"Birthday Problem"
~All students of statistics should be exposed to the very famous "Birthday Problem". It's interesting & defies common sense.
~It turns out, in a group of 23 people, it is likely that 2 people share the same birthday (maybe not the same year, however).
~The probability for n=23 is .505 (likely), for n=30 it increases to 70%, for n=40, it becomes 90%, at n=50 it's 97%, & at n=60, it's almost certainly the case at 99%. Here is the math behind it.
~P(2 people have the same BD) = 1 - P(no 2 have the same BD)
~Note: this is the "backdoor" approach to probability (see link).
~So, computing the right side, we have P(no 2 have the same BD) = P(choosing a person with a BD) times P(choosing a 2nd person whose BD is different from first) times P(choosing a 3rd person
whose BD is different from first & second) times so on & so on.
~P( 2 have the same BD) = 1 - (365/365)(364/365)(363/365)(362/365) & so on.
~Note: When there are 23 fractions, n=23, P=.505. Be careful computing, since trying to do 365! is too large (over flow) for your calculator, so piece by piece, multiplying & dividing will get it.
~Note: If time permits, I'll try it on our class.