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Math Help - Show event is completly independent of another event

  1. #1
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    Show event is completly independent of another event

    Question :
    Show that if the event A is completely independent of events B and C, then A is independent of logical sum B \cup C
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  2. #2
    MHF Contributor matheagle's Avatar
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    assume that

      P(AB)=P(A)P(B) and   P(AC)=P(A)P(C)

    the question is, does

    P(A(B\cup C))=P(A)P(B\cup C)????????

    I don't think so.
    Try tossing a fair die.
    Let A=2,4,6
    B=2,3
    and
    C=2,5

    I get 1/6 =1/6 for the first two
    but then I get 1/6 and 1/4 on the last one.
    Last edited by matheagle; December 2nd 2009 at 10:06 PM.
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  3. #3
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    Could u please explain me

    Quote Originally Posted by matheagle View Post
    assume that

      P(AB)=P(A)P(B) and   P(AC)=P(A)P(C)

    the question is, does

    P(A(B\cup C))=P(A)P(B\cup C)????????

    I don't think so.
    Try tossing a fair die.
    Let A=2,4,6
    B=2,3
    and
    C=2,5

    I get 1/6 =1/6 for the first two
    but then I get 1/6 and 1/4 on the last one.

    could u please explain a little bit properly?/
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  4. #4
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    Quote Originally Posted by zorro View Post
    Question :
    Show that if the event A is completely independent of events B and C, then A is independent of logical sum B \cup C
    I guess "completely independent" is what is also called "mutually independent" (by opposition to "pairwise independent"), i.e. it means that P(A\cap B)=P(A)P(B) (same with A,C and B,C), and P(A\cap B\cap C)=P(A)P(B)P(C).

    Then the property is proved as follows (for instance):
    P(A\cap(B\cup C))=P((A\cap B\cap C)\cup(A\cap B^c\cap C)\cup (A\cap B\cap C^c)), and the three (main) events on the right-hand side are disjoints, hence
    P(A\cap(B\cup C))=P(A\cap B\cap C)+P(A\cap B^c\cap C)+P(A\cap B\cap C^c),
    and you can compute the latter probabilities using A\cap B^c\cap C=(A\cap C)\setminus (A\cap B\cap C) (and similarly for the last term) and the assumptions (recalled above).

    You should get \cdots=P(A)P(B\cup C), or in fact P(A)(P(B)+P(C)-P(B)P(C)), which is the same...
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  5. #5
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    I am again stuck

    Quote Originally Posted by Laurent View Post
    I guess "completely independent" is what is also called "mutually independent" (by opposition to "pairwise independent"), i.e. it means that P(A\cap B)=P(A)P(B) (same with A,C and B,C), and P(A\cap B\cap C)=P(A)P(B)P(C).

    Then the property is proved as follows (for instance):
    P(A\cap(B\cup C))=P((A\cap B\cap C)\cup(A\cap B^c\cap C)\cup (A\cap B\cap C^c)), and the three (main) events on the right-hand side are disjoints, hence
    P(A\cap(B\cup C))=P(A\cap B\cap C)+P(A\cap B^c\cap C)+P(A\cap B\cap C^c),
    and you can compute the latter probabilities using A\cap B^c\cap C=(A\cap C)\setminus (A\cap B\cap C) (and similarly for the last term) and the assumptions (recalled above).

    You should get \cdots=P(A)P(B\cup C), or in fact P(A)(P(B)+P(C)-P(B)P(C)), which is the same...

    ,


    <br />
Pr(A \cap(B \cup C)) = Pr(A \cap B \cap C) + \frac{Pr(A \cap C)}{Pr(A \cap B \cap C)} + \frac{Pr(A \cap B)}{Pr(A \cap B \cap C)}
    I am stuck here ????
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  6. #6
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    Quote Originally Posted by zorro View Post
    ,


    <br />
Pr(A \cap(B \cup C)) = Pr(A \cap B \cap C) + \frac{Pr(A \cap C)}{Pr(A \cap B \cap C)} + \frac{Pr(A \cap B)}{Pr(A \cap B \cap C)}
    I am stuck here ????


    If B\subset A, P(A\setminus B)=P(A)-P(B), of course... not \frac{P(A)}{P(B)}...

    And then just use the assumption of independence.
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  7. #7
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    I did get u

    Quote Originally Posted by Laurent View Post


    If B\subset A, P(A\setminus B)=P(A)-P(B), of course... not \frac{P(A)}{P(B)}...

    And then just use the assumption of independence.

    I didnt understand your answer to my question ...sorry
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