I've got the random vector (X,Y), which is defined by the following density function: $$ f(x,y)= \begin{cases} 6x&~\text{ if}~ (x,y)\in T \\ 0&~\text{otherwise} \end{cases} $$
$T$ : a triangle with formed by the vertex on $(0,0)$, $(0,1)$ and $(1,0)$. We also have a new random variable called $Z$, where $Z=X+Y$ and $z$ is a real number between $(0,1)$.
My question is: how can I write $P(Z\le z)$ in function of $z$?
I've written $P(Z\le z)=(X+Y\le z)=P((X,Y)\in B)$, where $B={(X,Y) | y\le z-x}$ But know how can I calculate $P((X,Y) \in B)$ ?
Is it ok if I do the double integral (where $x$ goes from zero to $1$ and $y$ goes from zero to $z-x$), of $6x dy dx$ ? From where I reach this solution: $3z-(1/3)$.
Your approach is correct with one minor error. Obviously $x$ cannot be larger than $z$. Therefore the integral you need is: $$ \int_0^z 6x\,dx\int_0^{z-x}dy=z^3. $$