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Math Help - Poisson Distribution, Independent Events

  1. #1
    Junior Member Hasan1's Avatar
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    Poisson Distribution, Independent Events

    1. Suppose that X and Y are independent Poisson random variables with parameters 1 and 2 respectively, Find:

    a) P(X = 1 and Y = 2);
    b) P((X + Y) / 2) >= 1);
    c) P(X = 1 | (X +Y)/2 = 2)[/quote]I found a) by multiplying the individual probabilities that X=1 and Y=2 because X and Y are independant


    How can I use the independance of X and Y to calculate b) and c)?


    2). Radioactive substances emit alpha particles/ The number of such particles reaching a counter over a given time period follows the Poisson distribuiton. Suppose two substances emit alpha particles independantly of each other. The first substance gives out alpha particles according to the Poission(3.87) distribution while the second substance emits alpha particles which reach the counter according to the Poisson(5.41) distribution. Find the chance that the counter is hit by at most 4 particles[/quote]Again I need to know how to use the independence of each substances alpha particle emission frequencies to calculate this probability


    I was thinking along the lines of:

    X: denotes number of alpha particles from substance 1 reaching the counter
    Y: denotes number of alpha particles from substance 2 reaching the counter

    P(X + Y <= 4) = P(X <= 2) + P(Y <=2)

    Thanks.
    Last edited by mr fantastic; March 23rd 2010 at 09:25 PM. Reason: Removed questins from quote tags so that the questins can be easily quoted.
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  2. #2
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    Quote Originally Posted by Hasan1 View Post
    1. Suppose that X and Y are independent Poisson random variables with parameters 1 and 2 respectively, Find:

    a) P(X = 1 and Y = 2);
    b) P((X + Y) / 2) >= 1);
    c) P(X = 1 | (X +Y)/2 = 2)
    I found a) by multiplying the individual probabilities that X=1 and Y=2 because X and Y are independant


    How can I use the independance of X and Y to calculate b) and c)?

    [snip][/quote]
    b) Calculate 1 - P( (X + Y) / 2) < 1).

    c) Apply Bayes Theorem.
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  3. #3
    Junior Member Hasan1's Avatar
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    Yes but how do I actually calculate 1 - P((X+Y)/2 < 1) ?
    The answer for b is 0.8008, but I don't know how they get that.

    I tried breaking that into

    1 - P(X+Y < 2)
    1 - P(X < 1) + P(Y < 1) (which gave me the wrong answer)
    or 1 - P(X < 2) + P(Y < 2) ( which also gave me the wrong answer)

    since X ~ Poisson(1) shouldn't P(X<1) be (e^-1)(1^0)/(0!) and simiilarly
    since Y~ Possion(2) -> P(Y<1) (e^-2)(2^0)/(0!) ?
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    Quote Originally Posted by Hasan1 View Post
    Yes but how do I actually calculate 1 - P((X+Y)/2 < 1) ?
    The answer for b is 0.8008, but I don't know how they get that.

    [snip]
    What pairs (X, Y) satisfy (X+Y)/2 < 1 ....?
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  5. #5
    Junior Member Hasan1's Avatar
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    Those would be:

    (X,Y) = (0,0)
    (X,Y) = (0,1)
    (X,Y) = (1,0)

    So does that make

    P((X+Y)/2< 1)

    equivalent to:

    P(X=0 and Y=0) + P(X=1 and Y=0) + P(X=0 and Y=1) ?
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  6. #6
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    Quote Originally Posted by Hasan1 View Post
    Those would be:

    (X,Y) = (0,0)
    (X,Y) = (0,1)
    (X,Y) = (1,0)

    So does that make

    P((X+Y)/2< 1)

    equivalent to:

    P(X=0 and Y=0) + P(X=1 and Y=0) + P(X=0 and Y=1) ?
    Yes.
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