A waitress serves two tables. She reports their complaints to the boss. Complain

Question

A waitress serves two tables. She reports their complaints to the boss. Complaints are received according to independent Poisson processes with respective rates 1 = 1/60 per minute and 2 = 1/90 per minute.

(a) After she starts working, what is the probability that the first complaint is from table 1?

(b) Given that the first complaint came from table 1 at time t, what is the probability that the next complaint is from table 2?

(c) Assume the boss checks the table 1 if the waitress does not report any complaints from table 1 for 30 minutes. After she starts working, what is the expected time until the first check?

Explanation / Answer

(a) Let N(t) be a Poisson process with rate =1/60 +1/90 = 5/180 = 1/36 We split N(t) into two processes N1(t) and N2(t) in the following way. For each arrival, a coin with P(H)=(1/60)/(1/36) = 3/5 is tossed. If the coin lands heads up, the arrival is sent to the first process (N1(t)), otherwise it is sent to the second process. The coin tosses are independent of each other and are independent of N(t). Then

Thus, N1(t) and N2(t) have the same probabilistic properties as the ones stated in the problem. We can now restate the probability that the first arrival in N1(t) occurs before the first arrival in N2(t) as the probability of observing at least one heads in one coin toss, which is 3/5

So, the probability that the first complaint is from table 1 = 3/5

(b) This is equivalent to observing (as in part (a)), the probability of observing at one head and then one tail in two coin tosses. So, the probability is

p = (3/5) * (2/5) = 6/25

(c) The boss checking the table after 30 min.

In a poisson process, the interarrival time distribution is exponential with rate and the exponential distribution is memoryless, so the expected time until the first check would be 30 min.

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