Question 918207
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Let *[tex \Large P(\phi)] be the probability that no two people share a birthday.


If you have 2 people, then the probability the second person has a different birthday than the first person, given 229 possible birthdays, is *[tex \Large \frac{228}{229}].  For the third person, the probability that the third birthday differs from that of both the first and second is *[tex \Large \frac{228}{229}\ \cdot\ \frac{227}{229}]


Following the same line of reasoning, the probability that none of the first twenty people share a birthday is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(\phi)\ =\ \prod_{i=0}^{19}\,\frac{229\ -\ i}{229}]


Since "at least 2" is the complement of "none", the probability you want is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ P(\phi)]


You can do your own calculator work.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \