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# probablitity

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01 Aug 2003, 21:38
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

First question that I was asked in my first job :

There are 100 students in a class. What is the probability that any two have birthdays on the same day.
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abhay pathania
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01 Aug 2003, 22:04
abhaypathania wrote:
First question that I was asked in my first job :

There are 100 students in a class. What is the probability that any two have birthdays on the same day.

Just to make this a fair question, you should eliminate Feb 29th for simplicity. (or you can add it in, but it will make the problem unncessarily complex because the assumption that every day is equally likely to be someones birthday will no longer be true).

Second, the wording of the question is a tad ambiguous. Do you want the probability that EXACTLY two have the same birthday, or AT LEAST TWO have the same birthday, or is the question, if you are to pick two people at random from a sample of 100, what is the probability that they have the same birthday? All of the interpretations have much different answers.
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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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02 Aug 2003, 10:28
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..
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abhay pathania
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02 Aug 2003, 10:58
Let us consider the case EXACTLY TWO PEOPLE
Sanity check says that the probability for exactly two out of 100 students to have the same birthday should be very small.

Total outcomes: there are 100 students; each may have a birthday in one of 365 days. = 365^100

Favorable... Hmm...
any 2 of 100 = 100C2=4950
have 365 days to be born=365*4590
in order to provide a condition that EXACTLY two
the third person has 364 days
the fourth has 363
....................
the hundredth 267

Finally, 4950*(365*364*...*267)/365^100= 5.7E-6
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02 Aug 2003, 14:41
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.
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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

02 Aug 2003, 14:41
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