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  #1  
Old 04-12-2002, 03:04 PM
jasonsports jasonsports is offline
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Default Birthday question

I can't seem to figure this one out. Can anyone help?
"8 people were all born in January. What is the probability that at least 2 of them have the same birthday?"

I think the total # of ways 8 people can be born in January is 31 P 8, but that's as far as I can get.

Thanks!
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  #2  
Old 04-12-2002, 03:31 PM
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DukeBlue DukeBlue is offline
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Default Birthday question

Approach it as 1 - Prob(no matches)

prob(no matches) = prob(#1 can have any b-day) x prob(#2 doesn't match #1)x(#3 doesn't match #2 or #1)x(#4 doesn't match #3 or #2 or #1) ...


= (31/31)x(30/31)x(29/31)x(28/31)x...x(24/31)

or (31 P 8)/(31^8) = .627

This problem shows up a lot in textbooks as...How many people need to be in a room so that there is a greater than 50% chance that there is at least one shared birthday.

Need to find n such that (365 P n)/(365 ^ n) > .5

Turns out that n >= 23 !!!!
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Old 04-12-2002, 03:33 PM
Higher Authority Higher Authority is offline
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Default

Use 1 - P(no one shares a Birthday)
This is the same as 1 - 31P8/31^8
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  #4  
Old 04-12-2002, 03:35 PM
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DukeBlue DukeBlue is offline
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Default Birthday question

Two errors I made: (31 P 8)/(31 ^ 8) = .373, and 1 - .373 = .627 which is the answer to the question.

And my eights became emoticons somehow...
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  #5  
Old 04-12-2002, 03:37 PM
jasonsports jasonsports is offline
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Default Thanks!!

Thanks for your help!!
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