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Science Forum

  1. 18 Apr '15 10:36 / 5 edits
    This seems to me like a pretty simple maths problem and yet, somehow, this one has so far got me frustratingly completely stumped:

    Suppose we observed how often a completely random event happened in an arbitrary very 'large' number of seconds, say, a trillion seconds.
    So, using conventional probability terminology, here we are using that observed trillion seconds as the sample space and one second as our unit measure.
    There is no pattern to when the events occurs so, in any give one second period, there is a non-zero probability of no event occurring and there is a non-zero probability of the event occurring just once and there is a non-zero probability of the event occurring twice and so on for infinitum although lets say the event can only occur a non-infinite number of times in each second but with no definable finite upper limit.

    Suppose, on average, we observe the event occurring with some average observed frequency F of v events per second s. So we know the approximate value of F and:

    F = v/s.

    Suppose F > 0 but F may or may not be greater than 1.
    Now suppose we completely randomly pick just one of those one second periods.
    Even if F >> 1, there must still be a small but non-zero probability that there will be no event in that second.
    So what is the correct algebraic equation (or a reasonable numerical approach if no such algebraic equation exists ) for at least a good approximation of the probability of no event occurring in our randomly chosen second?
  2. 18 Apr '15 12:08
    Start here:
    http://en.wikipedia.org/wiki/Poisson_distribution
  3. 18 Apr '15 12:14 / 3 edits
    Originally posted by twhitehead
    Start here:
    http://en.wikipedia.org/wiki/Poisson_distribution
    Thanks for that That looks like the sort of thing I want.
    I will now, as soon as I get the time, slowly, carefully and systematically go through that all in due course and, hopefully, understand it all.
  4. 20 Apr '15 10:05
    Originally posted by twhitehead
    Start here:
    http://en.wikipedia.org/wiki/Poisson_distribution
    From that link: The equation in the link simplifies for the special case in my OP problem because I am only interested here in the probability of there being NO event in that second. So, simplifying and adapting the equation in the link to my OP problem:”

    let:
    P(X=0) = probability of a second interval having no event i.e. having 0 events.
    e = Euler's number = 2.71828... (this is just the base of the natural logarithm )

    then we simply have (for 0 events) :

    P(X=0) = e ^ ( – F )

    and, when I tried out a few F values, I always got results that intuitive seems very roughly about right so I am pretty sure I have got that right.

    But, if I DID want to know the probability of there being a specific non-zero number of events in that second …....

    For 1 event;
    P(X=1) = F * e ^ ( – F )

    For 2 events;
    P(X=2) = ( (F^2) * e ^ ( – F ) ) / 2

    For 3 events;
    P(X=3) = ( (F^3) * e ^ ( – F ) ) / 6

    For 4 events;
    P(X=4) = ( (F^4) * e ^ ( – F ) ) / 24

    ...etc.

    Simple!
  5. 20 Apr '15 11:48
    Originally posted by humy
    From that link: The equation in the link simplifies for the special case in my OP problem because I am only interested here in the probability of there being NO event in that second. So, simplifying and adapting the equation in the link to my OP problem:”

    let:
    P(X=0) = probability of a second interval having no event i.e. having 0 events.
    e = Euler's numbe ...[text shortened]... e ^ ( – F ) ) / 6

    For 4 events;
    P(X=4) = ( (F^4) * e ^ ( – F ) ) / 24

    ...etc.

    Simple!
    Or for n events... [It's always good to be general ]

    For n events;
    P(X=n) = ( (F^n) * e ^ ( – F ) ) / n!
  6. 20 Apr '15 19:08
    Originally posted by googlefudge
    Or for n events... [It's always good to be general ]

    For n events;
    P(X=n) = ( (F^n) * e ^ ( – F ) ) / n!
    yes, and that was the basic form of the general equation given in the link.