# Probability that entrance 2 has seen more cars than entrance 1

I am trying to do probability exercises but when it is not Binomial I am kinda lost.

A parking lot had two entrances. On average, a car arrives at entrance 1 of the parking lot every $20$ minutes and a car arrives at entrance 2 of the parking lot every $15$ minutes. We suppose that the number of cars arriving at entry 1 is independent of the number of cars arriving at entry 2. Given that $3$ cars entered the parking lot today between 14h and 15h, what is the probability that entrance 2 has seen more cars arriving during this period than entrance 1 ?

E(entrance 1)= 20
E(entrance 2)= 15
P(e2 > e1 knowing there are $3$ car between in $60$ min)


I don't even know where to start. Is it conditional? I thought about the Poisson distribution but because of the second entrance I don't see what to do. Is it really a Poisson distribution?

## 1 Answer

• Three cars enter through two entries in a hour.
• Cars independently make a choice of which to enter, and do so at rates of $3$ per hour and $4$ per hour.
• Therefore the probability that any particular car enters entry-one equals $3/7$, and the conditional distribution of the count of these (given that there are three total) is binomial. $$X_1\mid (X_1+X_2=3)~\sim~\mathcal{Bin}(3, 3/7)$$
• Find the probability that entry-two exceeds entry-one given a total of three cars.   That is $\mathsf P(X_1\leq 1\mid X_1+X_2=3)$
• Thank you very much. Now I know for sure that probability is not my thing. I have to work on it. – alison monroe Jul 28 '17 at 9:03