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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