Answer:
The probability of 2 or more errors is 0.018.
Step-by-step explanation:
Let X = number of typographical errors on a page of a certain magazine.
The expected number of errors is, λ = 0.2.
The random variable X follows a Poisson distribution with parameter λ = 0.2.
The probability mass function of X is:
[tex]P(X=x)=\frac{e^{-0.2}0.2^{x}}{x!};\ x=0,1,2,3...[/tex]
Compute the probability of 2 or more errors as follows:
P (X ≥ 2) = 1 - P (X < 2)
= 1 - P (X = 0) - P (X = 1)
[tex]=1-\frac{e^{-0.2}0.2^{0}}{0!}-\frac{e^{-0.2}0.2^{1}}{1!}\\=1-0.81873-0.16375\\=0.01752\\\approx0.018[/tex]
Thus, the probability of 2 or more errors is 0.018.
