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Math Help - show that events are independent

  1. #1
    Yan
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    show that events are independent

    Show that if events A and B are independent, then events A' and B are independent.
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  2. #2
    Member Last_Singularity's Avatar
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    Quote Originally Posted by Yan View Post
    Show that if events A and B are independent, then events A' and B are independent.
    If events A,B are independent, then we know that
    P(A \cap B) = P(A) P(B) by definition. That's our key.

    For the proof, we will use the following lemma:
    A^C \cap B = B \setminus (A \cap B)

    Proof of the lemma:
    A^C \cap B = (\Omega \setminus A) \cap B = (\Omega \cap B) \setminus (A \cap B) = B \setminus (A \cap B)

    So now we know that lemma is true, we can use it to conclude that:
    P(A^C \cap B) = P(B) - P(A \cap B)

    Looking at the right hand side, we have:
    =P(B) - P(A \cap B)
    =P(B) - P(A)P(B) (using our key assumption)
    =(1-P(A))P(B)
    =P(A^C)P(B)

    So we conclude that
    P(A^C \cap B) = P(A^C)P(B) as desired.

    QED
    Last edited by Last_Singularity; January 11th 2009 at 05:08 PM.
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  3. #3
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    \begin{array}{*{20}c}<br />
   {P\left( {A^c  \cap B} \right)} &  =  & {P\left( B \right) + P\left( {A \cap B} \right)}  \\<br />
   {} &  =  & {P\left( B \right) + P\left( A \right)P\left( B \right)}  \\<br />
   {} &  =  & {P(B)\left( {1 - P(A)} \right)}  \\<br />
   {} &  =  & {P(B)P\left( {A^c } \right)}  \\<br /> <br />
 \end{array}
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  4. #4
    Flow Master
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    Quote Originally Posted by Plato View Post
    \begin{array}{*{20}c}<br />
{P\left( {A^c \cap B} \right)} & = & {P\left( B \right) {\color{red} - } P\left( {A \cap B} \right)} \\<br />
{} & = & {P\left( B \right) {\color{red} - } P\left( A \right)P\left( B \right)} \\<br />
{} & = & {P(B)\left( {1 - P(A)} \right)} \\<br />
{} & = & {P(B)P\left( {A^c } \right)} \\<br /> <br />
\end{array}
    Small typos fixed (in red).

    A Karnaugh table is a visual way of making both the first line of Plato's proof and the conclusion of Last Singularity's lemma crystal clear.
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