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.

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    $\begingroup$ Are you familiar with the relationship between Poisson processes and Gamma waiting times? If not, review some of the posts at stats.stackexchange.com/…. $\endgroup$ – whuber Jan 11 at 17:53
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    $\begingroup$ 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? $\endgroup$ – BruceET Jan 12 at 4:58

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