A bank has 3 tellers, each occupied with 1 person, average time at each teller is 5 minutes. What is the probability of waiting less than 3 mins? This is an exponential distribution with $\lambda =\frac{1}{5}$
The way this goes is:
The first customer finishes before three minutes OR second customer finishes before three minutes Or third customer finishes before 3 minutes.
With this, it would 
$$3\int_0^3 \lambda e^{-\lambda x}dx =3(1-e^{\frac{-3}{5}})$$
Is this correct?
 A: These are not mutually exclusive, so you don't just add them up.
Instead, the customer is still waiting if:
The first customer is still going at 3 minutes AND the second customer is still going at 3 minutes AND the third customer is still going at 3 minutes.
These are independent.
A: Sadly not.  Our intuition suggests that the probability of waiting less than 3 minutes should be an increasing function of the number of tellers, but that this probability cannot increase without bound.  If you have one teller, the probability would be $$\Pr[T \le 3] = 1 - e^{-3/5} \approx 0.451188.$$  But if you simply multiply this by $3$, you get a probability that exceeds $1$, which is absurd.
Instead, let's consider the complementary probability; that is to say, what is the chance of needing to wait more than 3 minutes, $\Pr[T > 3]$?  In order for this event to occur, all three tellers must continue to be occupied beyond 3 minutes.  Since the service time of each teller is independent and identically distributed, this suggests that $$\Pr[T > 3] = \Pr[T_1 > 3] \Pr[T_2 > 3] \Pr[T_3 > 3] = (e^{-3/5})^3 = e^{-9/5} \approx 0.165299.$$  Then we find $$\Pr[T \le 3] = 1 - \Pr[T > 3] = 1 - e^{-9/5} \approx 0.834701.$$  Why did this approach work?  Why did we have to first consider the complementary probability?  Why couldn't we have calculated $$\Pr[T \le 3] = \Pr[T_1 \le 3]\Pr[T_2 \le 3]\Pr[T_3 \le 3],$$ where $T_1, T_2, T_3$ are the service times of each of the three tellers?  The reason is because our waiting time is the least of the individual waiting times of those already being serviced.  We can be seen by the next available teller, even if the other teller(s) remain occupied.  So it is not necessary for all tellers to become available for us to stop waiting.
So why did the complementary probability calculation work?  Because in order for us to keep waiting past 3 minutes, all of the tellers must continue to be occupied.  So each event $T_1 > 3$, $T_2 > 3$, and $T_3 > 3$, must occur.
