Question

Suppose that a book of 600 pages contains 40 printing mistakes. Assume that these errors are randomly distributed throughout the book and x, the number of errors per page has a Poisson distribution. What is the probability that 10 pages selected at random will be free of errors?
1. \(\frac{1}{3}e^{-1}\)
2. \(2e^{-\frac{1}{3}}\)
3. \(e^{-\frac{2}{3}}\)
4. \(\frac{1}{3}e^{-2}\)

Answer 1

Correct Answer - Option 3 : \(e^{-\frac{2}{3}}\)

Explanation:

Assuming errors are randomly distributed throughout the book and x, the number of errors per page has a poisson distribution.

\(P\left( {X = x} \right) = \frac{{{e^{ - λ }} \cdot {λ ^x}}}{{x!}}\;\;\;\;\;\;\left( {x = 0,\;1,\;2, \ldots .} \right)\)

Probability of mistakes = \(\frac{40}{600}=\frac{1}{15}\)

Mean λ = np ⇒ \(10\times \frac{1}{15} = \frac{2}{3}\)

\(P\left( {X = 0} \right) = \frac{{{e^{ - λ }} \cdot {λ ^0}}}{{0!}}\)

\(P\left( {X = 0} \right) = {{{e^{ - λ }} {λ ^0}}}{{}}\)

\(P\left( {X = 0} \right) = {{{e^{ - 2/3 }} {}}}{{}}\)