How would I calculate the probability of atleast 2 people having the same birthday given a group of 3 using counting principles. I know that using P(same bday) = 1 - P(not same bday). I calculated this to be $1- \frac{365!}{(365-3)!365^3}$. Is there a way to find P(same bday) using combinatorics principles only?
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$\begingroup$ yes of course it's possible! $\endgroup$– userCommented Dec 4, 2017 at 17:28
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$\begingroup$ Please, if you are ok, you can accept the answer and set it as solved. Thanks! $\endgroup$– userCommented Jan 20, 2018 at 22:35
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1 Answer
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yes of course it's possible!
HINT
start from $365^3$ overall possibilities and then evaluate all the possible favourable cases
EG
three distinct birthdays = ${365}\choose3$
same birthdays for three = ${365}\choose1$
etc.
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$\begingroup$ I'm not sure where to go from here, could you clarify what my initial step should be? $\endgroup$ Commented Dec 4, 2017 at 17:49
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$\begingroup$ for each person you can choose 1 day among 365, thus you have to consider $365*365*365=365^3$ overall possibilities $\endgroup$– userCommented Dec 4, 2017 at 17:54
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$\begingroup$ Would it be 365/365 for the first then 365/365*1/365 for the second? Then divided by 365^3 $\endgroup$ Commented Dec 4, 2017 at 18:10
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$\begingroup$ the way that I’ve suggested is the classical definition of probability: #fav.cases/#total cases, the #total cases is given by $365^3$ $\endgroup$– userCommented Dec 4, 2017 at 19:29
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$\begingroup$ I hope it's clear to you, if you need some other hint do not hesitate to ask, I'm glad to help you if I can. $\endgroup$– userCommented Dec 4, 2017 at 20:48