how to solve this type of poisson distribution

Question

Suppose that the number of typing errors per page has a Poisson distribution with average 4 typing errors. What is the probability that in a given page the number of typing errors will at least 2?

Answer 1

P(x) = λ^xe^-λ/(x!) where λ is here 4 because μ for a Poisson distribution is λ

In order to find P(x ≥ 2), find the equivalent 1 - P(x < 2) = 1 - P(0) - P(1) plugging in for 0 and 1 into the formula for the distribution

P(0) = e^-4 and P(1) = λe^-λ

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