Question

Consider a point on the trans-australian highway, where two old wombats live. arrivals of cars at this point follow a poisson distribution; the average rate of arrivals is 1 car per 12 seconds.

Answer 1

Given that the arrivals of cars at the trans-australian highway follow a poisson distribution with the average rate of arrivals as 1 car per 12 seconds.

This is a Poisson distribution with a mean of 1 car every 12 seconds.

Part A:

The probability of a Poisson distribution is given by:

$P(X=x)= \frac{e^{-\lambda}\lambda^x}{x!}$

Given tha one of the old wombats requires 12 seconds to cross the highway, and he starts out immediately after a car goes by.

The probability he will survive is the probability that no car passes through the point in the next 12 seconds.

$P(x=0)= \frac{e^{-1}(1)^0}{0!} =e^{-1}=0.368$

Therefore, the probability that he will survive is 0.368



Part B:

Given that at the same point of the highway there is another old wombat, slower but tougher than in the previous exercise. He requires 24 seconds to cross the road, but it takes two cars to kill him. (A single car won’t even slow him down. The arrival rate of cars is still one car per 12 seconds.)

Since the arrival rate of cars is one car per 12 seconds, then, in 24 seconds the arrival rate is 2 cars. (i.e. λ = 2)

The probability that he survives is the probability that 2 cars did not pass through the point.

$P(X\neq2)=1-P(X=2)=1- \frac{e^{-2}(2)^2}{2!} =1-0.135=0.865$