Question

 Assume that typographical errors committed by a typesetter occur completely randomly. Suppose that a book of 600 pages contains 600 such errors. Using the Poisson distribution, calculate the probability (a) that page 1 contains no errors (b) that page 1 contains at least three errors

Answer 1

Answer: (a) 0.3679

(b) 0.0803

Explanation:

Given : A book of 600 pages contains 600 such errors.

Then , the average  number of errors per page = $\dfrac{600}{600}=1$

$\text{i.e. }\lambda=1$

The Poisson distribution function is given by :-

$P(x)=\dfrac{e^{-\lambda}\lambda^x}{x!}$

Then , the probability that that page 1 contains no errors ( Put $x=0$ and $\lambda=1$) :-

$P(x=0)=\dfrac{e^{-1}1^0}{0!}=0.3678794411\approx0.3679$

Now, the probability that page 1 contains at least three errors :-

$P(x\geq3)=1-(P(0)+P(1)+P(2))\\\\=1-(\dfrac{e^{-1}1^0}{0!}+\dfrac{e^{-1}1^1}{1!}+\dfrac{e^{-1}1^2}{2!})=0.0803013970714\approx0.0803$