# Pick $2$ numbers from $[-1,1]$,what is the probability that their sum is greater than $1$?

Pick 2 numbers from $$[-1,1]$$, what is the probability that their sum is greater than 1?

It is equal to the probability that the sum of 2 uniform random variables on $$[-1,1]$$ is greater than 1?

so far,

I got $$f(x) = 1/2$$ $$(-1 < x < 1)$$ and $$f(y) = 1/2$$ $$(-1 < y < 1)$$, I need to calculate $$P(X + Y > 1)$$.

I plot the picture of the above convolution, it is a triangle with vertices on $$(-2,0),(2,0),(0,1/2)$$.

So $$P(X + Y > 1)$$ is the area to the right side of $$x = 1$$, which is $$1/2 * 1/4 * 1 = 1/8$$, is this correct?

Yes, that is right. An alternative way to get the same answer is to argue as follows. For the sum to exceed $$1$$, we need both variables to be positive, so $$P(X+Y>1)=P(X,Y>0)\times P(X+Y>1\mid X,Y>0).$$ Now $$P(X,Y>0)=\frac12\times\frac12$$. Conditioning on this, gives independent uniform variables on $$[0,1]$$. If $$A,B$$ independent uniform on $$[0,1]$$, $$A+B>1$$ if and only if $$(1-A)+(1-B)<1$$, and $$(1-A),(1-B)$$ are also independent uniform on $$[0,1]$$. It follows that $$P(X+Y>1\mid X,Y>0)=\frac12$$, giving an overall probability of $$\frac12\times\frac12\times\frac12$$.
I think the quickest way to see this is geometrically. Choosing two points uniformly on $$[-1,1]$$ is the same as choosing one point uniformly in the box $$[-1,1]\times[-1,1]\subset\mathbb{R}^2$$. Precisely $$\frac18$$ of this square is above the line $$x+y=1$$.