Probability that three independent uniform $(0,1)$ random variables can form a triangle I am preparing for a probability exam and while practicing stuck on this question. Do not even know how to begin.
Let $X_1$, $X_2$, $X_3$ be independent uniform $(0,1)$ random variables. What is the probability that we can form a triangle with three sticks of length $X_1$, $X_2$, $X_3$?
I am thinking of using $X_1 + X_2 > X_3$ for this to happen and there are three such combinations. But how to proceed next ?
 A: 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 $$1-3\cdot \frac16=\frac12.$$
A: 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 $1/2$.
A: Here is a slightly related answer. The key is to get the inequalities right to map it to the problem at hand. Note: The problem discussed in the blog does apply directly to your problem, but is almost there
A: Let $x,y,z$ be the three numbers. The probability that $z > x + y$ is given by
$$
\int_0^1 dz \int_0^z dy \int_0^{z-y} dx = 1/6.
$$
Any of the three variables could be the largest, so multiply by $3$ and we obtain $1/2$.
Here's a simpler geometric argument. Suppose the largest of the three is $a$. Then the other two must lie in the square with area $a^2$. In that square, the northwest-to-southeast diagonal gives the line where the other two sum to $a$. Below that line, the other two sum to less than $a$. Clearly that line cuts the square in half. Hence, for any $a$, the probability is $1/2$.
