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