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Probability

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Probability

by cypherskull » Sun Jun 10, 2012 9:23 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
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by eagleeye » Sun Jun 10, 2012 1:01 pm
Hi cypherskull:

The correct answer should be D. Let me explain:

We need to find the probability of at least 2 people having the same birthday. Whenever we need to find an "at least two" condition, we should look for the no two have the same condition. Then we can use p(at least 2) = 1-p(all different). Let's calculate.

We have 85 people. Total no. of ways 85 people can have birthdays =365^85. We need to find the probability of all 85 having different birthdays.
For the first person we have 365 options, for the second 364 (since 1 date has already been selected), third has 363 etc.
Total no. of ways people can have distinct birthdays= 365*364*363*362*361*.....281 ( since 85th term is 365-85+1).
Now 365*364*363........*281 = 365*364*363.....*281*280!/280! (multiplying and dividing by 280! to make the answer compact.
= 365*364*363......*2*1/(280!) = 365!/280!.

Therefore probability of at least 2 people having the same birthday = 1 - 365!/(365^85*280!).

Hence D.

Let me know if this helps :)[/spoiler]
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by ankita1709 » Sun Jun 10, 2012 10:21 pm
Nice explanation.. :)
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by parveen110 » Fri Mar 21, 2014 3:22 am
We need to find the probability of at least 2 people having the same birthday. Whenever we need to find an "at least two" condition, we should look for the no two have the same condition. Then we can use p(at least 2) = 1-p(all different).


I have approached the problem in the similar way, however, i feel it is a bit simpler to look at it the following way:

# of distinct ways in which b'days may be assigned to 85 students out of 365 days=365P85
Now, total # of ways in which b'days may be assigned without any restriction=365^85

Probablity= favourable outcome/total number of outcomes
= 365P85/365^85

Therefore, number of ways in which b'days may be assigned so that atleast two students have their b'days falling on the same day= 1- 365P85/365^85
= 1-(365!/(280!*365^85))--Answer
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