The CDF of the k-th order statistic is given by- \begin{align*} F_{X_{(k)}}(x) & = \sum_{i=k}^n {n \choose i} [F_X(x)]^i\cdot [1 - F_X(x)]^{n-i} \end{align*}
But I do not understand why there is an ${ n \choose i}$ term there? From what I understand, there cannot be different combinations as we require the first (not any) $r$ many order-statistics to be $\leq x$. Say, if I wanted to calculate $F_{X_{(n-1)}}(x)$ , then I could do the following- \begin{align*} F_{X_{(n-1)}}(x) &= P[X_{{(n-1)}} \leq x] \\ &= P[X_{{(n-1)}} \leq x < X_{(n)}] + P[X_{{(n)}} \leq x ] \\ &= [F_X(x)]^{n-1}\cdot[1 - F_X(x)] + [F_X(x)]^n \\ &= [F_X(x)]^{n-1} \end{align*} But if I use the formula written above, the answer comes out to be- \begin{align*} F_{X_{(n-1)}}(x) &= n[F_X(x)]^{n-1}\cdot[1 - F_X(x)] + [F_X(x)]^n \end{align*} It isn't clear to me why do we need the extra $(n-1)$ terms?
Suppose $n=4$. Then $P(X_{(3)} \le x < X_{(4)})$ does not equal $[F_X(x)]^{3} (1-F_X(x))$. What is true is that \begin{align} P(X_1 \le x, X_2 \le x, X_3 \le x, X_4 > x) \\ P(X_1 \le x, X_2 \le x, X_3 > x, X_4 \le x) \\ P(X_1 \le x, X_2 > x, X_3 \le x, X_4 \le x) \\ P(X_1 > x, X_2 \le x, X_3 \le x, X_4 \le x) \end{align} each have probability $[F_X(x)]^3 (1-F_X(x))$ (note that we have crucially used independence of $X_1,\ldots, X_4$, whereas $X_{(1)}, \ldots, X_{(4)}$ are not independent, nor are their CDFs the same as $F_X$). Summing the four probabilities above yields $4[F_X(x)]^3(1-F_X(x))$.