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Question Number 145136 by ArielVyny last updated on 02/Jul/21

Soit X une variable aleatoire de loi geometrique de parametre p∈]0.1[ calculer P({X≥4})

$${Soit}\:{X}\:{une}\:{variable}\:{aleatoire}\:{de}\:{loi} \\ $$$$\left.{geometrique}\:{de}\:{parametre}\:{p}\in\right]\mathrm{0}.\mathrm{1}\left[\right. \\ $$$${calculer}\:{P}\left(\left\{{X}\geqslant\mathrm{4}\right\}\right) \\ $$

Answered by Olaf_Thorendsen last updated on 03/Jul/21

P(X= k) = q^(k−1) p = (1−p)^(k−1) p P(X≥4) = 1−P(X=1)−P(X=2)−P(X=3) P(X≥4) = 1−(1−p)^0 p−(1−p)p−(1−p)^2 p P(X≥4) = 1−p−(1−p)p−(1−p)^2 p P(X≥4) = (1−p)(1−p−(1−p)p) P(X≥4) = (1−p)(1−2p+p^2 ) P(X≥4) = (1−p)(1−p)^2 P(X≥4) = (1−p)^3

$${P}\left({X}=\:{k}\right)\:=\:{q}^{{k}−\mathrm{1}} {p}\:=\:\left(\mathrm{1}−{p}\right)^{{k}−\mathrm{1}} {p} \\ $$$${P}\left({X}\geqslant\mathrm{4}\right)\:=\:\mathrm{1}−{P}\left({X}=\mathrm{1}\right)−{P}\left({X}=\mathrm{2}\right)−{P}\left({X}=\mathrm{3}\right) \\ $$$${P}\left({X}\geqslant\mathrm{4}\right)\:=\:\mathrm{1}−\left(\mathrm{1}−{p}\right)^{\mathrm{0}} {p}−\left(\mathrm{1}−{p}\right){p}−\left(\mathrm{1}−{p}\right)^{\mathrm{2}} {p} \\ $$$${P}\left({X}\geqslant\mathrm{4}\right)\:=\:\mathrm{1}−{p}−\left(\mathrm{1}−{p}\right){p}−\left(\mathrm{1}−{p}\right)^{\mathrm{2}} {p} \\ $$$${P}\left({X}\geqslant\mathrm{4}\right)\:=\:\left(\mathrm{1}−{p}\right)\left(\mathrm{1}−{p}−\left(\mathrm{1}−{p}\right){p}\right) \\ $$$${P}\left({X}\geqslant\mathrm{4}\right)\:=\:\left(\mathrm{1}−{p}\right)\left(\mathrm{1}−\mathrm{2}{p}+{p}^{\mathrm{2}} \right) \\ $$$${P}\left({X}\geqslant\mathrm{4}\right)\:=\:\left(\mathrm{1}−{p}\right)\left(\mathrm{1}−{p}\right)^{\mathrm{2}} \\ $$$${P}\left({X}\geqslant\mathrm{4}\right)\:=\:\left(\mathrm{1}−{p}\right)^{\mathrm{3}} \\ $$

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