How do I apply the poisson process to find out whether the deer will get hit?

Question

Traffic on a road follows a Poisson process with rate $6$ cars per minute. A deer runs out of the woods and tries to cross the road. If there is a car passing in the next five seconds, then there will be a collision. I know that every minute, there are $6$ cars so every second, there is $0.1$ car.

(a) Find the probability of a collision.

For this part, I denoted another random variable called $X \sim Poisson(0.1*5)$ because I thought to handle part (a), I needed another random variable that I could relate to by using Poisson distribution. Thus,

$P(\text{the probability of a collision}) = P(\text{there is at least a car every } 5 \text{ seconds})$, i.e $P(X \geq 1) = 1 - P(X=0)$.

(b) What is the chance of a collision if the deer only needs two seconds to cross the road?

I don't have a clear approach to this part but I think I still need a random variable called

$Y \sim Poisson(0.1*2)$

which denotes the times a deer can cross the street. After this, I am stuck. How do I find out the probability that a deer will get hit given that if every $5$ seconds there is a car, and a deer only needs $2$ seconds to cross the road?

Thank you!

Answer 1

The parts (a) and (b) are the same. Recall, that for a Poisson process $(N_{t})_{t \geq 0}$, the number of events (in your case the number of cars passing) in any given time interval is Poisson distributed, with parameter proportional to the length of the interval.

You were indeed able to work out the distribution for the number of cars in any given 5 second interval for part (a), yet your wording hints that you do not fully grasp what is going on:

P(the probability of a collision) = P(there is at least a car every 5 seconds)

I would recommend thinking about it more rigorously. Assume the deer enters the road at time $t_{0}$ (in seconds scale). Then you could write:

\begin{align} P(\text{The deer entering at $t_{0}$ collides with a car}) &= P(\text{At least one car arrives in time interval}\ [t_{0}, t_{0} + 5]) \\ &= P(N_{t_{0} + 5} - N_{t_{0}} \geq 1) \\ &= P(X \geq 1), \end{align} where $X$ denotes the random variable you defined in question.

Now think, what in the above argument changes in part (b)?