Poisson buses at the stop

Question

Buses arrive at a stop according to a Poisson process with rate $\lambda = 4$ per hour.

(a) What is the probability that exactly $2$ buses arrive in the next $30$ minutes? (b) What is the probability that no bus arrives in the next $15$ minutes?

Accepted Answer

(a) The count N in 30 minutes is Poisson with mean \mu = 4 	imes 	frac12 = 2: P(N = 2) = e^{-2}\frac{2^2}{2!} = 2e^{-2} \approx 0.271. (b) The count in 15 minutes has mean 4 	imes 	frac14 = 1: P(N = 0) = e^{-1} \approx 0.368. Equivalently: the waiting time to the next bus is exponential with rate 4/hour, so P(	ext{wait} 	frac14 	ext{ hr}) = e^{-4 \cdot 1/4} = e^{-1}, counts and waiting times are two views of the same process. Two facts worth quoting in a follow-up: the exponential wait is memoryless (having already waited 10 minutes doesn't shorten the remaining wait), and disjoint int…

Poisson buses at the stop

Your answer

More Probability questions