A Problem on Uniform Probability Distribution

Question

Consider three independent uniformly distributed (taking values between 0 and 1) random variables. What is the probability that the middle of the three values (between the lowest and the highest value) lies between a and b where 0 ≤ a < b ≤ 1 ?

Answer 1

Our probability space $\Omega$ is the unit cube $[0,1]^3$, and fate chooses a point $\omega=(x,y,z)\in\Omega$. That the three quantities $x$, $y$, $z\ $ "are independent and uniformly distributed" in $\Omega$ means the following: For any reasonable set $S\subset\Omega$ the probability that the point $\omega$ chosen by fate lies in $S$ is just the volume of $S$. (Note that ${\rm vol}(\Omega)=1$ in this example.)

We now consider a particular $S\subset\Omega$ as follows: $$ S:=\{(x,y,z)\>|\>x\leq y\leq z\ \ \wedge\ \ a\leq y\leq b\}\ .$$ Since the probability $P$ that the chosen point $\omega$ lies in $S$ is equal to ${\rm vol}(S)$ this probability can be written as a triple integral with $y$ as outmost variable: $$P={\rm vol}(S)=\int_a^b \Bigl(\int_0^y dx\Bigr)\Bigl(\int_y^1 dz\Bigr)\ dy=\int_a^b y(1-y)\ dy\ .$$ Now the six orderings of $x$, $y$, $z$ are equiprobable and mutually exclusive. Therefore the probability you are interested in is $\ 6P$.