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Happy Birthday (3) (Posted on 2006-06-09) Difficulty: 3 of 5
In Happy Birthday, the question was if there are N people in a room, what is the probability that there are at least two people in the room who share a birthday?

What if instead exactly two was required? If there are N people in a room, what is the probability that there are exactly two people in the room who share a birthday?

(Note: Assume leap year doesn't exist, and the birthdays are randomly distributed throughout the year.)

No Solution Yet Submitted by Sir Percivale    
Rating: 4.0000 (3 votes)

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Solution thoughts | Comment 1 of 12

So the first n-1 people in the room have different birthdays

365*364*363.../365^(n-1)

and the nth person has the same as one of the prior n-1

(n-1)/365

Therefore the probability of exactly two having the same birthday is

P = (365!/(365-n)!)*(n-1)/365^n


  Posted by Bob Smith on 2006-06-09 07:38:11
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