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Four yellow buses and two grey buses that could be in any order (with equal probability) are traveling together with the probability of a delay of at least $\mathit{t}$ seconds between any two consecutive buses being $\mathit{P((t,\infty)) = e^{-t} }$ $\forall$ $\mathit{t \ge 0}$ , independent of all other delays. What is the expected delay between grey buses?

I know that this problem deals with total expectation, so I will have to have several random variables, such as G for the grey buses and Y for the yellow, and then use the fact that I know the probability of the delay between any two buses to find the delay between the two grey buses.

Really, I'm not sure how to set this up so that it works. Is the distribution between any two buses an exponential RV? If that were the case, could we easily find conditional distributions and expectations to work with?

Any help is appreciated!

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Hints:

  • What is the expected length of a gap between consecutive buses? This is indeed an exponential random variable

  • What is the expected number of gaps between grey buses? You might find this easier if you consider the expected number of yellow buses between the two grey buses and perhaps the probability a given yellow bus comes between the grey buses, or you could just look at all the possible patterns

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Hint:

  1. There can be $0,1,2,3,4$ different yellow buses between two different grey buses, try to find probability of each case.

  2. Identify the distribution, then you can easily find expectation between two consecutive buses. All the delays are independent, so you can find answer.

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  • $\begingroup$ So, correct me if I'm wrong, but since the buses are called out in any order with equal probability, then the probability that 0 yellow are between the grey is the same as 2, etc., which would be 1/6? From there it would be easy to find the expected values and so forth. Sometimes these question seem like they're too easy (and then I'm usually doing something wrong!) $\endgroup$ – user641300 Feb 4 at 3:05
  • $\begingroup$ @user641300 It should be $\tfrac{240}{720}$ for $0$ yellow. Try to think about it for a while, 6 different things are assigned to 6 different positions. $\endgroup$ – Flashhh Feb 4 at 3:13
  • $\begingroup$ Sorry again but this is using combinations right? Where P(Y=0) is [(2 choose 1)(2 choose 1)(5 choose 1)(3 choose 1)(4 choose 1)]/ the entire possibilities. I'm having trouble understanding where you got your numbers, even after consulting my textbook and other sources. $\endgroup$ – user641300 Feb 4 at 4:01
  • $\begingroup$ [$5\choose1$$4!$$2\choose1$]/$6!$ $\endgroup$ – Flashhh Feb 4 at 4:13
  • $\begingroup$ The final answer is $\tfrac{7}{3}$, you can check it yourself. $\endgroup$ – Flashhh Feb 4 at 4:15

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