## Probability of someone watching a video and being in a certain place...

There is a video of a person attempting to hang themselves in a prison safe cell using a novel hanging point. The prison has had a request for the footage to be made public, and want to know the probability that releasing the video onto the internet might allow a person who views the footage and then ends up in the same prison and same kind of cell to imitate the person on the video.

Here are the figures involved:
Assume 2000 views per year world-wide if the footage were released.
There are 10 cells similar to the one depicted in the video in the prison in question.
The state in which the attempt took place has 7,790,000 inhabitants.
World population is 7,580,000,000 people.

To me there are two ways you could approach it (but I don't know probability that well, so I could be wrong there):

1. Take the number of cells (10), and divide it by world population (7,580,000,000), then multiply that by views per year (2000) to get the percentage of the population who might end up in those cells after having watched the video. But I think this misses some of the factors.

2. Take other factors that increase the unlikelihood into account: not only would the viewer need to be in the same cell, but the same prison, country, state, and be suicidal. My question is: do these events compound? Do I take, say, country population divided by world population (=result 1), then multiply this by state population divided by country population (=result 2) and then multiply result 1 by result 2, and so on?

I think I know how to do the calculations in either case. I just don't know which set of calculations would be more correct. Any help much appreciated. Thanks.