The arrivals of phone calls at a telephone switching office is a Poisson random

Question

The arrivals of phone calls at a telephone switching office is a Poisson random process N(t) with an arrival rate lambda = 4 calls per second. We monitor N(t) starting from t = 0 over a 10-second interval. Let Sn be the time of the arrival of the nth call. (You can leave the answer as a function of e). What is P(N(1) =0), the probability of no phone calls during the first second? What is P(N(4)-N(3)= 4), the probability of four phone calls arrive between the third and the fourth second ? What is E[S10], the expected time that the 10th phone call arrives ? Hint: consider intra arrival time What is V ar[S10], the variance of random variable S10 ?

Explanation / Answer

i am denoting lamda by L in the below discussion for my convinience.

Poisson distribution, P(r)= (e^-L )* (L^r)/L!

above P(r) signifies the probability of "time between two successive calls is r"

1) so for r=0
P(0)=e^-4

2) It is nothing but P(4)
P(4)= (32/3) *e^-4

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