What is the probability that the student who finishes last takes over twice as long as the student who finishes first? Three students independently attempt to solve a statistics problem. Assume the times taken (in minutes) by each student to solve the problem are identically distributed on $U(0,30)$. What is the probability that the student who finishes last takes more than twice as long as the student who finishes first? What if there are $n$ students and the distribution is $U(0,\theta)$?
I believe this is an ordered statistics problem. So, $T_1, T_2, T_3 \sim U(0,30)$ independently, where $T_i$ is the time is takes for the $i$th student to complete the problem. So, we want to find $P(T_{(3)} > 2T_{(1)})$. 
For the general case with $n$ students, we have $T_1, \cdots, T_n \sim U(0,\theta)$ independently. So what want to find the probability $P(T_{(n)}>2T_{(1)})$.
From here, I do not know how to evaluate $P(T_{(3)} > 2T_{(1)})$ or $P(T_{(n)}>2T_{(1)})$. How would you go about this calculation?
 A: Name the students Adam, Bob, Chris with finishing times $a, b, c$ respectively.
We will first compute the probability of the event  $E_{abc} = \{(a,b,c)| a < b < c, 2a < c \}$. Then, we will multiply the result by six since there are six permutations of $a,b,c$.
From your notation, $U(0,30)$, I will assume the times are distributed uniformly on $[0,30]$ for each student, so the 3-tuple of times $(a,b,c)$ is distributed uniformly on $[0,30]x[0,30]x[0,30]$ with density $\rho(a,b,c)=(1/30)^3=1/27000$
$P(E_{abc})= 1/27000 \int_{a=0}^{15} \int_{c=2a}^{30} \int_{b=a}^c dbdcda = 1/8$
$P(E)= 6*P(E_{abc}) = 3/4$
Now that I see the numerical result, I suspect there is a simpler argument.
A: In a more general case, we'd like to calculate $\mathbb{P}(T_{(n)}>kT_{(1)})=1-\mathbb{P}(T_{(n)}<kT_{(1)})$. Assume we're given $T_{(n)}$, then $\mathbb{P}(T_{(n)}<kT_{(1)}|T_{(n)}=a)=\left({k-1\over k}\right)^{n-1}$, because all the other $n-1$ RVs will fall in the range $\left[\frac{a}{k},a\right]$. Since this result is the same for all possible $T_{(n)}$, the unconditional probability is also equal to this. This can also be seen mathematically using total probability theorem:
$$\begin{align}\mathbb{P}(T_{(n)}<kT_{(1)})&=\int_0^\theta \mathbb{P}(T_{(n)}<kT_{(1)}|T_{(n)}=a)f_{T_{(n)}}(a)da\\&=\left({k-1\over k}\right)^{n-1}\underbrace{\int_0^\theta f_{T_{(n)}}(a)da}_1\end{align}$$
Then $$\mathbb{P}(T_{(n)}>kT_{(1)})=1-\left(\frac{k-1}{k}\right)^{n-1}$$
In our case, $k=2, n=3$, we have $\mathbb{P}(T_{(3)}>2T_{(1)})=3/4$
