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There being 365 days in a year and 50 being less than 365 therefore 2 even far less than that the chances are virtually 0.

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Assume that all 366 days (including leap day) are equally likely to be a person's birthday. The probability that none of them share a birthday is 1*P(second person selected doesn't share a birthday with first person selected)*P(third person selected doesn't share a birthday with first or second person selected)*...*P(fiftieth person selected doesn't share a birthday with the first, second, third,...,forty-ninth person selected).

P(second person selected doesn't share a birthday with first person selected) = 365/366

P(third person selected doesn't share a birthday with first or second person selected)=364/366

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P(fiftieth person selected doesn't share a birthday with the first, second, third,...,forty-ninth person selected)=317/366

P(none share a birthday)=(365/366)*(364/366)*...*(317/366), which is approximately .0299.

P(at least two share a birthday) = 1-(365/366)*(364/366)*...*(317/366)=1-.0299=.9701 = 97.01%.

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โˆ™ 2009-11-20 06:05:18
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Q: If 50 people are chosen at random what is the probability that at least two of them have their birthday on the same day?
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