Question

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

• A. $\frac{1}{5}\sqrt{e}$
• B. $\frac{1}{e^{0.1}}\left(\frac{221}{200}\right)$
• C. $\frac{1}{e^{0.1}}\left(\frac{111}{200}\right)$
• D. $\frac{1}{5}{e}^{0.1}$

Answer 1

Answer: (B)

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
$=P(x=0)+P(x=1)+P(x=2)$
$=e^{-0.1}\left[\frac{(0.1{)}^0}{0!}+\frac{(0.1{)}^1}{1!}+\frac{(0.1{)}^2}{2!}\right]$
$=e^{-0.1}[1+\frac{1}{10}+\frac{1}{200}=e^{-0.1}\left[\frac{200+20+1}{200}\right]$
$=e^{-0.1}\left(\frac{221}{200}\right)=\frac{1}{e^{0.1}}\left(\frac{221}{200}\right)$