Finding $P(X+Y>1/2)$ given joint density of $(X,Y)$

Question

I am unable to get the right answer for this problem:

Let $X,Y$ have joint probability density function:

$$f(x,y)=x+y \quad\text{ for } 0<x<1,0<y<1$$

Then $P\left(X+Y>\frac{1}{2}\right)$ equals ?

I drew the diagram and found that we have to add area of two regions, one before the point $x=1/2$ and one rectangle after. However I am not getting the correct answer. I even tried to subtract the area of the small triangle formed with no luck.

Thanks in advance :)

Answer 1

It is easier to find P(X+Y <1/2) and subtract it from 1 to get the answer. You just have to integrate f(x,y) such that x varies from 0 to 1/2 and y varies from 0 to (1/2) -x. This works out to be 1/24. So your answer is 1-1/24 = 23/24