#1
So people may remember my making a thread like this before. I have another interesting math problem for extra credit, and here it is:

There are 27 people in a class... what is the probability that two of them have the same birthday?

It seems to me like it would be a very low probability, but my teacher said that it was above 50%.

If you get an answer, please elaborate.
#3
Is the probability that only two, at least two, no more than two,etc. It makes a big difference. It binomialpdf/cdf in a TI-83 graphing calculator. n= 27 p= 1/365 x depends on the wording.
1992 Ibanez S540 sol
1988? Kramer Pacer Mutt
#7
Quote by helper1234
No way is it above 50%

It will be 1 in 365 x 1 in 365

No it won't. For only 2 people, it's 1- (364/365+(363/365)+(362/365)+ etc 26 times
1992 Ibanez S540 sol
1988? Kramer Pacer Mutt
#8
Quote by sk8boardbob2
Is the probability that only two, at least two, no more than two,etc. It makes a big difference. It binomialpdf/cdf in a TI-83 graphing calculator. n= 27 p= 1/365 x depends on the wording.



It was only worded "two" so more than two would also be ok I guess because two people would still have the same birthday.
#12
Alright i finally manually did it and i got the real answer i think. It's 1- the probability that no one has the same birthday which is (365x364x363....340)/(365^26). The final answer is .598 or a 59.8% chance two people have the same birthday.
1992 Ibanez S540 sol
1988? Kramer Pacer Mutt
#13
Quote by sk8boardbob2
Alright i finally manually did it and i got the real answer i think. It's 1- the probability that no one has the same birthday which is (365x364x363....340)/(365^26). The final answer is .598 or a 59.8% chance two people have the same birthday.


Yeah i think you did it right but the answer is actually 61.82 %
Last edited by LiquidTension99 at Apr 27, 2008,