How can I find $P(X+Y>4)$? I have the following density function:
$$ f(x, y)=\begin{cases} \dfrac{x^2+y^2}{50} & 0<x<2, 1<y<4, \\
\\ 0 & \text{elsewhere.} \end{cases}$$
How can I calculate $ P(X+Y>4)$ ?
 A: Guide:
Sketch the region $0 < x < 2, 1< y < 4$, and $x+y > 4$.
Evaluate $$\int_0^2 \int_{4-x}^4 f(x,y) \,dy\,dx$$
A: $$
\left( \iint\limits_{\left\{ (x,y) \, : \, \begin{array}{l} 0 \, \le \, x \, \le \, 2 \\ 1 \, \le \, y \, \le \, 4 \\ x \, + \, y\, > \, 4  \end{array} \right\}} 
\cdots\cdots \right) = \left( \iint\limits_{\left\{ (x,y) \, : \, \begin{array}{l} 0 \, \le \, x \, \le \, 2 \\ 1 \, \le \, y \, \le \, 4 \\ y\, > \, 4 \, -\, x \end{array} \right\}} \cdots\cdots \right) = \int_0^2 \left( \int_{4-x}^4 \cdots\cdots \, dy \right) dx
$$
Here's something you need to be careful about: Suppose for that last term we had written
$$
\int \left( \int \cdots\cdots \, dx \right) \, dy
$$
with the integral with respect to $x$ on the inside instead of the outside. Then we would need
$$
\int_2^4 \left( \int_{4-y}^2 \cdots\cdots \, dx \right) \, dy
$$
with $y$ going from $2$ to $4$ rather than from $1$ to $4.$ That is because when $y<2$ then $4-y>2,$ and with $x$ going from $4-y$ to $2$ we would have $x>2,$ which is not within the rectangle.
