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

Step-by-step explanation:

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

Then , the average number of errors per page =
`(600)/(600)=1`

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

The Poisson distribution function is given by :-

`P(x)=(e^(-\lambda)\lambda^x)/(x!)`

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

`P(x=0)=(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-((e^(-1)1^0)/(0!)+(e^(-1)1^1)/(1!)+(e^(-1)1^2)/(2!))=0.0803013970714\approx0.0803`