Let $U_1$, $U_2$ and $U_3$ be i.i.d. $\mathsf {Uniform}[0,1]$ random variables. Find $P(U_1+U_2 > U_3)$
Here the answer states that the probability is $\frac{5}{6}$. However I am quite confused as to how we obtain such an awnser. A hint is to use conditional expectation.
Any advice or explanation is apreciated.
I saw the following link :Probability that $\max(U_1,U_2) > U_3$ for independent uniform random variables $U_i$
And the key argument was the symetry of the system.
The PDF of $U_1+U_2$ is supported on $[0,2]$ and equals $x$ on $[0,1]$ and $2-x$ on $[1,2]$.
It follows that: $$\begin{eqnarray*}\mathbb{P}[U_1+U_2>U_3] &=& \int_{0}^{1}\int_{u}^{2}(1-|v-1|)\,dv\,du\\&=&\int_{0}^{1}\left(\frac{1}{2}+\int_{u}^{1}v\,dv\right)\,du \\&=&\int_{0}^{1}\frac{2-u^2}{2}\,du\\&=&1-\frac{1}{6}=\color{blue}{\frac{5}{6}}.\end{eqnarray*}$$