Lucas is a student who did a math competition against 2875 other people. His rank in his results will allow him to have a school. Everybody has distinct ranks. The nearest of 1 is better. The result shows that he is in rank 1947.
Because Lucas wants to know his chance of having the school of his dream, he looked the data from last year.
He found some elements :
- There were 2876 people in the competition.
- The school always takes 96 person.
- Last year, the median rank was 1060 (so normally 48 person were chosen lower or equal than 1060 and 48 bigger or equal than 1060).
- Because the school rounds up the number, the mean rank was 1051. Lucas suppose the real value was between 1051 and 1052 (excluded).
With those information, is it possible to find the probability that Lucas could be chosen if it was last year ? If it is then what is his probability ?
For me it is an exercise on combinatorics.
We know than the mean $m$ is in the interval $[1051, 1052[$. We let $x_1,\dots, x_{96}\in[1, 2876]$ distincts with :
- $x_1,\dots,x_{47}\in [1, 1059]$
- $x_{48} = 1060$
- $x_{49},\dots,x_{95}\in [1061, 2876]$
- $x_{96} \in [1, 2876]$
- $\sum x_i/96 = m$
So it should be possible if we find the number of ways to make that sum and if we find the number of ways to have an $x_n = 1947$ with $n \ge 49$. Then we can make ratio between the two allowing us to find the probability.
However I don't think it is possible to calculate it. I am questioning myself if it can be possible using Python.
I would prefer not think as combinatorics problem and I want to know if there is a better and normal way to solve it.
Thanks for your help !
EDIT : To explain this system of ranks : This system is not only just for one school. For example, the school will ask the first one if they want to come.
If he says yes then the remaining places is reduced by one and the school goes to the next best rank person. If he says no then the remaining places doesn't change and the school goes to the next best rank person.