Answers
If time between arrivals is exponentially distributed with parameter λ, you have a Poisson process with parameter λ.
Note that λ = 1/mean time between arrivals = 1/1 = 1.
a) The wording here is tricky. If you are talking about the probability that more than two aircraft arrive in a particular hour, this would be 1 - P(0)- P(1) - P(2) = 1 - exp(-1) - exp(-1)*1^1/1! - exp(-1)*1^2/2! =
1 - exp(-1)(2.5) = 0.0803013970713942
b) In general, the probability of x intervals of n having an event = C(n,x)p^x(1-p)^(n-x)
Thus, we have C(10,0) 0.0803013970713942^0*(1-0.0803013970713942)^10 =
(1-0.0803013970713942)^10 = 0.432967469811295
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