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.
