This is the scenario:
Packets arrive to a router according to a Poisson process with rate $\lambda$ and, indipendently from any other system variable, they are trasmitted to link 1, with capacity $C_1$, with probability $p_1$, or to link 2, with capacity $C_2$, with probability $p_2 = 1 - p_1$. Packets lengths are r.v. i.i.d. with negative exponential distribution and mean value $L$.
We adopt all the hypothesis to model the router like this:
We have $K = 10$. Suppose that at some point of time we have 9 packets in the router, and that link 1 is idle.
Question: What is the probability that the current trasmission ends before a new packet arrives?
I tried to answer in this way: We can say that the queueing system 1 is empty, because link 1 is idle, so this 9 packets are all in the queueing system 2 (1 in service and 8 in the queue). Due to the memoryless propriety of Poisson processes, at any time instant we observe the system, the probability that a transmission ends is still negative exponentially distributed, with average transmission time equal to $\bar x_2 = L/C_2$ seconds.
The probability that the current transmission ends before a new packet arrives can be thinked as the probability that 0 packets arrives in $\bar x_2$ seconds, i.e.:
$P[0$ packets in $\bar x_2$ seconds$] = e^{-\lambda t}(\lambda t)^k/ k!$, substituting $t = \bar x_2$ and $k=0$.
Do you think is correct? The information that I have 9 packets in the system is needless, right? Thank you all.