newgmattest wrote:If there are 85 students in a statistics class and we assume that there are 365 days in a year, what is the probability that at least two students in the class have the same birthday (assuming birthdays are distributed independently)?

Correct Answer: 1 - 365!/[(280!)(365^85)]

Can anyone explain solution for this please?

1st student can have its birthday on any of the 365 day, thus 1st student has 365 ways of having a birthday,
similarly second,(third,fourth and so) student can also have its birthday on any of the 365 days,

hence total no. of ways of having a birthday = 365^85;

probability that at least two students in the class have the same birthday= 1- none of the student in the calss have the same birthday;

now no. of having birthdays on different days= (365C85)*85!;
365C85 represents no. of ways of selecting 85 days from the 365 days of the day,, now the selected 85 days can be arranged among 85 students in 85! ways,

hence required probability is 1-(365C85)*85!/365^85;
=1-365!/280!*365^85

Hi,
Total Number of ways in which 85 students can have their birthdays in an year is (365)^85 because each student has 365 days to choose from.
Number of ways in which we can arrange 85 students on different birthday dates is 365P85 = 365!/280!
So, probability that each student has distinct birthday dates is p(E)=(365!/280!)/365^85.
Probability that at least two students have the same birthday = 1-probability that each student has distinct birthday = 1 - 365!/[(280!)(365^85)]
Please use spoiler for the OA.

Cheers!