What is the probability that a triangle can be constructed…

What is the probability that a triangle can be constructed from three randomly chosen segments whose lengths are not greater than $l$?

Step 1: The model

The lengths $x,y,z$ are independent and uniform on $[0,l]$: the point $(x,y,z)$ is uniform in the cube $[0,l]^3$ of volume $l^3$.

Step 2: When does the triangle fail?

A triangle exists iff $x\lt y+z$, $y\lt x+z$, $z\lt x+y$. It fails when one of the opposite inequalities holds, e.g. $x\ge y+z$. Two of them together, say $x\ge y+z$ and $y\ge x+z$, would give $x\ge x+2z$, so $z=0$ — a set of volume $0$. So the failure events are (practically) disjoint.

Step 3: Volume of one failure event

$\{x\ge y+z\}$ inside the cube: for each $x$, the pairs $(y,z)$ with $y+z\le x$ form a triangle of area $\frac{x^2}2$ (it fits in the square since $x\le l$). So

\[ V = \int_0^l\frac{x^2}{2}\,dx = \frac{l^3}{6}. \]

Step 4: The probability

\[ P(\text{no triangle}) = 3\cdot\frac{l^3/6}{l^3} = \frac12, \qquad \boxed{P(\text{triangle}) = 1-\frac12=\frac12.} \]

Step 5: Comparison with Problem 81

In Problem 81 the three pieces must add up to $l$, and the probability is only $\frac14$. Here the three lengths are free, and the chance is twice as large: $\frac12$.