I’m currently working on a problem where I add $3$ uniform independent random variables $x_1, x_2, x_3$, each from $(0,1)$, and find the probability that $x_1+x_2+x_3geq 1$ and $x_1+x_2 < 1$.

I approached it by finding the pdf $f_{x_1+x_2}(z)$ by convolution and then treating $x_1+x_2$ like a random variable from $(0,2)$ and using convolution to find the pdf $f_{(x_1+x_2)+x_3}(z)$.

There are many posts that establish

$$f_{x_1+x_2}(z) = begin{cases}0 & z< 0 \ z & 0 leq z leq 1 \ 2-z & 1 < z leq 2 \ 0 & z> 2 end{cases}$$

And I have worked that math out and am fine with the reasoning.

Then I do the following.

$$f_{(x_1+x_2)+x_3}(z) = int_{-infty}^{infty} f_{x_1+x_2}(t)f_{x_3}(z-t)dt$$

Since I want $P(x_1+x_2+x_3 geq 1) = 1 – P(x_1+x_2+x_3 < 1)$, all I need is the case where $0 leq z leq 1$. Well, $f_{x_1+x_2}(t)$ is nonzero when $t in (0,2)$, and $f_{x_3}(z-t)$ is nonzero when $0 leq z-t leq 1$ or $t leq z leq t+1$. We know $zleq t+1$ is redundant since $z leq 1$ and $t geq 0$, so by combining conditions we have $f_{x_1+x_2}(t)f_{x_3}(z-t)$ is nonzero iff $0<tleq z $. In fact, since $0 < t leq z leq 1$, $f_{x_1+x_2}(t) = t$. And, since $0 leq z-t leq 1$, $f_{x_3}(z-t) = 1$. Thus,

$$f_{(x_1+x_2)+x_3}(z) = int_{0}^{z} t dt = frac{z^2}{2} text{ for } 0 leq z leq 1$$

Also,

$$P(x_1+x_2+x_3 < 1) = int_0^1 f_{(x_1+x_2)+x_3}(z) dz = int_0^1 frac{z^2}{2} dz = frac{1}{6}$$

This implies $P(x_1+x_2+x_3 geq 1) = frac{5}{6}$. To find the probability that their sum is greater than or equal to $1$ and the sum of $x_1+x_2$ is not, we multiply the probabilities. Using $P(x_1+x_2 < 1) = int_0^1 f_{x_1+x_2}(z)dz = int_0^1 z dz = frac{1}{2}$, we should have that this probability is $frac{5}{6}*frac{1}{2} = frac{5}{12}$ or about $0.4166ldots$

I ran a Python simulation of this to see if it was about right, but ended up getting around $frac{1}{3} = 0.33ldots$. I’m confident the simulation is right, but I can link or add the code if that would be helpful. I believe that I possibly went wrong in treating $x_1+x_2$ like a random variable; is there some condition to convolution for finding pdfs that I violated? Or is it simply an error in the reasoning?

Thank you for all help, I really appreciate it.