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#1
04-12-2002, 03:04 PM
 jasonsports Member Join Date: Apr 2002 Posts: 30
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!
#2
04-12-2002, 03:31 PM
 DukeBlue Member Join Date: Jan 2002 Location: 34d 00m 25s, -83d 59m 43s Favorite beer: Augustinerbrau Posts: 98
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) &gt; .5

Turns out that n &gt;= 23 !!!!
#3
04-12-2002, 03:33 PM
 Higher Authority Member Join Date: Sep 2001 Location: every where Posts: 47

Use 1 - P(no one shares a Birthday)
This is the same as 1 - 31P8/31^8
#4
04-12-2002, 03:35 PM
 DukeBlue Member Join Date: Jan 2002 Location: 34d 00m 25s, -83d 59m 43s Favorite beer: Augustinerbrau Posts: 98
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...
#5
04-12-2002, 03:37 PM
 jasonsports Member Join Date: Apr 2002 Posts: 30
Thanks!!

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