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Thread: Sets and Probability

  1. #1
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    Sets and Probability

    Prove:

    $\displaystyle

    P\{ (A \cup B) \cap C\} = P\{ A \cap C\} + P\{ B \cap C\} - P\{ A \cap B \cap C\}


    $
    Last edited by kid funky fried; Aug 30th 2008 at 09:56 PM. Reason: add content
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  2. #2
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    The expression $\displaystyle P\left\{ {A \cap B} \right\} \cap C$ has no meaning.
    It like adding a matrix to a number: meaningless!
    So what are you working out?
    What is the problem?
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by kid funky fried View Post
    prove:

    $\displaystyle
    P\{ (A \cap B)\} {\color{red}\cap C} = P\{ A \cap C\} + P\{ B \cap C\} - P\{ A \cap B \cap C\}
    $

    So I did this:

    $\displaystyle
    \begin{gathered}
    P(A \cup B) = P(A) + P(B) - P(A \cap B)theorum \hfill \\
    by substitution, \hfill \\
    (P(A) + P(B) - P(A \cap B)) \cap C = \hfill \\
    P(A \cap C) + P(B \cap C) - P(A \cap B \cap C) \hfill \\
    \end{gathered}
    $
    If it is no included inside of the probability term, $\displaystyle P\left\{(A\cap B)\right\}\cap C$ has no meaning, as Plato mentioned.

    However, if it was $\displaystyle P\left\{(A\cap B)\cap C \right\}$, then we can work with it...maybe.

    From what I'm understanding, I think you are looking for $\displaystyle P\left\{(A\cup B)\cap C \right\}$

    This is the same as $\displaystyle P\left\{(A\cap C)\cup (B\cap C) \right\}={\color{red}P(A\cap C)+P(B\cap C)-P(A\cap B\cap C)}$

    I hope this helps

    --Chris
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  4. #4
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    As I have pointed out to you, $\displaystyle P\left( {A \cap B} \right) \cap C$ is totally meaning less.
    First $\displaystyle P\left( {A \cap B} \right)$ is a number while $\displaystyle C$ is a set in a probability space.
    We do not intersect numbers with sets: it is meaningless.
    I think that I have done this for you, did I not?
    You cannot pove what you have written!
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  5. #5
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    revsion

    Hopefully, my revision has meaning.

    I just want to make sure this is legit.

    $\displaystyle
    P(A \cap B) \cap P(C) = P(A \cap B \cap C)
    $

    Kid
    Last edited by kid funky fried; Aug 31st 2008 at 08:32 AM. Reason: error
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  6. #6
    Moo
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    Hi,
    Quote Originally Posted by kid funky fried View Post
    Hopefully, my revision has meaning.

    I just want to make sure this is legit.

    $\displaystyle
    P(A \cap B) \cap P(C) = P(A \cap B \cap C)
    $

    Kid
    Sorry to say, but this expression is not good either...
    Again, $\displaystyle \cap$ is usually used for sets, whereas P(robabilities) are values, numbers...
    You can't weigh apples and pears together when you buy fruits ^^
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  7. #7
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    Thx

    Thanks Moo, I think I understand now.
    I will revise-again.

    Kid
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  8. #8
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    Have I seen the light?

    $\displaystyle

    \begin{gathered}
    P\{ (A \cup B) \cap C\} = P(A \cap C) + P(B \cap C) - P(A \cap B \cap C) \hfill \\
    Dist. \hfill \\
    P\{ (A \cup B) \cap C\} = P\{ (A \cap C) \cup (B \cap C)\} \hfill \\
    Theor. \hfill \\
    P(A \cup B) = P(A) + P(B) - P(A \cap B) \hfill \\
    I\_got, \hfill \\
    P(A \cap C) + P(B \cap C) - P\{ (A \cap C) \cap (B \cap C)\} = \hfill \\
    P(A \cap C) + P(B \cap C) - P\{ (A \cap B \cap C). \hfill \\
    \end{gathered}

    $


    Now I understand Plato's disdain!
    (I think)
    Last edited by kid funky fried; Aug 31st 2008 at 08:43 AM.
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  9. #9
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    Quote Originally Posted by Moo View Post
    Hi,

    Sorry to say, but this expression is not good either...
    Again, $\displaystyle \cap$ is usually used for sets, whereas P(robabilities) are values, numbers...
    You can't weigh apples and pears together when you buy fruits ^^

    Moo, Do you think I could weigh apples and pears together if they are the same price?
    ( I do this sometimes, also!)lol

    Thx for ur help. I appreciate it. I (obviously)did not realize that the intersection was the issue at hand.
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