Suppose that a book of 600 pages contains 40 print mistakes. Assume that these errors are randomly…
Suppose that a book of 600 pages contains 40 print mistakes. Assume that these errors are randomly distributed throughout the book and the number of errors per page follows a Poisson distribution. The probability that all the 10 pages selected at random with no print mistakes is
$\frac{1}{3} e^{-1}$
$2 e^{-1 / 3}$
$e^{-2 / 3}$
$\frac{1}{3} e^{-2}$
Solution
Assuming errors are randomly distributed throughout the book and $x$, the number of errors per page has a poisson distribution
$P(X=x)=\frac{e^{-\lambda} \lambda^x}{x !}(x=0,1,2, \ldots$.
Probability of mistakes $=\frac{40}{600}=\frac{1}{15}$
Mean $\lambda=n p=10 \times \frac{1}{15}=\frac{2}{3}$
$p(X=0)=\frac{e^{-\lambda} \lambda^0}{0 !}=e^{-\lambda}=e^{-2 / 3}$