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.
Best Answer
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.