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Math Help - [SOLVED] Mutually Indpendent Question

  1. #1
    mtb
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    [SOLVED] Mutually Indpendent Question

    Hi,

    I just needed a bit of help with the following question.

    Show that if A, B and C are mutually independent, then A and (B U C) are independent.

    From what I understand if A, B and C are mutually independent then

    P(AnBnC) = P(A)P(B)P(C)
    P(AnB) = P(A)P(B)
    P(AnC) = P(A)P(C)
    P(BnC) = P(B)P(C)


    To prove A and (B U C) are independent do I have to prove P(A|(BUC) ) = P(A) because I don't know what to do after
    P((BUC)nA)/P(BUC) = P[(BnA)U(AnC)]/P(BUC)



    THANK YOU
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  2. #2
    MHF Contributor

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    Your definition of "mutual independence" is indeed correct. As you can see, it involves intersections. So you must reduce P(A\cap(B\cup C)) to an expression using intersections. The usual trick is to use the complement since (B\cup C)^c=B^c\cap C^c (where B^c=\Omega\setminus B if \Omega is the whole probability space).

    So here are the first steps (that you should justify): P(A\cap(B\cup C))=P(A)-P(A\cap(B\cup C)^c)=P(A)-P(A\cap B^c \cap C^c).

    You will need to know that if A, B, C are mutually independent, then A, B^c, C^c are mutually independent as well. Either you know it already or you'll have to prove it.

    Laurent.
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  3. #3
    mtb
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    Can I continue like this


    <br /> <br />
P(A)-P(A\cap B^c \cap C^c) = P(A)-P(A)P( B^c \cap C^c) = P(A)[1 -P( B^c \cap C^c)]<br /> <br />

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

    Therefore they are independent
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  4. #4
    MHF Contributor

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    I'd rather insert an intermediate equality P(A)-P(A)P(B^c)P(C^c)=P(A)-P(A)P(B^c\cap C^c) in what you wrote, since you don't know directly that P(A\cap B^c\cap C^c)=P(A)P(B^c\cap C^c) (this is almost what you were asked to prove!). Then this is OK.
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  5. #5
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    \begin{array}{rcl}<br />
   {P\left[ {A(B \cup C)} \right]} &  =  & {P\left( {AB \cup AC} \right)}  \\<br />
   {} &  =  & {P(AB) + P(AC) - P(ABC)}  \\<br />
   {} &  =  & {P(A)P(B) + P(A)P(C) - P(A)P(B)P(C)}  \\<br />
   {} &  =  & {P(A)\left[ {P(B) + P(C) - P(BC)} \right]}  \\<br />
   {} &  =  & {P(A)\left[ {P(B \cup C)} \right]}  \\ \end{array}
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  6. #6
    mtb
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    Thanks for all the help!!!
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