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In a book consisting of 600 pages, there are 60 typographical errors. The probability that a randomly chosen page will contain at most two errors, is
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The correct answer is:
$\frac{1}{e^{0.1}}\left(\frac{221}{200}\right)$
$\begin{aligned} & \text { (b) We have, } \lambda=\frac{60}{600}=0.1 \\ & \therefore \quad P(X=x)=\frac{e^{-\lambda}(\lambda)^x}{x !}=\frac{e^{-0.1}(0.1)^x}{x !}\end{aligned}$
$\begin{aligned} & \text { Required probability }=P(x=0)+P(x=1)+P(x=2) \\ & =e^{-0.1}\left[\frac{(0.1)^0}{0 !}+\frac{(0.1)^1}{1 !}+\frac{(0.1)^2}{2 !}\right] \\ & =e^{-0.1}\left[1+\frac{1}{10}+\frac{1}{200}\right]=e^{-0.1}\left[\frac{200+20+1}{200}\right] \\ & =e^{-0.1}\left(\frac{221}{200}\right)=\frac{1}{e^{0.1}}\left(\frac{221}{200}\right)\end{aligned}$
$\begin{aligned} & \text { Required probability }=P(x=0)+P(x=1)+P(x=2) \\ & =e^{-0.1}\left[\frac{(0.1)^0}{0 !}+\frac{(0.1)^1}{1 !}+\frac{(0.1)^2}{2 !}\right] \\ & =e^{-0.1}\left[1+\frac{1}{10}+\frac{1}{200}\right]=e^{-0.1}\left[\frac{200+20+1}{200}\right] \\ & =e^{-0.1}\left(\frac{221}{200}\right)=\frac{1}{e^{0.1}}\left(\frac{221}{200}\right)\end{aligned}$
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