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Question Number 224709 by solihin last updated on 28/Sep/25
Answered by mr W last updated on 30/Sep/25
$${number}\:{of}\:{possibilities}\:{that}\:{n} \\ $$$${people}\:{have}\:{different}\:{birthdays}: \\ $$$$\mathrm{365}×\mathrm{364}×...×\left(\mathrm{365}−{n}+\mathrm{1}\right) \\ $$$${total}\:{number}\:{of}\:{possibilities}: \\ $$$$\mathrm{365}×\mathrm{365}×...×\mathrm{365} \\ $$$${probability}\:{that}\:{n}\:{people}\:{have}\: \\ $$$${different}\:{birthdays}: \\ $$$$\frac{\mathrm{365}×\mathrm{364}×...×\left(\mathrm{365}−{n}+\mathrm{1}\right)}{\mathrm{365}×\mathrm{365}×...×\mathrm{365}} \\ $$$${probability}\:{that}\:{at}\:{least}\:{two}\:{people}\: \\ $$$${have}\:{the}\:{same}\:{birthday}\:{is} \\ $$$${p}=\mathrm{1}−\frac{\mathrm{365}×\mathrm{364}×...×\left(\mathrm{365}−{n}+\mathrm{1}\right)}{\mathrm{365}×\mathrm{365}×...×\mathrm{365}} \\ $$$${with}\:{n}=\mathrm{22}:\:{p}\approx\mathrm{0}.\mathrm{4757}<\mathrm{50\%} \\ $$$${with}\:{n}=\mathrm{23}:\:{p}\approx\mathrm{0}.\mathrm{5073}>\mathrm{50\%} \\ $$$${i}.{e}.\:{at}\:{least}\:\mathrm{23}\:{people}\:{are}\:{needed}. \\ $$