [0,1]$ and draw the area $x>2y$, you will see that you get the boundaries $2y< x <1$ and $0<y<\frac{1}{2}$, so you will get $$P(X >2Y)=\int_{0}^{\frac{1}{2}} \int_{2y}^{1} f(x,y) dx dy=\int_{0}^{\frac{1}{2}}1-2y\,dy=\frac{1}{4}.$$
Remark: If you look at the drawing, you will see that it's easier to calculate $P(X >2Y)$ (area is one triangle) than $P(X <2Y)$ (area is a trapezium, which you will need to split into two separate integrals).