Given a system with three servers that have service times which are independently exponentially distributed with rates $\lambda_1,\lambda_2,\lambda_3$. We arrive to find one person waiting with the three servers occupied, when one opens up a person waiting goes to the open server.
In need of finding the probability of being second to last leave the system, so that would mean no one else shows up or at least we are just considering being second to last of people currently present at time we arrive.
I am after an answer verification and also wondering if there was another way to think about this then what I have done.
So my solution is as follows: (I won't do the complete solution but rather part of it as the rest follows by a symmetrical argument.)
In order for us to be second to last we need four servers to finish where the $2^{nd}$ server to finish and the $4^{th}$ to finish are the same as we go to the $2^{nd}$ server and also need $2^{nd}$ and $3^{rd}$ to finish to be different, else we leave prematurely according to the event we are after.
That is if we let $1,2,3$ represent the servers and put them in sequences according to when they finish. Lets consider the case where we end up at server 1 then we get the following sequences:
$\{2,1,2,1\},\{3,1,3,1\},\{2,1,3,1\},\{3,1,2,1\},\{1,1,2,1\},\{1,1,3,1\}$.
That is, if we end up ate server 1, there are 6 sequences that correspond to being second to last to leave. Similarly if we end up at server 2 or 3 they will, respectively, have 6 cases as well.
And because we have independence and that the minimum will be exponentially distributed with rate $\lambda_1+\lambda_2+\lambda_3$ we get
$\displaystyle \frac{\lambda_1^2}{(\lambda_1+\lambda_2+\lambda_3)^4}(\lambda_2^2+\lambda_3^2+2\lambda_2\lambda_3+\lambda_1\lambda_2+\lambda_1\lambda_3)$
And we need a similar result for starting at server 2 and server 3.
Is there a more succinct way to go about this?