Probability that three independent uniform $(0,1)$ random variables can form a triangle

Solution 1:

The desired probability corresponds to the volume of the subset of the unit cube $[0,1]^3$ that is bounded by the three planes $x+y=z$, $x+z=y$, $y+z=x$. Each of these planes chops off a tetrahedron (e.g. the one with vertices $(0,0,0)$, $(1,0,1)$, $(0,1,1)$ and $(0,0,1)$ for the plane $x+y=z$) of volume $\frac 16$. These tetrahedra are disjoint (only the biggest number can be bigger than the sum of the other two numbers), hence the volume remaining is $$1-3\cdot \frac16=\frac12.$$

Solution 2:

I am too dumb to imagine cutting a tetrahedron in the 3D space, so here is a slight variation of Hagen von Eitzen's answer. Let the three sides be $x,y,z$. Suppose $z$ is fixed and it is the longest side. Then the probability that $x,y,z$ form side lengths of a triangle is the area bounded by $\left\{(x,y): 0\le x\le z,\ 0\le y\le z,\ x+y\ge z\right\}$, which is $z^2/2$. Integrate from $z=0$ to $z=1$, we get $1/6$. Multiply by $3$ (previously we have fixed one of the three sides as the longest one), we obtain the answer as $1/2$.