Let X and Y be two jointly continuous random variables with the given joint PDF; $\begin{equation} \nonumber f_{XY}(x,y) = \left\{ \begin{array}{l l} \frac{1}{3}x^3+\frac{1}{5}y^2 & \quad -2 \leq x \leq 1, 0 \leq y \leq 3 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}$
Could someone check my work here?
The marginal pdf of X; $f_{X}(x) = \int_{0}^{3}f_{XY}(x,y)dy$
And the marginal pdf of X is $ f_{Y}(y) = \int_{-2}^{1}f_{XY}(x,y)dx$
$P(X>0,y<2) = \int_{0}^{2} \int_{0}^{1} f_{XY}(x,y) dx dy$
$P(X<-1 or Y<2) = P(X<-1)+P(Y<2)$
Having trouble with last part of this problem which is;
$P(X+Y>0) = ?$
All I know is that $X+Y>0$ when for $X = a, Y>-a$ or when X = a & Y > -a also when X and Y are both greater than 0 i.e. $X>0,Y>0$.
Unsure where to go from here.
The answer to the "or" question is not set up correctly. Let $A$ be the event $X\lt -1$ and $B$ the event $Y\lt 2$. Then $\Pr(A\cup B)=\Pr(A)+\Pr(B) -\Pr(A\cap B)$. Each of the probabilities on the right is an easily set up integral.
For the last question, draw the line $x+y=0$. We want to find the probability that the pair $(X,Y)$ ends up in the part of our basic rectangle above the line $x+y=0$. The picture should make it not too hard to set up an appropriate integration. But it will be substantially easier to compute the probability that $X+Y\le 0$, since the region of integration is nicer.