Given $U_1$, ..., $U_n$ and $V$ with the same density function, find $P(V \le U_{(n)})$ This is exercise 3.81 from John Rice's Mathematical Statistics and Data Analysis, the complete problem is as follows:
Given $U_1$, ..., $U_n$ and $V$ with the same density function $f$ and the cdf $F$, with $F^{-1}$ uniquely defined, find $P(V \le U_{(n)})$ and $P(U_{(1)} < V < U_{(n)})$.
$U_{(i)}$ is the $i$th random variable ordered by size, so $U_{(1)}$ and $U_{(n)}$ are the min and max value of all $U_{i}$, respectively.
I was able to solve this using law of total probability (integrating over $v$), but what is nagging me is that in the problem it included a hint: "$F(U_i)$ has a uniform distribution." So this makes me wonder if I am missing a simpler solution, or if my solution is wrong and just managed to arrive at the same answer by a fluke. I would really appreciate it if anyone can enlighten me on how $F(U_i)$ might be useful in solving this problem. Thanks so much in advance!
 A: 
Rename $V$ as $U_{n+1}$ and May the Force (of Symmetry) be with you...

More seriously, defining $U_{n+1}=V$ produces an i.i.d. sample $(U_k)_{1\leqslant k\leqslant n+1}$ with a continuous distribution, hence there is no tie, almost surely, and every ordering of these $n+1$ values is equally probable. 
For example, $[V\leqslant U_{(n)}]$ means that $U_{n+1}$ is not the largest result in the sample. Each random variable produces the largest result with probability $\frac1{n+1}$, hence
$$
P[V\leqslant U_{(n)}]=1-\frac1{n+1}.
$$
Likewise, $[U_{(1)}\lt V\lt U_{(n)}]$ means that $U_{n+1}$ is neither the largest nor the smallest result in the sample. Each random variable produces the largest result or the smallest result with probability $\frac2{n+1}$, hence
$$
P[U_{(1)}\lt V\lt U_{(n)}]=1-\frac2{n+1}.
$$
More generally, define $U_{(0)}=-\infty$ and $U_{(n+1)}=+\infty$, then, for every $1\leqslant k\leqslant n+1$,
$$
P[U_{(k-1)}\lt V\lt U_{(k)}]=\frac1{n+1},
$$ 
hence, for every $0\leqslant i\leqslant j\leqslant n+1$,
$$
P[U_{(i)}\lt V\lt U_{(j)}]=\frac{j-i}{n+1}.
$$
These results are valid when there is no tie, that is, as soon as the distribution of the sample is continuous.
