I have a problem which states as below:
A service station handles jobs of two types, A and B. (Multiple jobs can be processed simultaneously.) Arrivals of the two job types are independent Poisson processes with parameters
λ1=3andλ2=4per minute, respectively. Type A jobs stay in the service station for exactly one minute. Each type B job stays in the service station for a random but integer amount of time which is geometrically distributed, with mean equal to 2, and independent of everything else. The service station started operating at some time in the remote past.
the question asked:
At time 0, no job is present in the service station. What is the PMF of the number of type B jobs that arrive in the future, but before the first type A arrival?
this is a question from a MIT open course where the teaching assistant has kindly provided an answer as below:
$P(k) = (\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}})^k (\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}) = (\frac{4}{7})^k (\frac{3}{7})$
it basically said the question was asking for the probability of the number of trials needed until the first success. i.e. a geometric distribution is what we used to describe this probability.
However my answer was different.
from the description we know B Arrival's PDF.
$P(k, \tau) = \frac{(\lambda_{2}\tau)^{k} e^{-\lambda_{2}\tau} }{k!}$
and given poisson process' interarrival time probability is described by expotential distribution, we could aslo derive from this the expected time of first arrival.
$E[T_{A}] = \frac{1}{\lambda} = \frac{1}{3}$
I think I can replace $\tau$ with expected arrival time of A and calculate the PDF of number of B arrivals.
$P(k, \frac{1}{3}) = \frac{\frac{4}{3}^{k} e^{-\frac{4}{3}} }{k!}$
I know it is asking for PMF and it makes sense to use geometric distribution under discrete time, but what's wrong with my PDF version?
could anyone explain to me if my answer is wrong and how it is wrong?