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Math Help - Poisson Probability

  1. #1
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    Poisson Probability

    The number of bugs in a computer is a Poisson random variable with a rate parameter of 3 bugs a year. When a bug occurs, a technician A is called to fix it with probability 0.5, a technician B is called to fix it with probability 0.2 and a technician C with probability 0.1. The probability that C won't be called to fix any bug during a year is:

    a. e^-2.7 b. e^-0.3 c. 1-e^-2.7 d. 1-e^-0.3

    cheers !!
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  2. #2
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    Start by finding the probability that c is not called, given that there are n bugs.

    P(C' | n) = 0.9^n

    Now, the required probability is:

    P(n=0) * 1
    +P(n=1)*0.9
    +P(n=2)*0.9^2
    ...

    We are told the distribution of N. P(N=n) = \frac{e^{-3} 3^n}{n!}

    So the sum you want to do is
    \sum  \frac{e^{-3} 3^n *0.9^n}{n!}

    \sum  \frac{e^{-3} (3*0.9)^n }{n!}

    \sum e^{-0.3} \frac{e^{-2.7} (2.7)^n }{n!}

    e^{-0.3} \sum \frac{e^{-2.7} (2.7)^n }{n!}

    You are supposed to recognise the above as
    e^{-0.3} * P(Y< \infty)
    where Y is a poisson(2.7) variable.

    So the answer is e^{-0.3}
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  3. #3
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    Talking about Simeon-Denis Poisson , today is the birthday of him (229 years old ) .
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  4. #4
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    thats a different kettle of fish entirely ^^
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