Question

The number of fires in a certain forest follows a Poisson process at a rate of 0.40 per week.

A) What is the probability that there are no more than 3 fires in a given week?

B) What is the the probability that the time between fires is at least 3 weeks?

Answer 1

Let X be the the number of fires in the forest. Then X~ Poi ( =0.40)

(a) We have to calculate the probability that there are no more than 3 fires in a given week, i.e we have to calculate probability

P (X <=3) = P (X=0) + P(X=1) + P (X=2) + P(X=3) where X~ Poi ( =0.40). the pdf for X is given by

P (X=k) = where =0.40, hence putting k=0,1,2,3 and adding these values we get

P (X<=3) = 0.6703+0.2681+0.0536+0.00715 = 0.9992

(b) The waiting times between fires follow exponential process with parameter (by properties of poisson process).

Thus if Y is the waiting times between fires, then Y~ exponential( =0.40) and pdf of Y is given by

f(Y,) = , y > 0. We have to calculate the probability that the time between fires is at least 3 weeks i.e

P (Y>=3) = = e^-1.2 = 0.3012 (We have used the result that for an exponential distribution with parameter k

P (Y >=k) = e^- )

Hence the probability that there are no more than 3 fires in a given week = 0.9992

and the probability that the time between fires is at least 3 weeks = 0.3012