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Question 331587: determine the probality that at least 2 people in a room of 13 people has the same birthday
a)i need to compute the probility that 13 people have different birthdays
b)the complement of 13 people have different birthdays is at least 2 people share the same birth days out of the 13 people
Found 2 solutions by solver91311, galactus:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
For the probability that 13 people have different birthdays, you need the probability that the 2nd person has a different birthday than the first, namely  , times the probability that the 3rd person has a different birthday than the first two, namely  , and so on:
\ =\ \frac{365!}{352!\,\cdot\,365^{13}})
Then for the probability that at least two share a birthday,
\ =\ 1\ -\ P(13,0))
In general for  people,
\ =\ 1\ -\ \frac{365!}{(365\,-\,n)!\,\cdot\,365^{n}})
As a sanity check on your arithmetic work, 23 people is roughly the break-even point, that is where the probability is very close to 50/50. If you have 50 people it is nearly certain that at least 2 will share a birthday. Your answer for 13 people should be somewhere in the 20% range.
John
My calculator said it, I believe it, that settles it
Answer by galactus(183)  (Show Source):
You can put this solution on YOUR website! The probability that no two share a birthday among n people is
Letting n=13 and we have about a 80.56% probability that no two out of the 13 share a birthday.
As stated, the complement of this is the probability that at least two of the 13 share a birthday.
1-.8056=.1944
There is about a 19.44% probability that at least two out of the 13 share a birthday.
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