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Q.
In a book consisting of $600$ pages, there are $60$ typographical errors. The probability that a randomly chosen page will contain atmost two errors, is
TS EAMCET 2020
Solution:
We have, $\lambda=\frac{60}{600}=0.1$
$\therefore P(X=x)=\frac{e^{-\lambda}(\lambda)^{x}}{x !}=\frac{e^{-0.1}(0.1)^{x}}{x !}$
Required probability
$=P(x=0)+P(x=1)+P(x=2)$
$=e^{-0.1}\left[\frac{(0.1)^{0}}{0 !}+\frac{(0.1)^{1}}{1 !}+\frac{(0.1)^{2}}{2 !}\right]$
$=e^{-0.1}\left[1+\frac{1}{10}+\frac{1}{200}=e^{-0.1}\left[\frac{200+20+1}{200}\right]\right.$
$=e^{-0.1}\left(\frac{221}{200}\right)=\frac{1}{e^{0.1}}\left(\frac{221}{200}\right)$