Calculate probability of joint PDF I'm given the following joint PDF and asked to calculate $P(X+Y>1)$
$f_X$$_Y$$(x,y)=2/5$ for $0<y<1$   & $0<x<5y$ 
and 
$f_X$$_Y$$(x,y)=$ $0$ else
I know I have to take the double integral, where the lower bound for the inner intergral is $1-x$ and the upper bound is $1$. But somehow I'm not getting the correct result, which I know is $5/6$ (from the solutions I've been given). 
Can anybody show me how to calculate $P(X+Y>1)$?
Thanks in advance
 A: We have that $(X,Y)$ is uniformly distributed over $S$, where
$$S=\{(x,y)\in\mathbb R^2 : 0<y<1, 0<x<5y\}. $$
Let $R\subset S$ be the region
$$\{(x,y)\in S: x + y > 1\}.$$
Then to find the probability $\mathbb P\{(X,Y)\in R\}$, we would integrate the density function over this region. But since the density function is constant, this boils down to finding the area of $R$. Notice that $S$ is the triangle bounded the the lines $x=0$, $y=1$, and $y=\frac15x$, and $R$ is the triangle bounded by the lines $y=1$, $y=\frac15x$, and $y=1-x$. The vertices of $R$ are the intersections of these lines, namely $(0,1)$, $(5, 1)$, and $\left(\frac56,\frac16\right)$. The area of this triangle is $\frac{25}{12}$, and hence $\mathbb P\{(X,Y)\in R\}$ is $\frac25\cdot\frac{25}{12} = \frac56$.
Alternatively, to find the bounds on the integral, we have $0<y<1$, $0<x<5y$, and $x+y>1$. So for $0<x<\frac56$, $y$ is between $1$ and $1-x$, and for $\frac56<x<1$, $y$ is between $\frac15 x$ and $1$. Hence
$$\mathbb P\{(X,Y)\in R\} = \int_R f = \int_0^{\frac56}\int_{1-x}^1\frac25\mathsf dy\mathsf dx + \int_{\frac56}^5\int_{\frac15 x}^1\frac25\mathsf dy\mathsf dx.$$
which evaluates to $\frac56$.
