Bus arrival probability... This question has two parts, 
Question one
A municipal bus system, when operating perfectly on time, provide at any given bus stop every 30 min. A person arrive at a random time at bus stop. What is the average waiting time until the next bus arrives? 
The Answer 
Since buses are following uniform distribution the average waiting time will be 15 min.
Question two
If the busses become totally disorganized and random following Poisson random process what will be the average waiting time given that number of buses remains content
I know the answer is 1/lambda, but how to find the lambda, is it still 15 min since the same number of busses are running. Can someone explain this to me!!!
Thanks 
 A: The number of buses remains constant, so on average we still have 1 bus / 30 minutes, or 1/30 buses per minute. This is $\lambda$ assuming minute as your time interval.
A: You are right, that you have to wait on average 15 minutes if the arrival time is uniformly distributed as $X\sim U(0,30)$
The expected value is $E(X)=\int_0^{30} \frac{x}{30} \, dx=\frac1{60}\cdot (900-0)=15$
Let the arrival time  poisson distributed where the expected number of buses is  $1$ per 30 minutes ($\lambda=\frac1{30}$). The pdf that we observe $k$ buses in  $t>0$ minutes is
$$P(N(t)=k)=e^{-\frac1{30}\cdot t} \frac{\left(\frac1{30}\cdot t\right)^k}{k!}$$
The expected value is $\frac{t}{30}$. That means that we have to wait on average $30$ minutes for one bus ($t=30$). We have to recognize that the waiting time between two bus stops can be larger than 30 minutes (t>0). There is no cut-off at $30$ minutes as in the case of the uniformly distributed variable since $P(X>30)=0$. For instance the probability that one bus does not arrive in the first 30 minutes but in the second half an hour is positive:
$$P(N(30)=0 \cap N(60)=1)=e^{-\frac1{30}\cdot 30}\cdot \frac{\left(\frac1{30}\cdot 30\right) ^0}{0!} \cdot e^{-\frac1{30}\cdot 30}\cdot \frac{\left(\frac1{30}\cdot 30\right) ^1}{1!}$$ 
$$e^{-1}\cdot \frac{\left(1\right) ^0}{0!} \cdot e^{-1}\cdot \frac{\left(1\right) ^1}{1!}\approx 13.53\%$$
