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    I've got this question, and I can't seem to find the answer.

    You expect 1.5 falls of snow a year. Assuming a Poisson model, after how many years will the probability of no falls of snow equal 0.0111?
    Assuming a Poisson distribution:
    X \sim P(1.5)

    Therefore, applying the general formula X \sim P(\mu) = \dfrac {e^{-\mu}*\mu^r}{r!} I get this:

    X \sim P(X=r) = 0.0111 = \dfrac {e^{-1.5}*1.5^r}{r!}

    Assuming that r = the number of years.


    So we get:

    e^{-1.5}*1.5^r = 0.0111*r!

    But I just don't know how to isolate the r or put it in a way that I can solve r.

    Any help would be much appreciated.

    Cheers
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    (Original post by SergioMZ)
    Any help would be much appreciated.

    Cheers
    I think you need a different line of attack.

    Assuming Poisson, as stated, what's the probability of no snow in the first year.

    Then treat no snow each year as independent events (which is what Poisson implies), ....

    Enough?
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    Uh-huhm...

    So basically I do P(X=0), which is e^{-1.5} = 0.22313.

    How should I carry on from here?
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    Well, I've done 0.22313^x = 0.0111, and by trying (as it is discrete data, you just have to try 1, 2, 3, etc.) I've found that the solution is 3 years.

    However, given that equation, how can I solve it? I think I already did it, but I can't seem to remember.

    Cheers.
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    (Original post by SergioMZ)
    Uh-huhm...

    So basically I do P(X=0), which is e^{-1.5} = 0.22313.

    How should I carry on from here?
    So the probability of no snow in any given year is 0.22313... (I assume you're correct, I've not checked.).

    So probability of no snow in "n" years is (0.22313...)^n
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    (Original post by SergioMZ)
    Well, I've done 0.22313^x = 0.0111, and by trying (as it is discrete data, you just have to try 1, 2, 3, etc.) I've found that the solution is 3 years.

    However, given that equation, how can I solve it? I think I already did it, but I can't seem to remember.

    Cheers.
    Use logs.

    0.22313^x = 0.0111

    So,

    x \log (0.22313) = log (0.0111)

    ....
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    Thank you
 
 
 
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