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Compliments Universe

  1. Apr 1, 2004 #1
    OK, if I have sets such as:

    |A| = 7
    |B| = 10
    |A intersection B| = 5
    (And the universe equaled 19)

    Then how do you find the compliment of |A intersection B|? And what would the answer be in this case? Thanks.
     
    Last edited: Apr 1, 2004
  2. jcsd
  3. Apr 1, 2004 #2

    matt grime

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    Do you really want to take the complement of a cardinality?
     
  4. Apr 1, 2004 #3

    ahrkron

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    You may find useful to make a drawing of the situation.
     
  5. Apr 1, 2004 #4
    OK, here is the problem:

    If Universe = 19 and:
    A = People who dislike NDP
    B = People who dislike Liberals
    C = People who dislike Conservatives
    D = People who dislike Canadian Alliance
    |A| = 7
    |B| = 10
    |C| = 11
    |D| = 6
    |B intersection A| = 5
    |A intersection C| = 5
    |B intersection C| = 6
    |A intersection D| = 3
    |B intersection D| = 4
    |C intersection D| = 5
    |C intersection B intersection A| = 3
    |B intersection A intersection D| = 2
    |C intersection A intersection D| = 3
    |C intersection B intersection D| = 4
    |A intersection B intersection C intersection D| = 2

    Are all given, then how many like all 4 parties (not dislike)? Can someone point me in the right direction for this? Thank you.
     
  6. Apr 1, 2004 #5

    Zurtex

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    I've never done this before so this may be a really stupid comment, but wouldn't it make it just that little easier if:

    |A intersection B intersection C intersection D| = 2

    Then you can delete this line and take 2 away from all these 4 lines:

    |A| = 7
    |B| = 10
    |C| = 11
    |D| = 6

    And the universe.
     
  7. Apr 1, 2004 #6
    Not sure...

    (Not really sure where to start with this myself...)
     
  8. Apr 2, 2004 #7

    matt grime

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    You want [tex]A^c\cap B^c\cap C^c \cap D^C[/tex] call this set E

    Let U denote the set of all people asked (the universe)

    By definition E = (U\A)n(U\B)n(U\C)n(U\D)

    can you work with the rules of sets to simplify that?

    Or can you think of a better way of doing it? Such as: (AuB)^c = (A^c)n(B^c)?
     
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