Probability of Sum of random variables with uniform distribution

Question

If $X\sim U[3,5]$ and $Y\sim U[2,6]$. Then what is the probability of $X+Y>7$, given $X$ and $Y$ are independent?

I'm confused on how does this get integrated for $X+Y>7$?

Answer 1

The distribution will be the convolution of the two distributions for $X$ and $Y$, a trapezoid (thanks to @kimchi lover for pointing out the shape of the distribution).

Then just find the area under the trapezoid from $7$ to $11.$

$$\textrm{Probability} = 1/2 + 1/4 = 3/4.$$

See here for another example of two uniform distributions convolved:

Find PDF of given random variable $Z=\frac{(X+Y)}{2}$

Answer 2

Defining a new pair $X^\prime:=\frac{X-3}{2}\sim U[0,\,1],\,Y^\prime:=\frac{Y-2}{2}\sim U[0,\,2]$ of independent variables, we want $P\left(X^\prime+Y^\prime=\frac{X+Y-5}{2}>1\right)$. Conditional on $X^\prime=x\in[0,\,1]$, this is equivalent to $P(Y^\prime>1-x)=\frac{2-(1-x)}{2}=\frac{1+x}{2}$. Now we average this over the distribution of $X^\prime$, viz.$$\int_0^1\frac{1+x}{2}dx=\frac34.$$

Answer 3

Draw a rectangle in the $x,y$ plane, with vertex set $\{(3,2),(5,2),(3,6),(5,6)\}$ and draw the line $x+y=7$ and shade the region in the rectangle above the line, that is, shade the convex hull of the points $\{(3,4),(5,2),(3,6),(5,6)\}$. The desired probability is the ratio of the shaded area to the area of the rectangle. Which of course is $(8-2)/8=3/4$.