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Thread: Mixed Poisson Distribution Help

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    Question Mixed Poisson Distribution Help




    Question:
    The number of accidents follows a Poisson distribution with mean 12. Each accident generates 1, 2 or 3 claimants with probabilities 1/6,1/3,1/2, respectively.
    Calculate the probability that there are at most 3 claimants.

    Attempt:
    E(N) = 12[\frac{1}{6}+2*\frac{1}{3}+3*\frac{1}{2}] = 12*\frac{7}{3}=28

    Step 2:
    P(N≤3)= \sum_{x=0}^3 \frac{28^x e^{-28}}{x!} = 2.8*10^{-9}

    That's the answer I got, but the number seems to be too small in value. Did I do anything wrong? Thank you!

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    Re: Mixed Poisson Distribution Help


    You made this mistake: The distribution of Total Claimants is NOT Poisson with mean=28.

    I'm not sure of a nice solution, but you can get T = Total Claimants <= 3 by:
    a) 0 accidents
    b) 1 accident
    c) 2 accidents: 1 claimant-2 claimants, or 2 claimants-1 claimant
    d) 3 accidents: 1 claimant-1claimaint-1 claimant

    We then calculate the probability of the above scenarios:
    a) P[0 accidents] = (12^0/0!) e^-12 = 6.1E-6
    b) P[1 accident] = (12^1/1!) e^-12 = 7.4E-5
    c) P[2 accidents] * P[1 claimant, and then 2 claimants] * 2 = (12^2/2!) e^-12 * (1/6) * (1/3) * 2 = 4.9E-5
    d) P[3 accidents] * P[1 claimant, then 1 claimant, then 1 claimant] = (12^3/3!) e^-12 * (1/6)^3 = 8.2E-6

    We then add up the above probabilities to finally determine P = 1.4E-4
    All things are known because we want to believe in them.

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