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Question Number 146927 by puissant last updated on 16/Jul/21

Answered by Olaf_Thorendsen last updated on 16/Jul/21

$$\lambda =\frac{P_{\stackrel{-}{V}}\left(M\right)}{P_V\left(M\right)}$$$$\lambda =\frac{P_{\stackrel{-}{V}}\left(M\right)P\left(\stackrel{-}{V}\right)P\left(V\right)}{P_V\left(M\right)P\left(V\right)P\left(\stackrel{-}{V}\right)}$$$$\lambda =\frac{P_M\left(\stackrel{-}{V}\right)P\left(M\right)P\left(V\right)}{P_M\left(V\right)P\left(M\right)P\left(\stackrel{-}{V}\right)}$$$$\lambda =\frac{P_M\left(\stackrel{-}{V}\right)P\left(V\right)}{P_M\left(V\right)P\left(\stackrel{-}{V}\right)}$$$$\lambda =\frac{(1-P_M\left(V\right))P\left(V\right)}{P_M\left(V\right)(1-P\left(V\right))}$$$$$$$$\mathrm{Pour}\mathrm{le}\mathrm{vaccin}\mathrm{A}:$$$${\lambda }_A=\frac{(1-P_M\left(A\right))P\left(A\right)}{P_M\left(A\right)(1-P\left(A\right))}$$$${\lambda }_A=\frac{(1-\frac{8}{1000})\frac{1}{4}}{\frac{8}{1000}(1-\frac{1}{4})}=\frac{124}{3}\approx 41,3$$$$$$$$\mathrm{Pour}\mathrm{le}\mathrm{vaccin}\mathrm{B}:$$$${\lambda }_B=\frac{(1-P_M\left(B\right))P\left(B\right)}{P_M\left(B\right)(1-P\left(B\right))}$$$${\lambda }_B=\frac{(1-\frac{6}{1000})\frac{1}{5}}{\frac{6}{1000}(1-\frac{1}{5})}=\frac{497}{12}\approx 41,4$$$$$$$$\mathrm{Plus}\lambda \mathrm{est}\mathrm{grand}\mathrm{et}\mathrm{plus}\mathrm{le}\mathrm{vaccin}\mathrm{est}$$$$\mathrm{efficace}:$$$$$$$${\lambda }_A\approx {\lambda }_B\mathrm{donc}\mathrm{les}\mathrm{deux}\mathrm{vaccins}\mathrm{B}\mathrm{ont}\mathrm{une}$$$$\mathrm{efficacite}\mathrm{comparable}$$

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