Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.



RE: A probability question...
Posted:
Nov 18, 1999 9:09 AM


A resampleing approache to finding the probability of "exactly two" with the same birthday, using the resampleing power of Fathom, Data Desk, Minitab or Excel (who did I leave out that will hate me). Put the numbers 1365 (forget leap year, I won't play) and pull a sample of 35 without replacement. Count how many duplicates of each number there are (this is harder in some programs than others, but I usually create a seperate variable with a count of how many equal this x value). Then if the max is 2 you have a winner, if the max is other than 2 something else happened. Create a measure for this (one for success, zero for failure), do it 1000 times and total the successes.
Pat Ballew, Misawa, Jp
"Statistics means never having to say you're certain."
Math Words & Other Words http://www.geocities.com/paris/rue/1861/etyindex.html
The Mathboy's page http://www.geocities.com/paris/rue/1861
Original Message From: Tim Erickson <tim@eeps.com> at EDUINTERNET Sent: Wednesday, November 17, 1999 2:46 PM To: MRowe14@aol.com at EDUINTERNET; apstatl@cln.etc.bc.ca at EDUINTERNET Subject: RE: A probability question...
>The question is: In a >class of 35 students, what is the probability of 2 students having the same >birthday.
This is one of those questions best answered by looking at the complement: what's the probability that, in a class of 35 students, all have different birthdays?
We look at them one at a time: The probability that the first one is different is 1. The probability that the second is different from the first is 364/365. That the third is different from the first two is 363/365.
and so forth, so that the 35th's probability of being different from all others is (36535+1)/365 = 331/365.
To get the probability that all of these are true together, we multiply, to get
P(all different) = (365 * 364 * .... * 331) / (365)^35 = 365!/( (36535)! 365^35) = about 0.186
which means that the probability that at least two have the same (note this is different from your "exactly two," but I suspect that this is what you mean) is 1  P, or about 0.814.
This is a famously surprising result. By the way, my usual ways of computing these things resulted (not surprisingly) in yournumberistoobig errors (since 365! is larger than the number of particles in the universe), so I computed it as if it were a spreadsheet, one fraction at a time. That let me also find the interesting tidbit that the probability of a match crosses 50% when you have 23 people.
Tim Erickson
Author's plug: since I'm the author of _Data in Depth_, the book that comes with Fathom (the new stats software from Key Curriculum Press), I should point out that a related problem appears on p 251 of that book:
Suppose people come into a room one at a time. Whenever a new person enters the room, he or she announces his or her birthday. If it matches that of anyone else in the room, the door is locked and no one else is let in.
On the average, how many people will get into the room? What's the distribution of this number?
======================================================================= The Advanced Placement Statistics List To UNSUBSCRIBE send a message to majordomo@etc.bc.ca containing: unsubscribe apstatl <email address used to subscribe> Discussion archives are at http://forum.swarthmore.edu/epigone/apstatl Problems with the list or your subscription? mailto://jswift@sd70.bc.ca =======================================================================
======================================================================= The Advanced Placement Statistics List To UNSUBSCRIBE send a message to majordomo@etc.bc.ca containing: unsubscribe apstatl <email address used to subscribe> Discussion archives are at http://forum.swarthmore.edu/epigone/apstatl Problems with the list or your subscription? mailto://jswift@sd70.bc.ca =======================================================================



