# Calculate gas-station probabilities

I would like to calculate probabilities for the next exercise:

Knowing the average amount of cars that drive per minute into a gas-station is 3.

** How can I calculate the probability of arriving at least 12 cars into the station in a period of 5 minutes?

** And the probability of a car arriving before two minutes go by?

Thank you all very much in advance.

• Don't you need some further assumption about distribution of arrivals? such as whether this is a Poisson process, or an exponential distribution, or uniform, or.... Commented May 16, 2013 at 13:06
• There was no initial assumption, just the average. I think an exponential Poisson distribution should be used but I can not figure out the exercise. Thanks. Commented May 16, 2013 at 13:14
• I think exponential distribution and Poisson distribution are two different things. Commented May 16, 2013 at 13:20
• A very similar problem is #2 at pat-rossi.com/MTH4451/Test2/Spring2009/spsol_09a.pdf Commented May 16, 2013 at 13:24

We use a Poisson model. So if $X$ is the number of cars in randomly chosen $1$ minute interval, then $X$ is assumed to have Poisson distribution with parameter $3$.
Thus by general theory, the number $Y$ of cars in a $5$ minute interval has Poisson distribution with parameter $\lambda=15$. thus the probability of fewer than $12$ cars in a $5$ minute interval is $$\sum_{k=0}^{11}e^{-\lambda}\frac{\lambda^k}{k!}.\tag{1}$$ This is a somewhat unpleasant calculation. Maybe compute from $k=11$ down. After a while, the terms you are adding become negligible. The probability of at least $12$ cars is $1$ minus the probability computed in $(1)$.
The number $Z$ of cars in a $2$ minute interval has Poisson distribution with parameter $6$. We want the probability of at least one car, which is $1$ minus the probability of no cars. The probability of no cars is $e^{-6}$. Alternately, for the no cars problem one can use the relationship between the Poisson and the exponential.