abhaypathania wrote:
yupp u are right. I got to be carefull. What I meant was ...
What is the probability that out of the hundred students ANY TWO have birthday on the same day .... ( or in other words it is = 1-P)
Where P is the probabiltiy that none of the students have birthday on the same day..
I hope I am clear..
Okay.
This would be more interesting if the number of people were smaller, say 20 to 25 people. But here goes...
In order to find the probability of ANY TWO (which to me means AT LEAST TWO), we best find the probability that NOBODY has the same birthday (let's call this P), then subtract from one.
Say we pick any one person, The probability that that next person will not have the same birthday is 364/365, then next 363/365, and so on until we get to 266/365.
So P = (1)(364/365)(363/365)....(266/365) = (364!/265!)/(365^99)
You would need a computer to calculate this, but if you think about it, this number will end up being very small. (we are multiplying a bunch of numbers less than 1 together 99 times)
So my answer at an interview would be: "It is almost certain that at least two people will have the same birthday in a group of 100"
Numerical answer using computer is P = 3 x 10^-7
and 1 - P = .9999997
Interesting result: If there were only 23 people, the answer would be
1 - P
= 1 - (364!/342!) / (365^22)
= 1 - .4927
= .5073
which means that there is a slightly more than 50-50 chance that there will be at least 2 people with the same birthday in a group of 23 people.
You could make money on this if you could get someone to give you odds.
_________________
Best,
AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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