Question

Suppose calls are arriving at a telephone exchange at an average rate of one per second, according to a Poisson arrival process. Find: a) the probability that the fourth call after time t = 0 arrives within 2 seconds of the third call; b) the probability that the fourth call arrives by time t = 5 seconds; c) the expected time at which the fourth call arrives.

Answer 1

Explanation has been given below

Step-by-step explanation:

a) inter arrival times are exponentially distributed with mean 1/n , where n = rate = 1/sec.

probability distribution function is F(t)=n*exp(-n*t).

reference to any kth packet and the (k-1)th packet

the answer is = integration of F(t).dt with limits 0 to 2 = 1 - exp(-2*n) = 1 - exp(-2)

b) t=5 , P(q) = exp(-5)*(5)^q/factorial(q)

probability of fourth call within t=5 seconds is =

that is P(4) P(5) ...... = 1 - ( P(0) P(1) P(2) P(3) ) ; put the values and get the answer.

c) number of calls/rate = 4/n = 4 seconds