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Probability P(A\B) = P(A) - P(B)

  1. Sep 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Let [itex]A \subseteq B \subseteq S[/itex] where [itex]S[/itex] is a sample space.
    Show that [itex]P(A \setminus B) = P(A) - P(B)[/itex]


    2. Relevant equations

    [itex]A \setminus B[/itex] denotes set difference; these are probability functions.

    3. The attempt at a solution
    I have,
    [itex]P(A \setminus B) = P(A \cap B^{C})
    = P(A) - P(A \cap B)
    = P(A) - [P(B) - P(A^{c} \cap B)]
    = P(A) - P(B) + P(A^{c} \cap B)[/itex]

    It seems like i'm close, but i've spent a while trying to figure out how to get rid of the [itex]P(A^{c} \cap B)[/itex].

    Any insight anyone?
    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 22, 2011 #2
    I think your inclusion is backwards: you mean [itex] B \subseteq A \subseteq S [/itex], right?

    Try writing A as the union of two disjoint sets--this will give you P(A) in terms of something that can be rearranged into what you're trying to prove.
     
  4. Sep 22, 2011 #3
    Yes, thank you.

    I've tried that... unless i'm missing something?
    A as the union of disjoint sets is [itex]A = (A \cap B) \cup (A \cap B^{c})[/itex].
    So, [itex]P(A) = P((A \cap B) + (A \cap B^{c})[/itex].
    When i plug this in and do some rearranging, i just get right back to where i ended up in the original post?
     
  5. Sep 22, 2011 #4
    Ah I see, your calculations just went off in an unexpected direction after the third equality. Take a look at your third equality: since [itex] B \subseteq A[/itex], then what is [itex] A \cap B[/itex]?
     
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