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 [tex]\dfrac1{365}[/tex] 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 [tex]\dfrac{364}{365}[/tex].
The probability, then, of having the next student born on 3 Januaray is [tex]\dfrac{363}{365}[/tex].
Continuing in this fashion, for 25 students you would find that no two share a birthday with probability
[tex]\dfrac{364}{365}\times\dfrac{363}{365}\times\cdots\dfrac{341}{365}=\dfrac{365!}{(365-25)!365^{25}}\approx0.4313[/tex]
