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Quick Help, Probability.

  1. Jan 27, 2006 #1
    Hi, I have a quick question. What does the triangle (Δ) mean? I was asked to prove this, but since it's not told in the book and I just wana get an idea of what the Δ means.

    Show that A Δ B^c = A^c Δ B

    Also after trying to prove the two sides, I got stuck here...

    For A Δ B^c =...=...=(A ∩ B^c) U (B^c ∩ A)
    and for A^c Δ B =...=...=(A^c ∩ B) U (B ∩ A^c)

    how do they equal?
  2. jcsd
  3. Jan 27, 2006 #2


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    [tex]A\Delta B = (A-B)\cup (B-A)[/tex] is the symmetric difference of the sets A and B and it contains only those points of A which are not in B and those points of B which are not in A.

    Example: If A={1,2,7,9,11} and B={3,5,7,11,13}, then the set difference, A-B, of A and B is the part of A not in B, namely A-B={1,2,9}; likewise, the set difference, B-A, of B and A is the part of B not in A, namely
    B-A={3,5,13}, and hence

    [tex]A\Delta B = (A-B)\cup (B-A)={1,2,9}\cup {3,5,13}[/tex]

    and this is equivalent to what you have since [tex]A-B=A\cap B^c[/tex]

    where [tex]B^c[/tex] is the complement of B in X (if X is the universal set containing A and B).
    Last edited: Jan 27, 2006
  4. Jan 28, 2006 #3
    Yes that is also given in the question but I can't make them equal. Like using your example, it doesn't equal. So does that mean it doesn't equal?
    I'm stuck at the same place as I posted in my first post.
  5. Jan 28, 2006 #4


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    Well, the two expressions you wrote aren't equal. (A ∩ B^c) U (B^c ∩ A) should really be (A ∩ B^c) U (B ∩ A^c), and you have a similar trouble in the second one.
  6. Jan 29, 2006 #5
    Could you please show me how you got that? I understand that it's correct but I just want to know how you got that.
  7. Jan 29, 2006 #6


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    Just use what Benorin said to expand (A - B) U (B - A): A - B = A n B^c
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