ykaiim 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)?

A) 85/365 x 84/365

B) 1/365 x 1/365

C) 1 - 85!/365!

D) 1 - 365!/(280! x 365^8^5)

E) 1 - 85!/365^8^5

This question was posted in PS forum, below is my solution from there:

The easiest way to solve this problem is to calculate opposite probability and subtract this value from 1:

The opposite probability is that all students have the birthdays on different days:

\frac{365}{365}*\frac{364}{365}*\frac{363}{365}*...*\frac{281}{365}=\frac{365*364*363*...*281}{365^{85}}=\frac{365!}{280!*365^{85}} total 85 birthdays (first student can have birthday on any day =1=365/365, the probability that the second student will have the birthday on another day is 364/365, the probability that the third student will have the birthday not on this two days is 363/365, and so on).

So, the probability that at least two students in the class have the same birthday is:

1-\frac{365!}{280!*365^{85}}.

Answer: D.

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