Birthday sharing math problem
Web$\begingroup$ @AndréNicolas : I think you missed a factor : P("n-1 don't share a birthday") = Nb of cases where n-1 don't share a birthday / $365^{(n-1)}$. P = Nb of cases where n-1 don't share a birthday * ${n \choose 2} / 365^{n}$ = P("n-1 don't share a birthday") * ${n \choose 2}$ / 365 Am I right? $\endgroup$ – WebMay 26, 2024 · What is the probability that two persons among n have same birthday? Let the probability that two people in a room with n have same birthday be P(same). P(Same) can be easily evaluated in terms of P(different) where P(different) is the probability that all of them have different birthday. P(same) = 1 – P(different)
Birthday sharing math problem
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WebMar 19, 2005 · The birthday problem asks how many people you need to have at a party so that there is a better-than-even chance that two of them will share the same birthday. … WebApr 14, 2015 · So from Albert’s statement, Bernard now also knows that Cheryl’s birthday is not in May or June, eliminating half of the possibilities, leaving July 14, July 16, Aug. 14, Aug. 15 and Aug. 17 ...
Web$\begingroup$ It looks as if the two calculations interpret distinct birthday differently. The homework solution sees it a day where at least one person has a birthday, distinct from other days where at least one person has a birthday. You see it as a day where exactly one person has a birthday distinct from all the other people's birthdays. WebMay 3, 2012 · The problem is to find the probability where exactly 2 people in a room full of 23 people share the same birthday. My argument is that there are 23 choose 2 ways times 1 365 2 for 2 people to share the same birthday. But, we also have to consider the case involving 21 people who don't share the same birthday.
WebNov 17, 2024 · The probability that Boris will share her birthday is 1 / 365. Likewise, the probability that Charlie will share Annie's birthday is 1 / 365. Since the dates of their birthdays are independent, the probability that both Boris and Charlie will have the same birthday as Annie is 1 ⋅ 1 365 ⋅ 1 365 = ( 1 365) 2 Share Cite Follow WebOct 13, 2016 · Cake-cutting is a metaphor for a wide range of real-world problems that involve dividing some continuous object, whether it’s cake or, say, a tract of land, among people who value its features...
WebOct 14, 2024 · The probability of NOT having the same birthday for a single pair is p b = 1 − 1 365 = 364 365 so for all the pairs we have: P ( # B ≥ 1) = 1 − P ( # B = 0) = 1 − ( 364 365) C k, 2 where C k, 2 is the number of possible pairs.
WebAnd we said, well, the probability that someone shares a birthday with someone else, or maybe more than one person, is equal to all of the possibilities-- kind of the 100%, the … greater western plumbing coloradoWebJul 25, 2024 · This probability is p 1 person 2 = 1 − 1 / 365 = 364 / 365, because all days have the same probability 1 / 365 to be the birthday of the second person except for one day, except for the day, when person 1 has his birthday, if we want to know the probability of different birthdays for all persons.This goes on and on and on for all n persons and we … flipchart tascheWebMay 30, 2024 · The probability that any randomly chosen 2 people share the same birthdate. So you have a 0.27% chance of walking up to a stranger and discovering that their birthday is the same day as yours. flipchart themenparkplatzWebOct 8, 2024 · The trick that solves the birthday problem! Instead of counting all the ways we can have people sharing birthdays, the trick is to rephrase the problem and count a much simpler thing: the opposite! P (At least one shared birthday) = … flip chart tablaWebJan 29, 2024 · Probability of no two people out of n sharing a birthday is N(umerator) D(enominator) where D = (1461)n. To calculate N you must consider two possibilities : exactly one of the people is born on Feb 29, or none of the people is born on Feb 29. Case-1 greater western mastiff batWebApr 22, 2024 · Download my Excel file: BirthdayProblem. By assessing the probabilities, the answer to the Birthday Problem is that you need a … flip chart \u0026 cdWebNov 21, 2015 · The number of ways to choose a pair of distinct birthdays is $\binom{365}{2}$. There are then $\binom{n}{2}$ ways to choose the pair who will have the earlier of these two birthdays, and for each such way there are $\binom{n-2}{2}$ ways to choose the pair who will have the later of the two birthdays. greater western flank phase 2