I would appreciate a little kickstart with this problem.
In a bank, there are 2 counters. The duration of serving a client at the first counter is a random variable ${X_1}$ ~ ${Exp(λ_1)}$, the duration of serving a client at the first counter is a random variable ${X_2}$ ~ ${Exp(λ_2)}$. Here ${λ_1 = 1.01}$ and ${λ_2 = 3}$.
When clients A and B arrived at the bank, both counters were occupied. Both clients want to be served at different counters, i.e., if A goes to the first one, B goes to the second one, and vice versa. On average, how long will the clients A and B will have to wait in order to get served, if A is first in the line and B is the second one?
All I know thus far is that since the problem involves about exponential distribution, we can make use of formulas
${P(T = t') = e^{-λ·t}}$ and ${E[X] = \frac{1}{λ}}$ since the problem asks how long on average they will have to wait.
So, it's possible to get the average serving times at each counter:
${E[X_1] = \frac{1}{1.01}} ≈ 0.99$
${E[X_2] = \frac{1}{3}} ≈ 0.33$
Where do I go from here though?
The exponential distribution is "memoryless" - past behavior has no effect on future expectations. What this means is that when someone walks up to the counter, the wait time is exponentially distributed with some expected wait time. After the person has been at the counter for some time, the further wait time has the exact same exponential distribution. The expected time for a customer to finish at counter A is 0.99, regardless of whether that person just walked up to the counter, or if they've been there for an hour already.
A will expect to wait 0.99, and B will expect to wait 0.33. It doesn't matter how long the customers already at the counters have been there, since their expected remaining time stays constant no matter how long they've been there. Furthermore, it doesn't matter how long A and B actually wait, so long as they haven't been served yet, they will always expect to wait another 0.99 and 0.33 units of time, respectively.
EDIT: This incorrectly assumes that there is one line for each counter, rather than one line serving both counters. B clearly must expect to wait longer than A, since A is first in line.