Question

Find the probability that of 25 students, no two have the same birthday

Answer 1

Assuming no student was born on Leap Day, and that birthdays are as likely to occur on one day as any other, each student has $\dfrac1{365}$ probability of being born on a particular day. So if one student's birthday was, say, 1 January, the probability that another student was born on 2 January is $\dfrac{364}{365}$.

The probability, then, of having the next student born on 3 Januaray is $\dfrac{363}{365}$.

Continuing in this fashion, for 25 students you would find that no two share a birthday with probability

$\dfrac{364}{365}\times\dfrac{363}{365}\times\cdots\dfrac{341}{365}=\dfrac{365!}{(365-25)!365^{25}}\approx0.4313$