probability - vehicle arriving and gap in between them Vehicle arriving at an intersection from one of the approach roads follow the Poisson distribution. The mean rate of arrival is 900 vehicles per hour. If a gap is defined as the time difference between two successive vehicle arrivals (with vehicle  assumed to be points), the probability that the gap is greater than 8 seconds is ____.
My approach for solving above question:
If 900 vehicles are passing per hour -> $\frac{1}{4}$ vehicles passing per second. So in 4 seconds 1 vehicle passes and in 8 seconds 2 vehicles can pass. 
The problem I am facing here is how that gap is related with the question.
It says gap is greater than 8 seconds -> now in 8 seconds 1 vehicle come but earlier in 8 seconds 2 vehicles used to come. so at this point I can say mean of event is 2. i.e. value of lambda is 2. so now are we finding P(X>1) => 1 - P(X=0) i.e. finding probability of more than 1 car will pass in 8 seconds having mean of 2. This last point I am not understanding properly.
 A: As you said, the mean rate of arrival is $\lambda = \frac{1}{4} \frac{\text{vehicles}}{s}$. By the problem statement, the probability of $n$ cars passing in a time interval $t$ is 
$$ P_n(t) = \frac{(\lambda t)^n}{n!}e^{-\lambda t}.$$
If $F_n(t)$ is the probability that $n$ cars have passed at time $t$, the probability that $n$ cars have also passed at time $t + \delta t$ is 
$$ F_n(t+\delta t) = Pr(\text{n cars have passed in t})Pr(\text{no cars have passed between } t \text{ and } t+ \delta t)$$
$$ F_n(t+\delta t) = F_n(t) (1-\lambda \delta t).$$ 
Taking $\delta t  \rightarrow 0$ and using the definition of the derivative gives the differential equation 
$$ \frac{d}{dt}F_n(t) = -\lambda F_n(t)$$
for the probability that only $n$ cars have passed at time $t$. This has the solution (it's separable)
$$ F_n(t) = \lambda e^{-\lambda t}.$$
Therefore, the distribution of times between cars passing is an exponential distribution with mean $\lambda$. 
The probability that the gap is greater than $8$ seconds is 
$$Pr(t>8) = \int_{8}^{\infty}\frac{1}{4} e^{-\frac{1}{4}t},$$
which you can easily evaluate. 
