Assume that vehicles arrive at a toll plaza according to a Poisson process. The average inter-arrival time between vehicles is $10$ minutes. One third of the vehicles are cars, and two thirds are trucks. (And the event that a vehicle is a car is independent of the arrival process.) The lunch hour is. Find:
$$P(\text{at least 8 cars arrived in [12:03, 12:07]} \mid 10 \text{ cars arrived during the lunch hour})$$
The answer is $$\sum^{10}_{i=8} {10 \choose i}\left(\frac{4}{60}\right)^i \left(\frac{56}{60}\right)^{10-i}$$ I know that the poisson distribution is the limiting case of a binomial distribution when number of trials is very large but here our n is small. Can anyone give me some hint as to why it is binomial in this example?
we can divide the hour in 1 min intervals in 60 sections
So every car has 60 options it can choose from
n(S) = $60^{10}$
Choosing 'i' car out of 10 = $\binom{10}{i}$
these 'i' cars only have 4 options available between [12:03, 12:07]
Total cases = $4^i$
The rest of the cars '10-i' has 56 options available
= $(56)^{10-i}$
Total no of ways to get exactly 'i' cars in interval [12:03, 12:07]
=$\binom{10}{i}$ $4^i$$(56)^{10-i}$
Probability = $\frac{\binom{10}{i} 4^i(56)^{10-i}}{60^{10}}$ =${10 \choose i}\left(\frac{4}{60}\right)^i \left(\frac{56}{60}\right)^{10-i}$