I am trying to figure out a solution to the following problem:
Let there be two groups of people, Group A and Group B. Group A represents x percent (e.g. 1%) of the world's population, and Group B represents y percent (e.g. 2%) of the world's population. What is the probability that a person from Group A will meet a person from Group B? Assume the following things:
- The average human being meets z people (e.g. 100,000) in a lifetime.
- The world's population is kept at a constant k people.
- Everyone in the world was born and will die at the same time.
Disclaimer:
I came up with this question myself, but I'm not a mathematician, so please feel free to clean this up if need be. Also, if there is not enough information in the problem to solve it, add assumptions and please indicate the reasons for adding them. The assumptions I wrote are my attempt at making the problem easier. If they are not necessary, and removing any produces a more accurate answer, then I encourage the removal of them.
Best Answer
Let $a$ be the number of people in group $A$ and $b$ be the number of people in group $B$. Then the chance that you never meet somebody from group $B$ is the chance that all your $z$ acquaintances are outside, $\frac {k-b}k\frac{k-b-1}{k-1} \ldots \frac{k-b-z+1}{k-z+1}=\frac {(k-b)!(k-z)!}{k!(k-b-z)!}$ If the chances that an $A$ individual meets a $B$ individual are independent, the chance they never meet is $\left(\frac {(k-b)!(k-z)!}{k!(k-b-z)!}\right)^a$. Although it is not obvious, this should be symmetric in $a$ and $b$.