p.199 ex.16 Assume that, during each second, a Dartmouth switchboard receives one call with probability .01 and no calls with probability .99. Use the Poisson ap- proximation to estimate the probability that the operator will miss at most one call if she takes a 5-minute coffee break. A Poisson probability distribution is useful for describing the number of events that will occur during a specific interval of time or in a specific distance, area, or volume p(x)μσ2=λxe−λx!x=0,1,2,3,…=λ=λ
where:
λ= the mean number of events in the time, distance, volume, or area
e= the base of the natural logarithm https://www.ck12.org/probability/poisson-probability-distributions/lesson/Poisson-Probability-Distributions-ADV-PST/
e <- exp(1)
lamda <- 0.01 * 300
c0 <- ((lamda^0)*(e^-3)) / factorial(0) # Chance of zero calls
c1 <- ((lamda^1)*(e^-3)) / factorial(1) # Chance of one call
c0c1<-c0 + c1 #the two added together
c0c1## [1] 0.1991483ppois(1, lambda=3)## [1] 0.1991483