What is the probability that 2 person in a 23 person room share same birthday?????
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We chose the 2 persons in 23C2 ways. They can have their birth day on any of the 365 days. The rest can have their birthday on the remaining 364.363.362....343 days. And the no. of total possible birthdates is 36523.
So the required probability is 23C2*365*364...*34436523
- Shaswata Roy But it's not said that only 2 persons share the same birthday.It is possible that 3 or 4 or 5 or... people share the same birthday.Even in those cases the above condition will hold.Upvote·0· Reply ·2013-03-08 09:34:23
- Sourup Nag I don't think so @Shaswata, see its clearly written there that 2 persons share the same person, if it had been atleat 2 persons, then your logic would have worked. U get my point? :)
- Sourup Nag sorry, 2 persons share same birthday, my mistake :P
\dpi{150} \fn_cs \textit{We try to find the probability that no 2 of the 23 people share the same birthday.}
\dpi{150} \fn_cs \textit{The first person can have his birthday on any of the 365 days.}
\dpi{150} \fn_cs \textit{The second person can have his birthday on any of the remaining 364 days.}
\dpi{150} \fn_cs \textit{...The }23^{rd}\textit{ person can have his birthday on any of the remaining 343 days.}
\dpi{150} \fn_cs \textit{Probability} =\frac{\textit{No. of favourable outcomes}}{\textit{ Total no. of possible outcomes}} = \frac{365\cdot 364\cdot 363\cdot ....\cdot 343}{365\cdot 365\cdot365\cdot ....\cdot 365 }
\dpi{150} \fn_cs \Rightarrow \textit{Probability}\approx 0.4968
\dpi{150} \fn_cs \therefore \textit{Probability of no 2 sharing birthdays = }1-0.4968 = 0.5032
- Shaswata Roy Last line- "probability that 2 share birthday".
- Soumyabrata Mondal thnks @Shaswata