# Poisson Process: a problem of customer arrival.

There is a question that I am not sure about the answer.

Customers arrive according to a Poisson process of rate 30 per hour. Each customer is served and leaves immediately upon arrival. There are two kinds of service, and a customer pays \$5 for Service A or$15 for Service B. Customers independently select Service A with probability 1/ 3 and Service B with probability 2/ 3

• a. During the period 9:30–10:30am, there were 32 customers in total. What is the probability that none of them arrived during 10:25–10:30am?

Attempt a: $P(N(60)-N(55)=0|N(60)=30)=\frac{P(N(60)-N(55)=0,N(60)=32)}{P(N(60)=32)}=\frac{P(N(55)=32)P(N(5)=0)}{P(N(60)=32)}$.

At this point, simply plug in the poisson process with $P(N(t)=n)=\frac{(\lambda t)^nexp(-\lambda t)}{n!}$

• b. What is the probability that the first two customers after 9:00am request Service B?

Attempt b: Treat this as two indep poisson process each has rate of $1/3 \lambda$ and $2/3 \lambda$. This is simply $(\frac {1/3 \lambda}{1/3\lambda +2/3 \lambda})^2$

• c. Determine the expected amount paid by all customers during a 10 minute period.

Attempt c: $E[N(t)]=E[N_1(t)]+E[N_2(t)]=5*1/3\lambda t+15*2/3\lambda t=35/3 \lambda t$. @10 min, we have $35/3*30/60*10=350/6$

• You should ask those as three separate questions. – 355durch113 Apr 9 '16 at 7:29

The (a) part could be simply solved by binomial. Although, Poisson is still an option, it just gets a little messy.

Binomial will simply be: Binomial(20,1/12). Try to check by solving. The answer will be (11/12)^32

a) Conditional on the number of arrivals in a given period, the arrivals are independently uniformly distributed over the period, so the probability that all of them arrived in $\frac{11}{12}$ of the hour is

$$\left(\frac{11}{12}\right)^{32}\approx6.2\%\;.$$

This is also what your more involved approach yields upon simplification.

b) This has nothing to do with the arrival times and thus with the Poisson process; each customer independently decides with probability $\frac23$ to choose service $B$, so the probability that the first two customers choose service $B$ is $\left(\frac23\right)^2=\frac49$. This is also what your more involved approach would have yielded upon simplification, except you seem to have accidentally applied it to service $A$ instead of $B$.