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Homework Help: Probability of rate x

  1. Aug 6, 2010 #1
    I want to find the probability that I will have to wait 2 seconds on average for event A to happen, given that the measured average wait time is 2.6 seconds.

    Attempt at a solution:

    The probability that event A will occur at any given second is 1/2.6. Then the probability of waiting 2 seconds on the average is (1/2.6 * 2) = 76.9%.
  2. jcsd
  3. Aug 6, 2010 #2
    Maybe I'm missing something. If the MEASURED average wait time is 2.6 seconds, then one should probably concede that 2.6 second is the actual average wait time, unless of course the measurements are biased or plain wrong.

    What is the probability that the average wait time will change to 2 seconds?

    That would require a fundamental change in the underlying system, which cannot be guaged by the information provided.

    Then again maybe I'm missing something here?
  4. Aug 6, 2010 #3
    Yes. I want to find the probability that say the next 6 occurances will happen on average 2 seconds apart.
  5. Aug 6, 2010 #4
    Could the formula be P(waiting 2 seconds)^6?
  6. Aug 6, 2010 #5
    To do this problem, we need more information. Is it a Poisson process? That's what I first thought of when I read the problem, but you don't mention that anywhere. Then you said that the probability of it occurring in any particular second was 1/2.6, which further makes it sound like a Poisson process. But then you say the "measured" average wait time, which sounds like a sampling question, where you're looking for error in the measurement. But there's no discussion of the variance of the sample, so we can't do much with that.

    Anyway, if it's a Poisson process with mean interarrival time 2.6s, then the underlying PDF for interarrival times is:

    [tex] f(x) = 2.6e^{-2.6x} (x > 0) [/tex]
  7. Aug 8, 2010 #6
    Thank you. So that would give me the probability that the wait time is greater than 0, right? So then maybe I can use the binomial dist formula to answer my question?
  8. Aug 8, 2010 #7
    Well, no. That's the probability density function for the interarrival wait times if the process is Poisson. A particular wait time is always greater than zero, right?

    It might be helpful if you gave us some context for this problem (ie, more of the problem if there is any, what class it's for).
  9. Aug 11, 2010 #8
    I think I found the answer to that question: e^-mean interval * desired times for event x to happen^mean interval / desired times for event A to happen!
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