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Math Help - Sets and Probability

  1. #1
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    Sets and Probability

    Prove:

    <br /> <br />
P\{ (A \cup B) \cap C\}  = P\{ A \cap C\}  + P\{ B \cap C\}  - P\{ A \cap B \cap C\} <br /> <br /> <br />
    Last edited by kid funky fried; August 30th 2008 at 09:56 PM. Reason: add content
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  2. #2
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    The expression 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:

    <br />
P\{ (A \cap B)\} {\color{red}\cap C} = P\{ A \cap C\} + P\{ B \cap C\} - P\{ A \cap B \cap C\} <br />

    So I did this:

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

    However, if it was 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 P\left\{(A\cup B)\cap C \right\}

    This is the same as 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, P\left( {A \cap B} \right) \cap C is totally meaning less.
    First P\left( {A \cap B} \right) is a number while 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.

    <br />
P(A \cap B) \cap P(C) = P(A \cap B \cap C)<br />

    Kid
    Last edited by kid funky fried; August 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.

    <br />
P(A \cap B) \cap P(C) = P(A \cap B \cap C)<br />

    Kid
    Sorry to say, but this expression is not good either...
    Again, \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?

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


    Now I understand Plato's disdain!
    (I think)
    Last edited by kid funky fried; August 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, \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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