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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 ?

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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. – Dilip Sarwate Dec 8 '12 at 18:08
up vote 2 down vote accepted

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 $1/2$.

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How did you get the probability of of the triangle as z^2/2 – user669083 Dec 9 '12 at 20:57
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 $z^2/2$. – user1551 Dec 9 '12 at 21:09
many thanks.... – user669083 Dec 9 '12 at 21:11
You're welcome. – 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 $$1-3\cdot \frac16=\frac12.$$

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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

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