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Math Help - some poisson probabilities

  1. #1
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    some poisson probabilities

    Hi! I would greatly appreciate if someone could look these over for me:

    Suppose Oils spills off the coast of BC in any given year has a Poisson Distribution with mean 3.

    A) what is the chance that there will be exactly 2 spills in 2010?

    Answer:

    P(x=2) = (e^-3)(3^2)(1/2!) = 0.22404

    B) what is the chance that there will be at least one spill over the next three years?

    Answer:

    our mean rate of 3 is for one year. For three years it is 9.

    P(X>=1) = 1 - P(x=0)
    = 1 - (9^0)*(e^-9)*(1/0!)
    = 1 - e^-9
    = 0.999876
    C) if there are 3 oil spills over the next two years, what is the chance that exactly two will occur in 2010?

    Answer:

    Let X1 = # of spills in 2010, X2 = # of spills in 2011.

    Then X1 ~ Poisson(3), X2~Poisson(3)


    P(X1 = 2 | X1 + X2 = 3) ~ Binomial (3,p) where p = mu1 / mu1 + mu2

    = (3 choose 2) (1/2)^2 (1/2)^1

    = 0.375



    Let X be the number of defects on a silicon wafer for computer chips manufactured by a high-tech company. Suppose defects occur randomly and uniformly over the surface of the wafer at a mean rate of 2 defects per wafer so that X follows a Poisson Distribution with mean 2. X ~ Poisson (2)

    A) Find the probability that a wafer has at least two defects

    Answer:

    P(X >= 2) = 1 - P(X=0) - P(X=1)
    = 0.59399415

    B) Find the probability that 3 wafers have each 1 defect.

    Answer:

    I found the probability that one wafer has one defect, and cubed it due to independence.

    I got 0.0198300174

    C) Find the probability that if we examine 3 wafers, the TOTAL number of defects is exactly one.

    Im thinking the following:

    find the probability of exactly one defect using poisson distribution

    then use this probability as "p" in

    3 choose 1 * p^1 * (1-p)^2

    Thanks!
    Last edited by jamesmartinn; January 15th 2010 at 05:49 PM.
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  2. #2
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    Problems A and B are correct. That is, the first two problems you listed. I haven't looked past those.

    Good job.
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  3. #3
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    Quote Originally Posted by jamesmartinn View Post
    Hi! I would greatly appreciate if someone could look these over for me:

    Suppose Oils spills off the coast of BC in any given year has a Poisson Distribution with mean 3.

    A) what is the chance that there will be exactly 2 spills in 2010?

    Answer:

    P(x=2) = (e^-3)(3^2)(1/2!) = 0.22404

    B) what is the chance that there will be at least one spill over the next three years?

    Answer:

    our mean rate of 3 is for one year. For three years it is 9.

    P(X>=1) = 1 - P(x=0)
    = 1 - (9^0)*(e^-9)*(1/0!)
    = 1 - e^-9
    = 0.999876
    C) if there are 3 oil spills over the next two years, what is the chance that exactly two will occur in 2010?

    Answer:

    Let X1 = # of spills in 2010, X2 = # of spills in 2011.

    Then X1 ~ Poisson(3), X2~Poisson(3)


    P(X1 = 2 | X1 + X2 = 3) ~ Binomial (3,p) where p = mu1 / mu1 + mu2

    = (3 choose 2) (1/2)^2 (1/2)^1

    = 0.375



    Let X be the number of defects on a silicon wafer for computer chips manufactured by a high-tech company. Suppose defects occur randomly and uniformly over the surface of the wafer at a mean rate of 2 defects per wafer so that X follows a Poisson Distribution with mean 2. X ~ Poisson (2)

    A) Find the probability that a wafer has at least two defects

    Answer:

    P(X >= 2) = 1 - P(X=0) - P(X=1)
    = 0.59399415

    B) Find the probability that 3 wafers have each 1 defect.

    Answer:

    I found the probability that one wafer has one defect, and cubed it due to independence.

    I got 0.0198300174

    C) Find the probability that if we examine 3 wafers, the TOTAL number of defects is exactly one.

    Im thinking the following:

    find the probability of exactly one defect using poisson distribution

    then use this probability as "p" in

    3 choose 1 * p^1 * (1-p)^2

    Thanks!
    It all looks OK method wise (I haven't checked the arithmetic) except for 2(c), which I would do by calculating Pr(X = 1) = a and Pr(X = 0) = b and then calculating 3 a b^2 since you want Pr(0, 0, 1) + Pr(1, 0, 0) + Pr(0, 1, 0).
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  4. #4
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    A big thanks to you guys for the feedback/suggestions, I really appreciate your taking the time to read this stuff over. (I know I posted quite a few problems)

    Cheers!
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