# Poisson process arrivals of buses

I would like to check if my calculations are correct: We have two buses, arrivals of bus (A) form a Poisson process of rate $1$ bus per hour and arrivals of bus (B) form an indep. Poisson process of rate $7$ buses per hour.

(1) Probability that exactly 3 buses of type (B) pass while I still wait for bus (A)

Well bus (B) has rate 7, bus (A) has rate 1, therefore $(\frac{7}{7+1})^3$

(2) Probability that exactly 5 buses of typ (A) pass in 1 hour.

My approach was to calculate $P(B=k|A+B=5)$ for $k\in[0,5]$ and add everything together, so I get $\binom 5 0 (\frac{7}{7+1})^0(\frac{7}{7+1})^5+...+\binom 5 5(\frac{7}{7+1})^5(\frac{7}{7+1})^0$, correct?

• For 1) you have to calculate the expected waiting time for a bus A. It is longer then 1/2 an hour as say by chance bus A comes after 10 min, then again after 1 hour 50 min. You are more likly to arrive at the bus stop in the 1 hour 50 min gap. In 2) nothing depends on bus B so it is P(k<6) for bus A Feb 10, 2014 at 12:26
• Thanks for your comment, would you mind to write it down formally as an answer? So I can give you the bounty. Feb 10, 2014 at 14:10
• ;) I'll try and find the answer for the expected waiting time. Feb 10, 2014 at 14:41
• The expected waiting time for bus A is 1 hour given a Poisson distribution. I.e twice what one may have naively assumed. Feb 11, 2014 at 19:47

• If you have a Poisson process then the interval arrival times are from an exponential distribution. You can of course have more general distribution with mean $\mu$ and variance $\sigma^2$ in which case the expected waiting time for a bus (given that you arrive at random) is given by $\frac{1}{2}\mu(1+(\frac{\sigma}{\mu})^2)$. So the greater the variance the longer you have to wait. This is because you are more likely to arrive in a big gap between buses. The maths is the same as for the "Inspection Paradox" of which google will tell you more. Feb 17, 2014 at 15:00