An elementary Probability Problem This maybe simple but I can not see the answer:
Choose 3 points in the unit interval $[0,1]$ (randomly) to get 4 subintervals. What is the probability that at least one of these 4 intervals has length greater than $\frac 13$ ?
 A: We are going to compute the probability $q$ that none of the four subintervals (called parts in the following) has a length $>{1\over3}$.
Choosing three points independently and uniformly in the interval $[0,1]$ is the same thing as choosing a point $(x,y,z)$ uniformly in the unit cube $C:=[0,1]^3$. The relevant probability measure on $C$ is ordinary volume. Therefore we have to compute the volume of a certain subset $Q$ of $C$ where certain inequalities involving $x$, $y$, and $z$ are true.
In order to simplify the combinatorics we are going to compute the volume of the set $Q_<:=\{(x,y,z)\in Q\>|\>x\leq y\leq z\}$, which makes up one sixth of $Q$.
A point $(x,y,z)\in C$ lies in $Q_<$ (in particular: gives not rise to a part of length $>{1\over3}$) iff the following inequalities are satisfied:
$$x\leq y\leq z,\quad 0\leq x\leq{1\over3},\quad {2\over3}\leq z\leq 1, \quad z-{1\over3}\leq y\leq x+{1\over3}\ .\tag{1}$$ 
It turns out that the first of these inequalities, $(1_1)$, is a consequence of the three others. Therefore we may choose $x$ and $z$ independently in the  intervals given by $(1_2)$ and $(1_3)$, and then $y$ in the (maybe empty) interval $I_{x,z}$ defined by $(1_4)$.
To compute ${\rm vol}(Q_<)$ we begin with
$$\int\nolimits_{I_{x,z}} dy=\left({2\over3}-(z-x)\right)^+\ , $$
where $t^+:=\max\{0,t\}$, 
and proceed with
$$\int_{2/3}^1 \left({2\over3}-(z-x)\right)^+\ dz=\int_{2/3}^{2/3+x}\left({2\over3}+x-z\right)={x^2\over2}\ .$$
In this way we finally obtain
$${\rm vol}(Q_<)=\int_0^{1/3}{x^2\over2}\ dx={1\over 6\cdot 27}\ .$$
It follows that $$q={\rm vol}(Q)=6\>{\rm vol}(Q_<)={1\over27}\ .$$ Therefore the probability $p$ that at least one part has a length $>{1\over3}$  is equal to $$p=1-q={26\over27}\ .$$
