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Math Help - [SOLVED] Joint /Conditional Probability Distributions

  1. #1
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    [SOLVED] Joint /Conditional Probability Distributions

    A candy company distributes boxes of chococlates with a mixture of creams, toffees, and cordials. Suppose that the weight of each box is 1 kg, but the individual weights of the creams, toffees and cordials vary from box to box. For a randomly selected box, let X and Y represent the weights of the creams and the toffees, respectively, and suppose that the joint density function of these var is

    f(x,y) = 24xy , 0<= x<=1, 0 <=y<=1, x+y<=1
    0 elsewhere

    a) find the prob that in a given box the cordials acount for more than 1/2 of the weight.
    b) find the mariginal density of the weight of the creams.
    c) Find the prob that the weight of the toffees in a box is less than 1/8 of a kg if it is known that creams constitute 3/4 of the weight.

    ---------------

    a) I integrate 24xy from 0 to 1/2 to dy. and my answer= 3/8
    however, the correct ans = 1/16
    where have i gone wrong ?

    b) g(x) = 12xy^2
    but the correct answer is 12x(1-x)^2.
    why?

    c) P(y<1/8 | x=3/4)
    = f(x,y) / g(x)
    and i got 2 but the correct answer is 1/4.

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  2. #2
    MHF Contributor matheagle's Avatar
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    3.41 from walpole?

    you cannot ignore x+y<1 and you need to draw the region (triangle)
    also, a probability cannot exceed 1.
    Last edited by matheagle; November 14th 2009 at 09:28 PM.
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  3. #3
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    Oh yes ! this is qn 3.41

    Sorry i don't get what you mean. i understand that prb cannot exceed 1 and that x+y <1
    which is given. so what should i do ?
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  4. #4
    MHF Contributor matheagle's Avatar
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    since x+y<1, these are DEPENDENT rvs

    P(X+Y\le 1/2)= \int_0^{1/2}\int_0^{1/2-y}24xy dxdy=1/16

    f_X(x)= \int_0^{1-x}24xy dy=12x(1-x)^2 where 0<x<1 which is a BETA RV

    f(y|x)= {f(x,y)\over f_X(x)}={2y\over (1-x)^2}

    so P(Y<1/8|X=3/4)=32\int_0^{1/8}y dy=1/4

    you should be happy that I teach out of walpole
    in fact, look at page xxiii, I'm the first reviewer.
    Last edited by matheagle; November 14th 2009 at 10:56 PM.
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  5. #5
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    Quote Originally Posted by matheagle View Post
    since x+y<1, these are DEPENDENT rvs

    P(X+Y\le 1/2)= \int_0^{1/2}\int_0^{1/2-y}24xy dxdy=1/16

    f_X(x)= \int_0^{1-x}24xy dy=12x(1-x)^2 where 0<x<1 which is a BETA RV

    f(y|x)= {f(x,y)\over f_X(x)}={2y\over (1-x)^2}

    so P(Y<1/8|X=3/4)=32\int_0^{1/8}y dy=1/4

    you should be happy that I teach out of walpole
    in fact, look at page xxiii, I'm the first reviewer.
    but why is it for :
    P(Y<1/8|X=3/4)=32\int_0^{1/8}y dy=1/4

    I can't just substitute 1/8 and 3/4 into the equation
    {2y\over (1-x)^2}
    which i got 4.
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by hazel View Post
    but why is it for :
    P(Y<1/8|X=3/4)=32\int_0^{1/8}y dy=1/4

    I can't just substitute 1/8 and 3/4 into the equation
    {2y\over (1-x)^2}
    which i got 4.
    YOU substitute the X=3/4 and you integrate Y from -infinity to 1/8
    note that all these densities are 0 when the variables are negative
    so you integrate from 0 to 1/8.

    IN GENERAL P(Y<1/8|X=3/4)=\int_{-\infty}^{1/8}f(y|X=3/4) dy

    Just like a marginal density, you integrate it over the appropriate region.


    It's 1 am, I'm tired
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  7. #7
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    ok, let me read thru again.

    thanks.
    Last edited by mr fantastic; November 15th 2009 at 02:44 AM. Reason: Deleted gibberish.
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  8. #8
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    i got 27/512.

    could you show the step by step workings for
    <br />
P(Y<1/8|X=3/4)=\int_{-\infty}^{1/8}f(y|X=3/4) dy<br />
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  9. #9
    MHF Contributor matheagle's Avatar
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    It's very basic calc 1

    f(y|X=x)=2y/(1-x)^2

    So f(y|X=3/4)=32y where 0<y<1/4

    AND not for 0<y<1, that's not a valid density

    This is why I told you to draw the region.
    y>0 and x+y<1, with x=3/4 that means that y<1/4 now.

    Thus P(Y<1/8|X=3/4)=32\int_0^{1/8} y dy =16(1/8)^2=16/64=1/4
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