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 ?
Best Answer
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 $$1-3\cdot \frac16=\frac12.$$