# Probability that exactly 12 buses will arrive within 3 hours

Let's suppose there are two buses $$A$$ and $$B$$. They draw up at the bus stop under the Poisson distribution with intensities $$3$$ and $$5$$ times per hour. (a) What's the expected length of time after the $$15$$th bus will arrive?, (b) What's the probability that exactly $$12$$ buses will arrive within $$3$$ hours?

If bus $$A$$ arrives on average $$3$$ times per hour and bus $$B$$ arrives on average $$5$$ times per hour, then a bus comes to the bus stop on average $$8$$ times per hour.

Poisson distribution: $$P(N(t)=j)=\frac{(\lambda t)^j}{j!}e^{-\lambda t}$$.

(b) $$P(N(t)=j)=\frac{(8*3)^{12}}{12!}e^{-8*3}\approx 0,00288$$

(a) $$P(N(t)=j)=\frac{8t^{15}}{15!}e^{-8t}$$, but I don't think it's a proper approach. I suppose the proper answer is $$\frac{15}{8} \approx 1,9$$, but I'm not sure how to show it.

I'll be thankful for any tips and help.

• Are you familiar with the relationship between Poisson processes and Gamma waiting times? If not, review some of the posts at stats.stackexchange.com/…. – whuber Jan 11 at 17:53
• The arrival rate per 3 hrs is $\lambda = 24.$ So why isn't the probability of exactly 12, found by using the Poisson PDF: 0.00288? – BruceET Jan 12 at 4:58