I'm doing this problem from Carol Ash's The Probability Tutoring Book
Let $X_1, X_2, ... , X_7$ be iid and have common density $f(x)$ and common distribution $F(x)$. Find the probability that the largest is less than 9 and the second largest is less than 5
My attempt:
We want $P(\max \leq 9 \cap \text{2nd largest} \leq 5)$
$$P(\max \leq 9 \cap \text{2nd largest} \leq 5) = 7P(\text{max} \leq 9 )P(\text{2nd largest} \leq 5)^6 \\= 7F(9)F(5)^6$$
My reasoning:
The $7$ comes from choosing 1 of the 7 to be the max
The first equality comes from independence and the second comes from the definition of the CDF.
However according to the back of the book the answer is $$F(5)^7 + 7(F(9) - F(5))(F(5))^6$$
Thir answer makes sense to me, but I don't know why my answer is wrong(logically it makes sense to me). The way their answer works (I believe) is they did
$$P(\text{all} \leq 5) + P(\text{1 X between 5 and 9 while the rest are less than 5}) $$
Where is my reasoning wrong? I choose one X to be anything less than $9$ and the others are less than $5$
That's the intuitive approach. It matches the integration result:
$$\begin{align}f_{\small X_{(6)},X_{(7)}}(y,z) &= \dfrac{7!}{5!} F(y)^5 f(y) f(z)\\[4ex] F_{\small X_{(6)},X_{(7)}}(y,z)&=\dfrac {7!}{5!}\int_{-\infty}^{5}\int_y^9 F(y)^5f(y)f(z)\mathrm d z\mathrm d y\\&=42\int_{-\infty}^5 F(y)^5f(y)(F(9)-F(y))\mathrm d y\\&=42\int_0^{F(5)} x^5 (F(9)-x)\mathrm d x&&{x=F(y)\\ \mathrm d x=f(y)\mathrm d y}\\&=\int_0^{F(5)} 7\cdot 6 x^5 F(9)- 6\cdot 7x ^6~\mathrm d x\\[1ex] &= 7 F(5)^6 F(9)-6F(5)^7 \\[3ex]&= F(5)^7 -7(F(9)-F(5))F(5)^6\end{align}$$
Which are "the others" when they are in the same category?
The constants in the expression derive from counting distinct arrangements. When the one you choose is less than 5 it becomes indistinguishable from the other six described that way. So you must partition the events between "all are birds of a feather" and "one stands out from the crowd" to properly account for this.
$$P\left(\bigcap_{i}\{X_i \leqslant 5\}\right) + P\left(\bigcup_j \left(\{5<X_j\leqslant 9\}\cap\bigcap_{i\neq j}\{X_i\leqslant 5\}\right)\right)\\{\huge=}\\ \mathsf P(X_1\leqslant 5)^7+ 7\mathsf P(5<X_1\leqslant 9)\mathsf P(X_2\leqslant 5)^6\\{\huge=}\\ F(5)^7-7\,(F(9)-F(5))\,F(5)^6 $$