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Lets say there are $N$ bus-stops and there could be either 0 or 1 bus at any station at any given time. The probability of a bus being on the station is $P_{rel}$. What can one say about the total buses on $N$ stations? Lets assume there are total 1000 observations.

PS : I am not sure if the problem is well-posed.

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  • $\begingroup$ I believe it would help to know much time each bus spends on a stop compared to how much time it spends driving between them. $\endgroup$
    – Arthur
    Oct 3, 2013 at 11:44
  • $\begingroup$ Lets assume its a discrete event system. I appear magically on all bus stations every 10 minutes at the same time and note down if bus is present of absent. $\endgroup$
    – Dilawar
    Oct 3, 2013 at 11:48
  • $\begingroup$ Oh, I'm sorry, I was wrong. I thought we were estimating the total number of busses on the road. Then assuming that each bus stop's variable is independant of the others', the sum over all stops will be binomial, and the expectation value is $N\cdot P_{rel}$. $\endgroup$
    – Arthur
    Oct 3, 2013 at 12:07
  • $\begingroup$ The 1000 observations are irrelevant, the problem is well-posed, and I am not sure if the OP knows why the result is what is stated in a post below. $\endgroup$
    – Did
    Oct 3, 2013 at 13:11

1 Answer 1

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I believe the expected number of buses at bus stops is $N\cdot P_{rel}$.

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  • $\begingroup$ I also thought so. But the solution seems to be on the line of binomial probability formula. This is an old 'math club' problem and either the problem or solution may not be correct but I am not sure $\endgroup$
    – Dilawar
    Oct 3, 2013 at 12:00
  • $\begingroup$ That is binomial. See comment on the original question. $\endgroup$ Oct 3, 2013 at 12:30
  • $\begingroup$ Binomial assumes independence but independence is neither given as a hypothesis nor necessary to reach the result. By the way, @Dilawar, do you know how to prove the assertion in the answer? $\endgroup$
    – Did
    Oct 3, 2013 at 13:09
  • $\begingroup$ Yes, I agree, independence was an assumption. $\endgroup$ Oct 3, 2013 at 13:11
  • $\begingroup$ @Did Yes, I worked it out. This was very helpful stat.berkeley.edu/~bradluen/stat2/lecture19.pdf $\endgroup$
    – Dilawar
    Oct 3, 2013 at 16:05

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