# Probability of sum of two continuous is greater than 1

I am given a two-dimensional absolute continuous random variable, whose density function is defined as followed:

$f_X,_Y(x,y)=1/2$ if $0<x<1$ and $0<y<4x$.

I have found the marginal density functions of X and Y to be:

$f_X(x)=2x$ if $0<x<1$

$f_Y(y)=\frac{1}{2}(1-\frac{y}{4})$ if $0<y<4$

The thing is now, that I have to find the probability that $P(X+Y>1)$. However, I'm lost here. Looking around it seems as though you able to find $P(Z\le z) = P (X+Y \le z)$ by intregrating:

$\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{z-y}f_X,_Y(x,y)dxdy$

And then I figured I could just minus the result of this integration from 1 to find the desired result. However, I'm not getting the correct result (for whatever reason):

1-$\int\limits_{0}^{1}\int\limits_{0}^{1-y}f_X,_Y(x,y)dxdy=$

Hope you can help, I might just be on a completely wrong track.

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Hint:

To be found is: $$\iint_{x+y>1}f_{X,Y}\left(x,y\right)dxdy=\frac{1}{2}\int\int1_{S}\left(x,y\right)dxdy=\frac{1}{2}\lambda(S)$$ where $\lambda$ denotes the Lebesguemeasure on $\mathbb{R}^{2}$ and $S$ denotes the set: $$\left\{ \left(x,y\right)\mid x+y>1\wedge0<x<1\wedge0<y<4x\right\}$$

Have a good look at that set. Finding $\lambda(S)$ is the same as finding the surface of that area.

$1_{S}$ denotes the characteristic function of $S$. It is prescribed by $\left(x,y\right)\mapsto1$ if $\left(x,y\right)\in S$ and $\left(x,y\right)\mapsto0$ otherwise.

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