Question

Assume that books from a certain publisher have an average of one misprint every 20 pages. (a) What is the probability that a given page has two or more misprints? (b) What is the probability that a book of 200 pages has at least one page with two or more misprints?

Answer 1

Ans Given problem says that on an average there are one misprint every 20 pages.

Let X denote the random variable which is the number of misprints in every page.

Thus X will follows poisson distribution and since there are on any average one misprint every 20 pages so on an average there are 1/20 misprints every page. Thus the average or mean for X is 1/20 per page.

Thus X is a poisson random variable with   = 1/20 per page.

First we need to calculate the probability that on a given page there are two or more misprints which is

P(X >=2) = 1 - P(X < 2) = 1 - P(X=0) - P(X=1)

Now for X which follows Poisson distribution P(X=k) =

Therefore , P(X >=2) = 1 - P(X=0) - P(X=1) = 1 -

= 1 - 0.99879 = 1.21 * 10^-3

So the probability that a page has two or more misprints is 1.21 * 10^-3 .

Let the number of pages in a 200 pages book with two or more misprints be denoted by Y. So here Y will follow Binomial Distribution with n = 200 and p = 1.21 * 10^-3 .

And for Y being a binomial distribution P(Y=k) =

We need to calculate P(Y>=1) = 1 - P(Y=0)

= 1 -

= 1 - 0.7851 = 0.215 aprroximately