Telephone calls arrive at an exchange according to the Poisson process at a rate λ= 2/min. Calculate the probability that exactly two calls will be received during each of the first 5 minutes of the hour.
the given solution is:
Assume that “N” is the number of calls received during 1 minute.
Therefore,
P(N= 2) = (e-2. 22)/2!
P(N=2) = 2e-2.
Now, “M” is the number of minutes among 5 minutes considered, during which exactly 2 calls will be received. Thus “M” follows a binomial distribution with parameters n=5 and p= 2e-2.
P(M=5) = 32 x e-10
P(M =5) = 0.00145, where “e” is a constant, which is approximately equal to 2.718.
