Stadium Seating Probability Problem Problem thought of at an actual football game: 
Say we are in a football stadium with each successive row elevated (slightly) above the previous row. The rows are numbered starting with $1$ at the bottom and continue upward to some number $n$ at the top. If one person decides to stand up, then every person directly behind that first person (in the same "column", if you will) will stand up in a domino-like fashion. 
Let's say that the probability that a given person will stand up is a fixed probability $p$ in the interval $[0, 1]$. Given $p$, what is the probability that a person sitting at row $k$ (in $[1, n]$) will stand up? 
Concretely, if the probability for each person of standing up is $0.02$, what is the probability that the person in row $37$ will stand up? 
For the person in row $1$, the answer would obviously be $0.02$. And it clearly becomes more and more likely that you will stand up the higher you sit. But beyond that, I don't know how to think about this problem. My probability has gotten a bit rusty; is there a type of random variable that models this problem? 
 A: This problem setup seems like you could model it with a geometric probability distribution.
And the trick is to consider the probability that someone in the $k$th row doesn't stand up. Let $p=0.02$, the probability that someone spontaneously decides to stand up.


*

*The probability that the person in the first row doesn't stand up is $1-p$.

*The probability that the person in the second row doesn't stand up is the probability that the person in the first row doesn't stand up and  the person in the second row doesn't decide to stand up anyways. $(1-p)^2$.

*And in general, the probability that the person in the $k$th row doesn't stand up is the probability that none of the people in rows 1 through $k$ decide to stand up. Overall, this results in a probability $(1-p)^k$.


Hence, if each person will spontaneously stand up with probability $p$, and everyone behind you will automatically stand up after you in a domino effect, the probability that the person in the $k$th row does stand up is:
$$p_k = 1-(1-p)^k$$

Concretely, when $p=0.02$, and $k=37$, we have that the person will stand up with probability $1-(1-0.02)^{37} \approx 0.526$.
