In a telephone exchange, it is believed that the durations of calls are exponentially distributed with mean length 10 min. Three calls start simultaneously at 1:00:00 PM; assume that the duration of these calls are independent.
i) What is the probability that the longest call last beyond 1:15:00 PM. hint: under what circumstances are all calls finished by 1:15:00PM.
ii) Suppose that the second longest call ends at 1:08:30 PM, what is the probability that the longest call is still going on at 1:23:30 PM
I have started answering the problem in the following way:
Xi - exponential(lambda i) for i = 1, 2, … n. and X1, …, Xn mutually independent r.v. then min{X1,…,Xn} follows Exp(Σ lambda i)
So the for the shortest call to last beyond 1:15:00 pm, I would have:
Z = X1 + X2 + X3 Z - Exp(3*1/10) = Exp(3/10)
P(Z > 15) = 1 - P(Z < 15) = e^(-15*3/10)
However, this leads to the minimum, not the maximum.
I am kind of stuck here.
Can anyone give me a path for a solution ?
$\lambda = 1/10 = 0.1$
Any of the calls could be the longest one.
Exponential cdf is $1- exp(-\lambda*x)$.
So the probability that the longest call last beyond 1:15:00 PM = probability that at least one of the calls is longer than 15 mins = $(1 -exp(-0.1*15))^2*exp(-0.1*15) + (1-exp(-0.1*15)) *exp(-0.1*15)^2 + exp(-0.1*15)^3$
As the calls are independent.
probability that longest call is longer than 23.5 minutes = $(exp(-0.1*23.5))$ = 0.095
probability that the second longest call and least longest calls are less than 8.5 minutes = $(1 - exp(-0.1*8.5))^2$
So the joint probability = $(1 - exp(-0.1*8.5))^2*(exp(-0.1*23.5)) $