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 $35/3*30/60*10=350/6$