Given a group of n students, and given the fact that there are 365 days in one year, the probability that there are no students with the same birthday is given by
[tex]p_{no} (n) = \frac{364}{365} \cdot \frac{363}{365} \cdot ... \cdot \frac{365-n+1}{365} = \frac{364!}{365^{n-1} (365-n)!} [/tex]
Therefore, the probabilty that there are at least 2 students with same birthday is the complementary of this value:
[tex]p_{yes} (n) = 1 -p_{no}(n) =1- \frac{364!}{365^{n-1} (365-n)!}[/tex]
And if we use n=36 (number of students), we find
[tex]p=1- \frac{364!}{365^{36-1} (365-36)!} \sim 0.83[/tex]
So, the probability that there are at least 2 students in a group of 36 with same birthday is approximately 83%.
