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If there are $12$ strangers in a room , what is the probability that no of them celebrate their birthday in same month .

I tried but not able to start .

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There are $12$ persons and $12$ months, hence all the birthdays should cover all the months and we obtain $$p=\frac{12!}{12^{12}}$$

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Think of it like this. The first guy can be born on any of the twelve months and the second guy can then be born on any of the eleven remaining months, so the probability that two of them are born in different months is $\frac{12}{12}*\frac{11}{12}$. The third guy can be born on any of the remaining ten months, so the probability is $\frac{12}{12}*\frac{11}{12}*\frac{10}{12}$. You should be able to continue from there. Also, you should look up the Birthday Paradox for a similar question.

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