# Suppose that X_1,…,X_n are independent, geometric_1(p) random variables. Evaluate P(min(X_1,…,X_n)>l) and P(max(X_1,…,X_n) le l).

Evaluating min/max probability with Geometric random variables
Suppose that ${X}_{1},\dots ,{X}_{n}$ are independent, ${\text{geometric}}_{1}\left(p\right)$ random variables. Evaluate $P\left(\text{min}\left({X}_{1},\dots ,{X}_{n}\right)>l\right)$ and $P\left(\text{max}\left({X}_{1},\dots ,{X}_{n}\right)\le l\right)$
The textbook solution is as follows:

and
I'm having trouble understanding pretty much the entire solution. If I were to summarize what's particularly puzzling me:
1) How was the first line of each solution derived (i.e. $P\left(\text{min}\left({X}_{1},\dots ,{X}_{n}>l\right)=P\left({\cap }_{k=1}^{n}\left\{{X}_{k}>l\right\}\right)$ and the corresponding equation for max)?
2) The PMF for ${\text{Geometric}}_{1}\left(p\right)$ is given as ${p}_{X}\left(k\right)=\left(1-p\right){p}^{k-1}$. How is $P\left({X}_{k}>l\right)={p}^{l}$? The same applies for the max case as well.
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Step 1
1) This is because the minimum of a set of numbers is greater than l if and only if all the numbers are greater than l. So $min\left({X}_{1},\dots ,{X}_{n}\right)>l$ if and only if ${X}_{1}>l$ and ${X}_{2}>l$ and ... and ${X}_{n}>l$. Note that the event "${X}_{1}>l$ and ${X}_{2}>l$ and ... and ${X}_{n}>l$" is written as $\bigcap _{k=1}^{n}\left\{{X}_{k}>l\right\}$. Thus $P\left(min\left({X}_{1},\dots ,{X}_{n}\right)>l\right)=P\left(\bigcap _{k=1}^{n}\left\{{X}_{k}>l\right\}\right).$.
The max case is similar reasoning; it's because the maximum of a set of numbers is less than or equal to l if and only if all the numbers are less than or equal to l.
Note that what I said above holds true regardless of the distribution of the ${X}_{k}$'s.
Step 2
2) Note that if X has PMF $\left(1-p\right){p}^{k-1}$ for $k\ge 1$, then for any integer $l\ge 1$, we have

(And so $P\left(X>l\right)=1-P\left(X\le l\right)={p}^{l}$.)