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 ?
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 $$1-3\cdot \frac16=\frac12.$$
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 $1/2$.
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 $1/2$.
If you describe an elephant to a kid but never show them the picture of one, they won't truly appreciate it.
What the (very good) answer by Hagen Von Eitzen is missing is a picture.
See the green triangles emnating from the origin, O? Those form the base of the three tetrahedra he's referring to.
Consider one of those three Terahedra, OABC. This represents the region, $u_3 > u_1+u_2$.
We can choose its base, ABC. This is a right triangle and has an area of $\frac{1}{2}$. The height of this tetrahedron is the red line, OC. And that is $1$. The volume of a Tetrahedron in 3-d space is one third of the area of its base multiplied by the height. So, the volume of OABC is $\frac{1}{3} \times \frac{1}{2} \times 1 = \frac{1}{6}$. Its also easy to see that the three Terahedra don't intersect.
The answer by user1551 is very nice but it feels somewhat out of the blue. Here is a more intuitive variation where we directly condition on $Z$.
Let $z\in (0,1)$ be fixed. We look for $$P(X < Y + z, Y< X + z, z < X+Y).$$ Since $(X,Y)$ has uniform distribution on the unit square, this probability is the area of the region $R = \{(x,y)\in [0,1]^2: x < y + z, y< x + z, z < x+y\}$. The region $R$ is easily sketched, see below when $z=3/4$.
The area of $R$ is found by subtracting the area of the complement: $$1-\frac {z^2}2 - 2\cdot \frac {(1-z)^2}2 = 2z - \frac{3z^2}2$$ and the final probability is $$E\left[2Z - \frac{3Z^2}2\right] = \frac 12.$$
