# Bursts in Poisson arrival process

Assume bursts arrive as a Poisson process and the number of instances from a bursts follow a certain distribution $$G$$ (Assume it's a geometric if it really matters). Now, I need to calculate the probability of having three instances in a time span $$[0,T]$$. There are multiple scenarios where how this may happen:

1. One arrival and 3 instances bursts from that single arrival.
2. Two arrivals, where two instances burst from a one arrival and only one instance bursts from the other.
3. Three arrivals, where a one instance bursts from each arrival.

I have no problem in calculating the probability of the first one, however I am struggling to calculate the second and the third scenarios. The reason is that the probability of having exactly two arrivals is $$(\lambda t)^2 e^{\lambda T} \over 2!$$ and I can't handle each event separately. For instance:

P[one arrival] * P[two bursts] * P[one arrival] * P[one bursts]

How to handle the second and third cases?

Say that $$a$$ is the number of arrivals (the average number of arrivals in the given interval is $$\lambda$$), $$b$$ is the number of bursts for one arrival, $$G(b)$$ is the probability of having $$b$$ bursts during one arrival: for each arrival you have to extract a number $$b$$ from $$G$$.
You want to calculate the probability that the total number of bursts observed in the given interval sum up to a number $$b_{tot}=3$$.
I assume that arrivals with no bursts are not possible. So, to have $$b_{tot} =3$$ you have three possibilities: $$(a=1, b_1=3)$$ or $$(a=2, b_1+b_2=3)$$ or $$(a=3, b_1+b_2+b_3=3)$$. Then,
$$P(a=3,b_1=1,b_2=1,b_3=1) = e^{-\lambda} \frac{\lambda^3}{3!} G(1)^3$$ $$P(a=2,b_1=1,b_2=2) = e^{-\lambda} \frac{\lambda^2}{2!} G(2)G(1)$$ $$P(a=2,b_1=2,b_2=1) = e^{-\lambda} \frac{\lambda^2}{2!} G(2)G(1)$$ $$P(a=1,b_1=3) = e^{-\lambda} \frac{\lambda^3}{3!} G(3)$$
$$P(b_{tot}=3) = P(a=1,b_1=3) + 2 P(a=2,b_1=2,b_2=1) + \\ +P(a=3,b_1=1,b_2=1,b_3=1)$$