If you could explain how to do this, the answer is to be rounded to 4 decimal places
Summation as k goes from 0 to N of e^(-lambda) (lambda)^k / k!
Prob X = 0. e^(-lambda) * (lambda)^0 / 0! = e^(-lambda)
Prob X = 1. e^(-lambda) * (lambda)^1 / 1! = lambda * e^(-lambda)
Answer: Prob (X = 0) + Prob(X = 1) with lambda = 2:
e^(-2) * (1 + 2) = 3e^(-2)
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Summation as k goes from 0 to N of e^(-lambda) (lambda)^k / k!
Prob X = 0. e^(-lambda) * (lambda)^0 / 0! = e^(-lambda)
Prob X = 1. e^(-lambda) * (lambda)^1 / 1! = lambda * e^(-lambda)
Answer: Prob (X = 0) + Prob(X = 1) with lambda = 2:
e^(-2) * (1 + 2) = 3e^(-2)