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Math Help - set proof

  1. #1
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    set proof

    Could someone please just give me a hint as to how I could prove this:

    {(A∩B)’} = 1+P{AB} – P{A}– P{B}

    Thanks.
    Last edited by kid funky fried; August 24th 2008 at 03:23 PM. Reason: complement misplaced
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  2. #2
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    \begin{array}{l}<br />
 P\left\{ {\left( {A \cap B} \right)^c } \right\} = P\left( {A^c  \cup B^c } \right) = P\left( {A^c } \right) + P\left( {B^c } \right) - P\left( {A^c  \cap B^c } \right) \\ <br />
  = 1 - P(A) + 1 - P(B) - \left[ {1 - P\left( {A \cup B} \right)} \right] \\ <br />
  = 1 + P\left( {A \cup B} \right) - P(A) - P(B) \\ <br />
 \end{array}
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  3. #3
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    Yes..

    Yes, a big THANK YOU, Plato!

    Kid
    Last edited by kid funky fried; August 24th 2008 at 06:48 PM. Reason: found the tab
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  4. #4
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    Quote Originally Posted by Plato View Post
    \begin{array}{l}<br />
P\left\{ {\left( {A \cap B} \right)^c } \right\} = P\left( {A^c \cup B^c } \right) = P\left( {A^c } \right) + P\left( {B^c } \right) - P\left( {A^c \cap B^c } \right) \\ <br />
= 1 - P(A) + 1 - P(B) - \left[ {1 - P\left( {A \cup B} \right)} \right] \\ <br />
= 1 + P\left( {A \cup B} \right) - P(A) - P(B) \\ <br />
\end{array}
    Can you (or anyone) explain what rules or laws got you to this? I follow the first step, because it is DeMorgan's Law. But from there I'm lost. I'm looking through all my notes and my texbook, but I'm not seeing the rules that allow you to make those changes.
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  5. #5
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    Quote Originally Posted by ban26ana View Post
    Can you (or anyone) explain what rules or laws got you to this? I follow the first step, because it is DeMorgan's Law. But from there I'm lost. I'm looking through all my notes and my texbook, but I'm not seeing the rules that allow you to make those changes.
    There's a property (I don't know the name...) that says :

    P(A \cup B)=P(A)+P(B)-P(A \cap B)

    You can see it on a Venn diagram. If you add A and B, you'll add twice A \cap B


    This is why P(A^c \cup B^c)=P(A^c)+P(B^c)-P(A^c \cap B^c)
    ^c denotes the complement - see here : Complement (set theory - Wikipedia, the free encyclopedia)
    And we have P(A^c)=1-P(A)

    So :
    P(A^c)+P(B^c)-P\underbrace{(A^c \cap B^c)}_{(A \cup B)^c ~, ~\text{by de Morgan's law}}=(1-P(A))+(1-P(B))-(1-P(A \cup B))

    ...
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  6. #6
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    Thanks Moo. I found the theorem in my book that says P(A) = 1 - P(A'). Is that the same one that makes P(A' *intersect* B') = 1 - P(A U B)?
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  7. #7
    Moo
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    Quote Originally Posted by ban26ana View Post
    Thanks Moo. I found the theorem in my book that says P(A) = 1 - P(A'). Is that the same one that makes P(A' *intersect* B') = 1 - P(A U B)?
    Yes, because A' \cap B'=(A \cup B)', which is de Morgan's rule.
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  8. #8
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    Thanks again Moo. I get it!
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