Question

A writer makes on average one typographical error every page. The writer has landed a 3-page article in an important magazine. If the magazine editor finds any typographical errors, they probably will not ask the writer for any more material. What is the probability that the reporter made no typographical errors for the 3-page article? Use the Poisson distribution and round your answer to 4 decimal places.

Answer 1

Answer:

The probability that the reporter made no typographical errors for the 3-page article is 0.7165.

Step-by-step explanation:

For Poisson Distribution:

p(k:λ) = [(λ^k)(e^-λ)]/k!

where k is the number of outcomes and λ is the rate at which the outcomes occur.

λ=np

where n=number of errors on one page

           p=probability that an error appears on a given page in the 3 page article

From the question we have n=1 and p=1/3. Computing these values:

λ=np

λ=1 x 1/3

λ=1/3

The question is asking for the probability that no error occurs in the 3-page article i.e. k=0. Using the Poisson formula mentioned above:

P(0:1/3) = (1/3)^0 e^(-1/3)/(0!)

            = (1)(0.7165)/1

P(0:1/3) = 0.7165

The probability that the reporter made no typographical errors for the 3-page article is 0.7165.