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提高液晶显示器（LCD）的生产率

如果每台 LCD 显示器的坏点少于或等于15，则它是合格的. 现在若要生产 4000×2000 像素的显示器，且要保证至少90%的合格率，求需要的最大坏点率.

    
Wolfram Language code: prob = Probability[n ≤ 15, nPoissonDistribution[4000 2000 λ]]
Wolfram Language code: LogLogPlot[prob, {λ, 10 ^ -7, 10 ^ -5}, PlotStyle -> Thick, ImageSize -> Medium, GridLines -> Automatic, Frame -> True, BaseStyle -> {FontFamily -> "Verdana"}, FrameLabel -> {Style["Pixel Failure Rate", Bold], Style["LCD Acceptance Rate", Bold]}]
Wolfram Language code: Maximize[{λ, prob ≥ 0.90 && 10 ^ -6 < λ < 2 10 ^ -6}, λ, Reals]