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Birthday Probability

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Birthday Probability

by kannan007 » Sun Aug 05, 2012 9:11 am
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 * 84/364
b)1/365 * 1/364
c)1-(85!/365!)
d)1-(365!/(280!*(365^85))
e)1-(85!/(365^85))

I came across this problem in a gmat question paper.I couldnt solve it:-(
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by coolhabhi » Sun Aug 05, 2012 1:03 pm
probability that at least two students = 1 - P(probability that students have the birthdays on different days )
To find P(probability that students have the birthdays on different days ), We need the probabilities of
1st student = 365/365 (Because 1st student bday can be any of 365 days)
2nd student = 364/365 (Because 1st student bday can be any of 364 days.Since 1 day is already taken by 1st student)
3rd student = 363/365
.
.
.85th student = 281/365 (281 because 365-85+1 = 281)

=>P(probability that students have the birthdays on different days ) = (365/365)(364/365)...(281/365)
=>P(probability that students have the birthdays on different days ) = 365!/280!(365^85)

=>probability that at least two students = 1- 365!/(280!(365^85))
Answer D
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