Find-the-probability-that-n-people-n-365-selected-at-random-will-have-n-different-birthdays-

Question Number 179545 by cortano1 last updated on 30/Oct/22

$$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{n}\:\mathrm{people}\: \\ $$$$\:\:\left(\mathrm{n}\leqslant\mathrm{365}\right)\:\mathrm{selected}\:\mathrm{at}\:\mathrm{random}\:\mathrm{will} \\ $$$$\:\:\mathrm{have}\:\mathrm{n}\:\mathrm{different}\:\mathrm{birthdays}. \\ $$

Commented by mr W last updated on 30/Oct/22

for 2 people i got p≈99.5% for 10 people i got p≈78.1%

$${for}\:\mathrm{2}\:{people}\:{i}\:{got}\:{p}\approx\mathrm{99}.\mathrm{5\%} \\ $$$${for}\:\mathrm{10}\:{people}\:{i}\:{got}\:{p}\approx\mathrm{78}.\mathrm{1\%} \\ $$

Commented by AST last updated on 30/Oct/22

For 2 people, p=((364)/(365))≈99.7% For 10 people,p≈88.3%

$${For}\:\mathrm{2}\:{people},\:{p}=\frac{\mathrm{364}}{\mathrm{365}}\approx\mathrm{99}.\mathrm{7\%} \\ $$$${For}\:\mathrm{10}\:{people},{p}\approx\mathrm{88}.\mathrm{3\%} \\ $$

Commented by MJS_new last updated on 31/Oct/22

this is the old Birthday−Problem in other words: how many people do we need to get a probability p greater than 50% for at least 2 people having the same birthday? the answer is 23 to get this we calculate the probability for the next person having a different birthday: p^− =((365)/(365))×((364)/(365))×((363)/(365))×... p=1−p^− for n people having different birthdays we get p^− =((365!)/((365−n)!×365^n )) and for at least 2 people with same birthday we get p=1−p^− =1−((365!)/((365−n)!×365^n ))

$$\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{old}\:\mathrm{Birthday}−\mathrm{Problem}\:\mathrm{in}\:\mathrm{other} \\ $$$$\mathrm{words}: \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{people}\:\mathrm{do}\:\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{get}\:\mathrm{a}\:\mathrm{probability} \\ $$$${p}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{50\%}\:\mathrm{for}\:\mathrm{at}\:\mathrm{least}\:\mathrm{2}\:\mathrm{people}\:\mathrm{having} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{birthday}? \\ $$$$\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{23} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{this}\:\mathrm{we}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{for} \\ $$$$\mathrm{the}\:\mathrm{next}\:\mathrm{person}\:\mathrm{having}\:\mathrm{a}\:\mathrm{different}\:\mathrm{birthday}: \\ $$$$\overset{−} {{p}}=\frac{\mathrm{365}}{\mathrm{365}}×\frac{\mathrm{364}}{\mathrm{365}}×\frac{\mathrm{363}}{\mathrm{365}}×… \\ $$$${p}=\mathrm{1}−\overset{−} {{p}} \\ $$$$\mathrm{for}\:{n}\:\mathrm{people}\:\mathrm{having}\:\mathrm{different}\:\mathrm{birthdays}\:\mathrm{we}\:\mathrm{get} \\ $$$$\overset{−} {{p}}=\frac{\mathrm{365}!}{\left(\mathrm{365}−{n}\right)!×\mathrm{365}^{{n}} } \\ $$$$\mathrm{and}\:\mathrm{for}\:\mathrm{at}\:\mathrm{least}\:\mathrm{2}\:\mathrm{people}\:\mathrm{with}\:\mathrm{same}\:\mathrm{birthday} \\ $$$$\mathrm{we}\:\mathrm{get} \\ $$$${p}=\mathrm{1}−\overset{−} {{p}}=\mathrm{1}−\frac{\mathrm{365}!}{\left(\mathrm{365}−{n}\right)!×\mathrm{365}^{{n}} } \\ $$

Commented by mr W last updated on 31/Oct/22

$${thanks}! \\ $$

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