I would like to determine the probability that at least two out of $n$ stochastic variables are above r, when $X_i=\{X_1, X_2,…,X_n\}$ are $n$ independent draws from a uniform distribution $U(0,1)$.
Some notation that I've come up with: I let $\min_{(1)}X_i$ denote the lowest of the $X_i$'s, $\min_{(2)}X_i$ is the second to lowest of the $X_i$'s and so on. Hence, the n'th lowest $X_i$ is the same as the highest of the $X_i$'s: $\min_{(n)}X_i=\max_{(1)}X_i$.
I get that the result must be something like:
$\mathbb{P}\left[\left(\min_{(1)} X_i>r\right) \cup \left(\min_{(2)} X_i>r, \min_{(1)} X_i<r \right)\cup … \cup \left( \max_{(3)} X_i>r, \max_{(4)} X_i<r\right) \cup \left( \max_{(2)} X_i>r, \max_{(3)} X_i<r\right) \right]$
I have solved the probability when $n=2$ to be: $\mathbb{P}\left[X_1>r,X_2>r \right]=\left(1-\frac{r-0}{1-0}\right)\left(1-\frac{r-0}{1-0}\right)=(1-r)^2$
and when $n=3$:
$\mathbb{P}\left[(X_1>r,X_2>r,X_3>r)\cup (X_1>r,X_2>r,X_3<r) \cup (X_1>r,X_2<r,X_3>r) \cup (X_1<r,X_2>r, X_3>r) \right] = \left(1-\frac{r-0}{1-0}\right)\left(1-\frac{r-0}{1-0}\right)\left(1-\frac{r-0}{1-0}\right)+3\left(1-\frac{r-0}{1-0}\right)\left(1-\frac{r-0}{1-0}\right)\frac{r-0}{1-0}=(1-r)^3+3r(1-r)^2$ \
and I could do the same for $n=4$ or perhaps even $n=5$, but after that it gets quite messy.
Can anyone help me come up with a more general solution or another approach that doesn't take up too much space?
Help of any sort is greatly appreciated!
Best Answer
Consider the complementary event: the event that at most one of the variables exceeds $r$. The probability that none of the exceeds $r$ is $r^{n}$. The probability that exaclty one of them exceeds $r$ is $nr^{n-1}(1-r)$. The required probability is $1-[r^{n}+nr^{n-1}(1-r)]$