Problems with formulating notions for a probability exercise about bank tellers and service time. I am currently working on a probability exercise, whose instructions could be found in the picture that is attached to this question.

Given the instructions, I would like now to introduce what problems I have with the question and its parts:
On 1a), I am pretty sure that given two service times X and Y for each of the two customers occupying a teller and a third person (i.e. me) about to occupy one of the tellers with service time Z, then I assume that the probability in this case is described by the expression P(X+Z>Y|P=p)P(P=p)+P(Y+Z>X|P=p)P(P=p), where P is the r.v. for indicating which teller was chose. Assuming that the choice of teller is random and unifrom, then P(P=p)=1/2, leading to the simplified expression 1/2x(P(X+Z>Y|P=p)+P(Y+Z>X|P=p)). However, I have some problems with how to compute each probability term to give a more final answer that is based only on the parameters that were specified (at least I think that how it should be done like).
On 1b), I think I understand the question, but I am less sure about how to begin here with the symbols here. I assume that provided that e.g. lambda>mu, then one wants to calculate the probability that the teller with parameter lambda is selected by the third person (i.e. you as specified in the instructions). But how exactly should I formulate this using the standard probability notion? How should I incorporate the parameter values into the probability notion about to be used here to stress this?
On 1c), I suppose that the correct probability expression to be calculated is P(Y>X) and one just calculates the double integral, where the interval of the outer integral is [0, \infty] for Y=y and the interval of the inner integral is [0, y] for X=x. I hope that this is the procedure, so it would be good to get some feedback on this one.
For your information: I have been searching after solutions to similar problems as these, but they did not help me well to grasp the concept that good at all.
Thank you very much for all help I get!
 A: A key observation to make here is that the exponential distribution has the memoryless property. Thus, we don't need to worry about the fact that we don't know how long the customers have been there.
1a) I am going to assume that when the exercise says that you randomly choose a teller, this is shorthand for saying that you pick either the teller with probability 1/2. In that case,
$$P(\mbox{You're Last}) = (1/2)P(\mbox{You pick teller $\lambda$ and are last})+(1/2)P(\mbox{You pick teller $\mu$ and are last})$$
Now,
$$P(\mbox{You pick teller $\lambda$ and are last})$$
is the probability that teller $\mu$ serves his customer before teller $\lambda$ serves two customers. It is actually easier to compute the complement of this event, the probability that $\lambda$ serves both customers before $\mu$.
We can think of the time to serve a customer as a race between the two tellers. The finish times are $T_{\lambda}$, $T_{\mu}$ exponential random variables with the parameters given by the subscript. A quick calculation shows that
$$P(T_{\lambda} < T_{\mu}) = \frac{\lambda}{\lambda+\mu}$$
By the memoryless property, when one service happens, a new race begins, regardless of how long has elapsed.
So the probability that $\mu$ serves his customer last is $P(T_{\lambda} < T_{\mu})^2$, since teller $\mu$ must lose the race twice. Thus
$$P(\mbox{You pick teller $\lambda$ and are last}) = 1-\left(\frac{\lambda}{\lambda+\mu}\right)^2$$
and similarly
$$P(\mbox{You pick teller $\mu$ and are last}) = 1-\left(\frac{\mu}{\lambda+\mu}\right)^2$$
and hence we get the answer
$$P(\mbox{You're Last}) = (1/2)\left[1-\left(\frac{\lambda}{\lambda+\mu}\right)^2\right]+(1/2)\left[1-\left(\frac{\mu}{\lambda+\mu}\right)^2\right].$$
1b and 1c can be solved using the same type of argument, so that you can avoid all the messy integration. 
