Answer:
The probability that the reporter made no typographical errors for the 3-page article is 5%.
Step-by-step explanation:
Let X = number of typographical errors made by the writer.
The average umber of mistakes mad by the writer every page is, E (X) = 1.
The random variable X is defined as finite number of occurrence of a certain activity in a fixed interval of time.
A Poisson distribution is used to describe the distribution of occurrences in a certain interval.
Thus, the random variable X follows a Poisson distribution.
It is provided that the writer has landed a 3-page article in an important magazine.
Then the average number of mistakes in the 3 pages is:
λ = 3 × E (X) = 3 × 1 = 3.
The probability mass function of the Poisson random variable X is:
[tex]P(X=x)=\frac{e^{-3}3^{x}}{x!};\ x=0,1,2,3...[/tex]
Compute the probability that the writer makes no mistake in a 3-page article as follows:
[tex]P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times 1}{1}=0.0498\approx 0.05[/tex]
The probability that the writer makes no mistake in a 3-page article is 0.05.
Thus, the probability that the reporter made no typographical errors for the 3-page article is 5%.
