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Birthday probability question
Message posted by Brian (via 207.68.66.130) on December 28, 2001 at 2:14 PM (ET)
I am by no means a probability guru, but I was posed the following questions:
Our company has 1400 employees. What is the probability that a day exists on which none of our employees has a birthday? What is the probability that the day will occur tomorrow?
I came up with ...
[(364/365)^1400] = 0.021475
or 2.15% chance that such a day exists and...
[(364/365)^1400]*(1/365)=0.000058 or 0.0059% chance that day is tommorow.
Would someone please comment and maybe share their findings? I would greatly appreciate it.
READERS RESPOND:
(In chronological order. Most recent at the bottom.)
Re: Birthday probability question
Message posted by JG (via 128.8.22.30) on December 28, 2001 at 6:56 PM (ET)
For next day or any specific day we have 364/365 to the power of 1400.
Re: Birthday probability question
Message posted by JG (via 128.8.22.30) on December 28, 2001 at 6:58 PM (ET)
For at least one day we have 364/365 to the power 1400 times 365.
Re: Birthday probability question
Message posted by JG (via 128.8.22.30) on December 28, 2001 at 7:02 PM (ET)
However, I am just guessing but I know that this is a standard textbook problem which at the moment I am too tired to look up.
Re: Birthday probability question
Message posted by Brian (via 207.68.66.130) on December 31, 2001 at 7:53 AM (ET)
Thanks for the response JG. I don't have any reference material to use, this was just kind of an office teaser. I appreciate the help. Maybe now I can actually get some WORK done today.
Re: Birthday probability question
Message posted by Jack Tomsky (via 12.144.103.66) on January 2, 2002 at 7:31 PM (ET)
The answer to this occupancy problem isP = Sum[(-1)^i {365!/(i!(365-i)!}*(1-(i/365))^1400]
where the sum is over i = 0, ..., 365.
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