If there is a room of 20 people what is the probability that at least two of them will share the same bday? I know that i can do 364/365 * 363/365, etc... and then subtract that answer from one (to find the complement). However, my calculator will not show such large numbers. Can i some how use factorials to figure this out? I don't know how to make it stop after about 19 or 20 numbers though. I don't want 364!
Any simple way to figure out this prob would be greatly appreciated. Thanks!
2006-06-17 09:30:38 · 4 answers · asked by earthchick 3 in Science & Mathematics ➔ Mathematics
Here's the formula:
The probability that all birthdays are different is
P(365,r)/365^r, where r is the number of people and P stands for permutation.
Thus, the probability that at least two are the same is
1 - P(365,r)/365^r
r = 20
1 - P(365,20)/365^20
= 1 - .588561616419
= .411438383581
So there is a .4114 chance that at least 2 people in a group of 20 people will have the same birthday.
The birthday problem is pretty interesting. If there are 23 people, the probability that at least two have the same birthday is .507. If there are 50 people, it is .970.
2006-06-17 09:36:40 · answer #1 · answered by MsMath 7 · 12⤊ 1⤋
Just look at the pattern
1-(365/365)(364/365)(363/365)...((365-n+1)/365) (if n is the number of people)
=1-(365!/(365-n)!)/(365^n)
= 1 - (365 nPr n)/(365^n)
2006-06-17 09:37:28 · answer #2 · answered by blahb31 6 · 0⤊ 0⤋
This has been worked out many places. Here is one:
2006-06-17 14:22:34 · answer #3 · answered by fcas80 7 · 0⤊ 0⤋
Do you EVER answer your own homework? Maybe a good question to ask would be "Where do airheads get answers?" Ooops, I suspect that's one question you could answer all of your little self?
2006-06-17 09:58:24 · answer #4 · answered by Anonymous · 0⤊ 2⤋