What is the probability of the third-highest order statistic? I'm interested in finding the probability that the third-highest order statistic ($X_3$) is greater than x, when the order statistics $X_1,X_2,...,X_n$ are i.i.d. $U(0,1)$.
I assume $n\geq 3$
So far I get that the probability of this happening is the same as one less the complimentary event, where the complementary event is that (i) the highest order statistics ($X_1$) is below x, (ii) only the highest order statistics is above x, and (iii) only the highest and second-highest order statistic are above x.
Thus, the probability that at least the third-highest order statistic is above x is:
$ \mathbb{P}(X_3>x) = 1 - \left(\underbrace{\mathbb{P}(X_1<x)}_\text{(i)} + \underbrace{\mathbb{P}(X_1>x, X_2<x)}_\text{(ii)} + \underbrace{\mathbb{P}(X_2>x, X_3<x)}_\text{(iii)}\right)\\
= 1 - \left(\mathbb{P}(X_1<x) + \mathbb{P}(X_1>x) \mathbb{P}(X_2<x) + \mathbb{P}(X_2>x)\mathbb{P}(X_3<x)\right) \\
= 1- \left(\mathbb{P}(X_i<x)^n + n (1-\mathbb{P}(X_i<x))\mathbb{P}(X_i<x)^{n-1} + n (1-\mathbb{P}(X_i<x))^2\mathbb{P}(X_i<x)^{n-2}  \right)\\
= 1- \left( r^n + n(1-r)r^{n-1}+ n(1-r)^2r^{n-2}\right)
$
But I'm not entirely sure if I got it right, and I haven't been able to find any sources that confirm my equation.
So please, if someone will help me understand if I'm on the right track, or doing it wrong, I would really appreciated it :)
EDIT: and I assume $X_i$ is just any of the $X_1,X_2,...,X_n$ not necessarily in order.
 A: We can use a similar method to your previous question, as you probably suspect. :)
Probability that at least two of $X_1,X_2,...,X_n$ iid uniform variables are above r?
To clear up notation, suppose $X_i \sim U[0, 1], i = 1, …, n$ are uniform random variables, and let us denote the $i^{th}$ order statistic as $X_{(i)}$. Since you’re looking for the third highest order statistic, in this notation we’re looking for the event $X_{(n-2)} > x$. But then the complementary event $X_{(n-2)} < x$ is precisely the event that at most $2$ elements exceed $x$! This gives us
$$
\begin{align*}
P(X_{(n-2)} > x) &= 1 - P(X_{(n-2)} < x) \\
&= 1 - [P(2 \ X_i’s \text{ exceed } x) + P(1 \ X_i’s \text{ exceed } x) + P(0 \ X_i’s \text{ exceed } x)] \\
&= 1 - \left[\binom{n}{2}(1-x)^2x^{n-2} + \binom{n}{1}(1-x)x^{n-1} + \binom{n}{0}x^n\right] \\
&= \boxed{1 - \frac{n(n-1)}{2} (1-x)^2x^{n-2} - n(1-x)x^{n-1} - x^n}.
\end{align*}
$$
As for your approach, it’s very nearly correct - the only issue is the computation of (iii). In event (iii), you want 2 $X_i$’s above $x$ and the other $n-2$ $X_i$’s below $x$. The probability that  $(X_1 > x, X_2 > x, X_3 < x, …, X_n < x)$ is $(1-x)^2 x^{n-2}$, but there are $\binom{n}{2}$ different arrangements of the $X_i$’s that give you a legal sample satisfying the conditions of event (iii), giving you a total probability of $\binom{n}{2} (1-x)^2 x^{n-2}$.
Hope this helps!
