Question 329406
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There are 365 ways out of 365 ways to choose the first person's birthday. Then, if you don't want the second person to have the same birthday, there are 364 ways out of 365 to choose the second person's birthday.  And then 363 out of 365 for the 3rd person.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(0)\ =\ \frac{364}{365}\ \cdot\ \frac{363}{365}]


With 3 people there are only 3 possibilities, none have the same birthday, two have the same birthday, or all three have the same birthday.  So *[tex \Large P(\geq 2)\ =\ 1\ -\ P(0)].


I'll leave you alone to spend some quality time with your calculator.


By the way, the probability of no shared birthdays in a selection of *[tex \Large n] people is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(0)\ =\ \frac{365!}{365^n(365\ -\ n)!}]


And *[tex \Large P(\geq 2)\ =\ 1\ -\ P(0)].


The break even point is about *[tex \Large n\ =\ 23] and *[tex \Large P(\geq 2)] is near certainty when *[tex \Large n\ =\ 50]


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