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Prove using three basic probability axioms

  1. Oct 7, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that [itex]P(A \cap B)≥1-P(\bar{A})-P(\bar{B})[/itex]

    for all [itex] A, B \subseteq S[/itex]


    using only these axioms:
    1) [itex]0 \leq P(A) \leq 1[/itex], for any event [itex]A \subseteq S[/itex]
    2) [itex]P(S) = 1[/itex]
    3) [itex] P(A \cup B) = P(A) + P(B)[/itex] if and only if [itex] P(A \cap B) = 0[/itex]


    2. Relevant equations
    None.


    3. The attempt at a solution

    My proof:

    First rearrange the shizzle:

    [itex]P(\bar{A}) + P(A \cap B) + P(\bar{B}) \geq 1[/itex]

    Now using the fact that the first two terms are disjoint, use axiom 3 to obtain:

    [itex]P(\bar{A} \cup (A \cap B)) + P(\bar{B}) \geq 1[/itex]

    Next note that the first term is equals to P(S), hence we get:

    [itex]P(S) + P(\bar{B}) \geq 1 [/itex]

    which holds for all [itex] B \subseteq S[/itex], because [itex]P(S) = 1[/itex] and [itex] P(\bar{B}) \geq 0 [/itex] for all B.

    Is this proof correct?
     
  2. jcsd
  3. Oct 7, 2011 #2
    In proof you can't start out claiming what you are trying to prove - you can only claim it isn't true and arriving at a contradiction. So you would say

    Suppose [itex]P(A \cap B)<1-P(\bar{A})-P(\bar{B})[/itex] then try to arrive at a contradiction.
     
  4. Oct 9, 2011 #3
    Got it! Thanks!
     
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