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Math Help - [SOLVED] (Simple?) Discrete joint function help

  1. #1
    Lok
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    [SOLVED] (Simple?) Discrete joint function help

    Hello, I have small problem with this problem, here it is

    Question :
    We pick 3 persons at random from a group of 13 in which exactly 5 have blue eyes.

    Let Xi = { 1 if the i ème person has blue eyes
    0 if not

    What is the joint mass function of X1 and X2?

    My actual attempt to a solution
    I know that when X1=0 that P(X1=0)=8/13, P(X2=0)=7/12 and P(X2=1)=5/12

    and when X1=1 that P(X1=1)=5/13, P(X2=0)=8/12 and P(X2=1)=4/12

    I also know that the joint mass function is p(x1, x2) = P(X1 = x1 and X2 = x2)

    My question is : so do I write P(X1=0 and X2=0) = P(X1=0) * P(X2=0) = 8/13*7/12
    or do I simply write P(X1=0 and X2=0) = (8/13,7/12) or something else???

    Thanks a lot.
    Last edited by Lok; February 15th 2009 at 09:42 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    Incorrect
    P(X2=0)=7/12 is false
    P(X2=0|X1=0)=7/12
    while P(X2=0)=8/13.
    It doesn't matter when you select.
    If you don't have any info on the first selection the probability is the same.

    The probability that the first card from a deck is the ace of spades is 1/52.
    The probability that the tenth card is the aces of spades is also 1/52.
    If you give me some info on the first nine cards, then the probability changes.

    So P(X_1=0, X_2=0)=P(X_2=0|X_1=0)P(X_1=0)= (7/12)(8/13)
    In general
    P(X_1=a, X_2=b)=P(X_2=b|X_1=a)P(X_1=a).
    Last edited by matheagle; February 15th 2009 at 10:13 PM.
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  3. #3
    MHF Contributor matheagle's Avatar
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    Likewise if we have an urn with 3 red and 4 blue balls and we select without replacement,
    the probability the second ball is red is STILL 3/7.
    It's the same probability as the first selection, since we have no info on the first pick.
    P(second ball is red)=P(first is red and second is red)+P(first is blue and second is red)
    Using conditional probabilities we get..
    =P(second is red|first is red)P(first is red)+
    P(second is red|first is blue)P(first is blue)
    =(2/6)(3/7)+(3/6)(4/7)
    =3/7.
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  4. #4
    Lok
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    Ok excellent so,

    P(X_1=0)=8/13 and P(X_1=1)=5/13

    P(X_2=0|X_1=0)=7/12 P(X_2=1|X_1=0)=5/12
    P(X_2=0|X_1=1)=8/12 P(X_2=1|X_1=1)=4/12

    so for my join mass function of X_1 and X_2 I have:
    P(X_1=0, X_2=0)=P(X_2=0|X_1=0)P(X_1=0) = 7/12 * 8/13 = 56/156 = 28/78 = 14/39
    P(X_1=0, X_2=1)= 5/12 * 8/13 = 40/156 = 20/78 = 10/39
    P(X_1=1, X_2=1)= 4/12 * 5/13 = 20/156 = 10/78 = 5/39
    P(X_1=1, X_2=0)= 8/12 * 5/13 = 40/156 = 10/39

    Right?

    Thanks.
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  5. #5
    MHF Contributor matheagle's Avatar
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    I'm too tired to check if you multiplied correctly.
    But, yes these fractions are correct.
    Make sure all 4 probabilities sum to one.

    You were confusing conditional probabilities with marginals.
    I need to go to bed.
    I teach two prob and stat classes in the morning.
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  6. #6
    Lok
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    Thanks a ton
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