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Math Help - conditional probability w/ boolean operators

  1. #1
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    conditional probability w/ boolean operators

    Hi, Can someone help me out? I just want to make sure i'm right since the solution doesn't spell out each step....

    As far as my understanding p(a | b) = [p(a /\ b)]/ p(b)

    but, i'm looking at problems that have
    p(a /\ b /\ c | p(a /\ b) = [p(a /\ b /\ c)] / p(a /\ b)

    or
    p(a V b V c | p(a V b) = [p(a V b V c)] / p(a V b)


    is this because of the identity law? no need to include p(a /\ b) in the numerator because they are already included in the first clause? in other words,
    p(a /\ b /\ c)] /\ p(a /\ b) is redundant and can be reduced to p(a /\ b /\ c) -as is the same for p(a V b V c) /\ p(a V b V c) ?

    Can someone confirm this? just wanna make sure i'm not misunderstanding the solution step... thanks!
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  2. #2
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    I am not sure that I understand the question.
    But this is true: \left( {A \cup B \cup C} \right) \cap \left( {A \cup B} \right) = \left( {A \cup B} \right).

    Now I don't think that is what you have. Is it?
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  3. #3
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    yeah i'm still trying to get a handle on the cool math app you guys use so my fault for not using it yet.
    mainly the solution has p(a /\ b /\ c | p(a /\ b) which could be rewritten a p(x | y). and p(x|y) = p(x /\ y) / p(y)

    That being said, if If x =a /\ b /\ c and y = a /\ b then shouldn't the equation be:
    [p(a /\ b /\ c)] /\ p(a /\ b) ] / p(a /\ b) ?

    But, the solution has it as:
    [p(a /\ b /\ c)] / p(a /\ b) instead....
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  4. #4
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    It you want to use symbols it makes is easy to read.
    [tex]P(A\cap B\cap C)| A\cap B) [/tex] you will get P(A\cap B\cap C)| A\cap B) .

    Here are some more.
    [tex] \cap [/tex] gives  \cap is intersection
    [tex] \subseteq [/tex] gives  \subseteq is subset
    [tex] \cup [/tex] gives  \cup is union
    [tex] \in [/tex] gives  \in is element
    [tex] \emptyset [/tex] gives  \emptyset is emptyset
    Last edited by Plato; July 31st 2009 at 04:11 PM.
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  5. #5
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    sorry i seem so slow. this is self-learning....
    (A\cap B\cap C) \cap (A \cap B) = (A\cap B) seems like a mystery to me.

    i'm looking at the distributive law and the absorption law as those that most closely relate but i can't see either applying in this case. sorry, but can you spell it out for this layman?
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  6. #6
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    Quote Originally Posted by uneakbreed View Post
    sorry i seem so slow. this is self-learning....
    (A\cap B\cap C) \cap (A \cap B) = (A\cap B) seems like a mystery to me. i'm looking at the distributive law and the absorption law as those that most closely relate but i can't see either applying in this case. sorry, but can you spell it out for this layman?
    I see why it is mystery to you: IT IS WRONG.
    It should be (A\cap B\cap C) \cap (A \cap B) = (A\cap B\cap C).

    Sorry about that. But I was working off your OP that used union rather than intersection.
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  7. #7
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    AHHH! ok. so it is the identity law. cool.

    so since they are equal, why don't i need to multiply it against itself? in other words, although they are equal, since the original equation is
    P(A|B) = \frac{P(A \cap B) \cap P(B)}{P(B)}

    then why not multiply P(A \cap B \cap C) to itself? In other words, if P(A \cap B) =P(A \cap B \cap C) making the equation be

     P(A \cap B \cap C) | A \cap B) = \frac{P(A \cap B \cap C) \cap P(A \cap B \cap C)}{P(A \cap B)}

    also:
    is there a web resource that you use that lists all the identities of boolean algebra? i found a couple but i never know how complete they are....
    Last edited by uneakbreed; July 31st 2009 at 04:30 PM. Reason: fixed formula
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  8. #8
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    Quote Originally Posted by uneakbreed View Post
     P(A \cap B \cap C) | A \cap B) = \frac{P(A \cap B \cap C) \cap P(A \cap B \cap C)}{P(A \cap B)}
    Given that all we have is this
     P(A \cap B \cap C) | A \cap B) = \frac{P(A \cap B \cap C) }{P(A \cap B)}
    ____
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