on distribue au hasard 8 boules b_1 ...b_8 dans 6 tiroirs t_1 ...t_6 .soit A_i l′evenement “le tiroir t_i est vide” les evenements A_1 et A


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Question Number 149471 by ArielVyny last updated on 05/Aug/21

$o n d i s t r i b u e a u h a s a r d 8 b o u l e s b_{1} . . . b_{8}$ $d a n s 6 t i r o i r s t_{1} . . . t_{6} . s o i t A_{i} l^{'} e v e n e m e n t$ $‘ ‘ l e t i r o i r t_{i} e s t {v i d e}^{″} l e s e v e n e m e n t s A_{1} e t$ $A_{2} s o n t - i l s i n d e p e n d a n t s ?$
	Answered by Olaf_Thorendsen last updated on 05/Aug/21

	$L^{'} evenement A_{1} a pour probabilite$ $p (A_{1}) = {(\frac{5}{6})}^{8} car le tiroir t_{1} vide$ $revient a distribuer les 8 boules dans$ $les 5 autres tiroirs que t_{1} .$ $Idem pour A_{2} . p (A_{2}) = {(\frac{5}{6})}^{8} .$ $L^{'} evenement A_{1} \cap A_{2} est realise quand$ $les 2 tiroirs t_{1} et t_{2} sont vides avec la$ $probabilite p (A_{1} \cap A_{2}) = {(\frac{4}{6})}^{8}$ $Or p (A_{1} \cap A_{2}) \neq p (A_{1}) p (A_{2}) .$ $(5, 41 % \neq 3, 90 %) . Les evenements A_{1}$ $et A_{2} sont donc dependants .$
	Commented by ArielVyny last updated on 05/Aug/21

	$m e r c i M r$
	Commented by LESSENGUI last updated on 06/Aug/21

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