Let $X\sim U[0,20]$ and $Y\sim U[5,15]$ two independent random variables.
Calaculate $P(X+Y\leq t)$.
Can anyone help me in this question, I get here:
$\left\{ X+Y\le t\right\} \Rightarrow\left\{ Y\le t-X\right\} \Rightarrow\left\{ X\in\mathbb{R}\right\} \cap\left\{ Y\le t-X\right\} \Rightarrow\\\int\limits _{0}^{20}\frac{1}{20}\left(\int\limits _{5}^{t-x}\frac{1}{10}dy\right)dx=\\\frac{1}{20}\cdot\frac{1}{10}\int\limits _{0}^{20}1\left(\int\limits _{5}^{t-x}1dy\right)dx=\\\frac{1}{200}\int\limits _{0}^{20}1\left(\int\limits _{5}^{t-x}1dy\right)dx$
But I don't know how to continue now. Thank you!
Do a drawing of the domain (a rectangle) and pass through the line $X+Y=t$.
The corresponding CDF is the area under the line multiplied by the joint density
$$F_T(t)=\begin{cases} 0, & \text{if $t<5$} \\ \frac{1}{200}\cdot\frac{(t-5)^2}{2}, & \text{if $5\leq t<15$} \\ \frac{1}{4}+\cdot\frac{(t-15)\cdot10}{200}, & \text{if $15\leq t<25$} \\ 1-\frac{(35-t)^2}{2\cdot200}, & \text{if $25\leq t<35$} \\ 1, & \text{if $t\geq 35$} \end{cases}$$
The 3 areas are obviously
a triangle
the triangle with area $\frac{1}{4}$ plus a parallelogram
1 minus the remaining triangle
Hint: if you find too difficult this exercise try first with the same exercise where $X,Y$ are independent uniform on $[0;1]$ then solve your exercise with this rectangle
Here is the drawing