Question

A book has 200 pages. The number of mistakes on each page is a Poisson random variable with mean 0.01, and is independent of the number of mistakes on all other pages. a. What is the expected number of pages with no mistakes

Answer 1

Answer:

198 pages have no mistakes.                                      

Step-by-step explanation:

We are given the following in the question:

The number of mistakes on a page follows a poison distribution with

$\mu = \lambda = 0.01$

Formula:

$P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}$

We have to find the probability that there are no mistakes on a page.

$P(X =0) = \displaystyle\frac{(0.01)a^0 e^{-0.01}}{0!} = 0.9901$

Thus, approximately 99.01% of the pages have no mistakes.

Number of pages =

$\text{Total number of pages}\times 99.01\%\\\\200\times \dfrac{99.01}{100}\\\\\approx 198$

Thus, 198 pages have no mistakes.