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If there are 85 students in a statistics class and we assume

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If there are 85 students in a statistics class and we assume [#permalink] New post 05 Jun 2010, 06:53
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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 * 84/365
B. 1/365 * 1/365
C. 1 - 85!/365!
D. 1 - 365!/(280!*365^85)[/m])
E. 1 - 85!/365^85
[Reveal] Spoiler: OA

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Last edited by Bunuel on 31 Mar 2012, 01:04, edited 2 times in total.
Edited the question and added the OA
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Re: Same birthday probability [#permalink] New post 05 Jun 2010, 06:59
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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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Re: Same birthday probability   [#permalink] 05 Jun 2010, 06:59
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