Vtyjkyuk

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:
$P(N(60)-N(55)=0|N(60)=30)=fracP(N(60)-N(55)=0,N(60)=32)P(N(60)=32)=fracP(N(55)=32)P(N(5)=0)P(N(60)=32)$.

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 $(frac 1/3 lambda1/3lambda +2/3 lambda)^2$

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

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

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 $frac1112$ of the hour is

$$left(frac1112right)^32approx6.2%;.$$

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 $left(frac23right)^2=frac49$. 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 $B$.

c) Your answer is correct.