Consider a bank with two tellers. Three people, Anne, Betty, and Carol enter the

Question

Consider a bank with two tellers. Three people, Anne, Betty, and Carol enter the bank at almost the same time and in that order. Anne and Betty go directly into service while Carol waits for the first available teller. Suppose that the service times for two servers are exponentially distributed with mean three and six minutes (or they have rates of 20 and 10 per hour).

(a)What is the expected total amount of time for Carol to complete her businesses?

(b) What is the expected total time until the last of the three customers leaves?

(c) What is the probability for Anne, Betty, and Carol to be the last one to leave?

Explanation / Answer

(a) Suppose TA; TB; TC are the service time for Alice, Betty and Carol. Then the expected waiting
time for Carol is E(minfTA; TBg), and the total amount of time for Carol will be E(minfTA; TBg) + TC =
1=(:25 + :25) + 4 = 6.
(b) We look at the two time segments, which are time for Carol waiting and starting
service. The expect time for Carol waiting is E(minfTA; TBg) = 2 as shown above. Now we only need to
know the expect time for Carol starting service. Since this is a Poisson process, it is memoryless. When
the 2nd time segment starts, for A or B whoever has not nished the service will have the same probability
distribution as when the 1st time segment starts. Thus the expected time for the 2nd time segment is
E(maxfTA or B; TCg). So the expected total time until all of them leave
is 8.
(c) The idea is the same as (b). Since it is Poisson process and it is memoryless, when Carol starts service,
whoever A or B left will have the same probability distribution as from beginning. So P(TC > TA or B) =
:25=(:25 + :25) = :5.

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