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Math Help - [SOLVED] Probability Proof

  1. #1
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    Exclamation [SOLVED] Probability Proof

    Hi i am new to this so can you please help with the following

    Let A,B,C be events. Explain why

    1. P(AuBuC)= P(A) + P(B/A) + P (C/(AUB)).

    Deduce that

    2. P(AUBUC)< P(A)+P(B) + P(C)

    3. Formulate and prove a version of part (2) for n events.

    Thank you
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  2. #2
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    First, note that you mean B\backslash A which is set difference.
    What you typed looks like conditional probability.
    The first one is just an exercise in sets.
    Looking at the diagram, it is easy to see three disjoint sets. Their union is the whole.
    A \cup \left[ {B\backslash A} \right] \cup \left[ {C\backslash \left( {A \cup B} \right)} \right] = A \cup B \cup C.
    So by the additive property P(A) + P(B) + P(C) = P(A) + P\left[ {B\backslash A} \right] + P\left[ {C\backslash \left( {A \cup B} \right)} \right]

    For #2, use the monotone property of probability.
    T \subseteq S \Rightarrow \quad P(T) \leqslant P(S) (here again you typed < when it should be \le).
    \left[ {B\backslash A} \right] \subseteq B \Rightarrow \quad P\left[ {B\backslash A} \right] \leqslant P(B).
    A similar statement for the third term. Make the substitutions it follows.
    Attached Thumbnails Attached Thumbnails [SOLVED] Probability Proof-venn_-prob.gif  
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  3. #3
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    Thumbs up

    Quote Originally Posted by Plato View Post
    First, note that you mean B\backslash A which is set difference.
    What you typed looks like conditional probability.
    The first one is just an exercise in sets.
    Looking at the diagram, it is easy to see three disjoint sets. Their union is the whole.
    A \cup \left[ {B\backslash A} \right] \cup \left[ {C\backslash \left( {A \cup B} \right)} \right] = A \cup B \cup C.
    So by the additive property P(A) + P(B) + P(C) = P(A) + P\left[ {B\backslash A} \right] + P\left[ {C\backslash \left( {A \cup B} \right)} \right]

    For #2, use the monotone property of probability.
    T \subseteq S \Rightarrow \quad P(T) \leqslant P(S) (here again you typed < when it should be \le).
    \left[ {B\backslash A} \right] \subseteq B \Rightarrow \quad P\left[ {B\backslash A} \right] \leqslant P(B).
    A similar statement for the third term. Make the substitutions it follows.
    Thanks for the Help.

    Sorry for typing the notations wrong but i didn't know how to post them
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  4. #4
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    "For #2, use the monotone property of probability.
    (here again you typed < when it should be ).
    .
    A similar statement for the third term. Make the substitutions it follows"

    Sorry to be a pain but i don't really understand what you mean by the monotone property of probability. Can you please exlain it to me.
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  5. #5
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    If G is a subevent of H then the P(G) is less than or equal to P(H).
    So P(B\backslash A) \leqslant P(B)\,\& \,P\left( {C\backslash (A \cup B)} \right) \leqslant P(C).
    Thus P(A \cup B \cup C) = P(A) + P(B\backslash A) + P\left( {C\backslash (A \cup B)} \right) \leqslant P(A) + P(B) + P(C).
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