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Another proof using the axioms of probability

  1. Mar 14, 2013 #1
    1. The problem statement, all variables and given/known data

    If A and B are events, use the axioms of probability to show that:

    [itex]if B \subset A, then P(B) \leq P(A)[/itex]

    2. Relevant equations

    Axiom 1: [itex]P(n) \geq 0[/itex]

    Axiom 2: [itex]P(S)=1[/itex]

    Axiom 3: If A1,A2,... are disjoint sets, then [itex]P(\bigcup _{i} A_{i}) = \sum_{i} P(A_{i})[/itex]

    3. The attempt at a solution

    I start with using the law of total probability to define the set A:

    [itex]A= (A \cap B) \cup (A \cap B^{C})[/itex]

    Then I use axiom 3 to get turn it into a probability:

    [itex]P(A) = P(A \cap B) + P(A \cap B^{C})[/itex]

    Since [itex]B \subset A, P(A \cap B) = P(B)[/itex]

    So

    [itex]P(A) = P(B) + P(A \cap B^{C})[/itex]

    [itex]P(B)=P(A) - P(A \cap B^{C})[/itex]

    And as axiom 1 states that a probability must be greater than or equal to 0,

    [itex]P(B) \leq P(A)[/itex]

    As for proving the equality case, this means that [itex]P(A \cap B^{C}) = 0[/itex], but then doesn't that just mean that A=B. Since the question states that B is a *proper* subset of A, am I incorrect in thinking that it might be a typo?
     
    Last edited: Mar 14, 2013
  2. jcsd
  3. Mar 14, 2013 #2

    vela

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    It says the probability must be greater than or equal to 0.
     
  4. Mar 14, 2013 #3
    Ok, fixed. But suppose that [itex]P(A \cap B^{C}) = 0[/itex], doesn't this mean that A=B?
     
  5. Mar 14, 2013 #4

    vela

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    Not necessarily. For example, say you have a continuous random variable X that's uniformly distributed on [0,1]. Let A=[0,1] and B=(0,1). Both P(A)=P(B)=1, but A≠B.
     
  6. Mar 14, 2013 #5
    [itex][/itex]I see, interesting..so basically, using axiom 1, I will get [itex]P(B) \leq P(A)[/itex], which completes my proof. Thanks :)
     
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