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On-dispose-de-N-1-urnes-l-urne-U-k-contient-k-boules-blanches-et-N-k-boules-noires-on-tire-successivement-sans-remise-n-boules-de-l-urne-et-on-note-An-l-evenement-choisir-n-boules-noires-lors-des




Question Number 145134 by ArielVyny last updated on 02/Jul/21
On dispose de N+1 urnes.l′urne U_k   contient k boules blanches et N−k boules  noires.on tire successivement sans   remise n boules de l′urne et on note   An l′evenement ′′choisir n boules noires  lors des n premiers tirages′′. Determiner  P(An). on notera U_k =′′choisir l′urne k′′
$${On}\:{dispose}\:{de}\:{N}+\mathrm{1}\:{urnes}.{l}'{urne}\:{U}_{{k}} \\ $$$${contient}\:{k}\:{boules}\:{blanches}\:{et}\:{N}−{k}\:{boules} \\ $$$${noires}.{on}\:{tire}\:{successivement}\:{sans}\: \\ $$$${remise}\:{n}\:{boules}\:{de}\:{l}'{urne}\:{et}\:{on}\:{note}\: \\ $$$${An}\:{l}'{evenement}\:''{choisir}\:{n}\:{boules}\:{noires} \\ $$$${lors}\:{des}\:{n}\:{premiers}\:{tirages}''.\:{Determiner} \\ $$$${P}\left({An}\right).\:{on}\:{notera}\:{U}_{{k}} =''{choisir}\:{l}'{urne}\:{k}'' \\ $$
Commented by Olaf_Thorendsen last updated on 03/Jul/21
P(A_n ) = ((N−k)/N)×((N−k−1)/(N−1))×...×((N−k−(n−1))/(N−(n−1)))  P(A_n ) = ((((N−k)!)/((N−k−n)!))/((N!)/((N−n)!)))  P(A_n ) = (((N−k)!(N−n)!)/((N−k−n)!N!))
$${P}\left({A}_{{n}} \right)\:=\:\frac{{N}−{k}}{{N}}×\frac{{N}−{k}−\mathrm{1}}{{N}−\mathrm{1}}×…×\frac{{N}−{k}−\left({n}−\mathrm{1}\right)}{{N}−\left({n}−\mathrm{1}\right)} \\ $$$${P}\left({A}_{{n}} \right)\:=\:\frac{\frac{\left({N}−{k}\right)!}{\left({N}−{k}−{n}\right)!}}{\frac{{N}!}{\left({N}−{n}\right)!}} \\ $$$${P}\left({A}_{{n}} \right)\:=\:\frac{\left({N}−{k}\right)!\left({N}−{n}\right)!}{\left({N}−{k}−{n}\right)!{N}!} \\ $$

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