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 ?

  • 1
    $\begingroup$ If $X_1+X_2 < X_3$, how do you form a triangle? Don't the lengths of any two sides need to sum to more than the length of the third side? Hint: compute the conditional probability of being able to form a triangle given the length of the third side. Then compute the unconditional probability by applying the law of total probability. $\endgroup$ – Dilip Sarwate Dec 8 '12 at 18:08

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 $1/2$.

  • $\begingroup$ How did you get the probability of of the triangle as z^2/2 $\endgroup$ – user669083 Dec 9 '12 at 20:57
  • $\begingroup$ The region $\left\{(x,y): 0\le x\le z,\ 0\le y\le z,\ x+y\ge z\right\}$ on the $xy$-plane is bounded by the $x$-axis, $y$-axis and the line $x+y=z$. It is a right angled triangle with both width and height equal to $z$. Hence the area (or probability) is $z^2/2$. $\endgroup$ – user1551 Dec 9 '12 at 21:09
  • $\begingroup$ many thanks.... $\endgroup$ – user669083 Dec 9 '12 at 21:11
  • $\begingroup$ You're welcome. $\endgroup$ – user1551 Dec 9 '12 at 21:12

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 $$1-3\cdot \frac16=\frac12.$$


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.