Question

Books from a certain publisher contains an average of 2 misprint per page. What is the probability that on at least one page in a 25-page book from this publisher there will be at least 5 misprint?

Answer 1

= 2

X: Number of misprints

X follows a Poisson distribution.

PMF of Poisson distribution

P(X 5) = 1 - P(X 4)

P(X 4) = P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)

P(X 4) = P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0) = 0.0902 + 0.1804 + 0.2707 + 0.2707 + 0.1353 = 0.9473

P(X 5) = 1 - P(X 4) = 1 - 0.9473 = 0.0527

P(no pages with 5 or more errors)
= 1 - (1 - 0.0527)^25 = 0.7413

P(at least one page with 5 or more errors) = 1 - 0.7413 = 0.2587