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Thread: sets

  1. #11
    Within my grasp! TwinnSplitter's Avatar
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    Re: sets

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    realized I screwed up, ignore this post
    Last edited by TwinnSplitter; 04-30-2005 at 04:43 AM.

  2. #12
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    Re: sets

    realized I screwed up, ignore this post
    Last edited by TwinnSplitter; 04-30-2005 at 04:44 AM.

  3. #13
    An Urch Guru Pundit Swami Sage scoot's Avatar
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    Re: sets

    I was looking for a way to draw a venn diagram. I do not know how draw that here. I got this link which has venn diagram. Please check this link.

    http://www.urch.com/forums/showthrea...85%25+surveyed (Set Theory Q (consumer survey))

  4. #14
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    1 out of 1 members found this post helpful. Good post? Yes | No

    Re: sets

    After checking out Scoot's link (and following the links from there), I have found the following general formulas for three-component set problems:

    If there are three sets A, B, and C, then
    P(AuBuC) = P(A) + P(B) + P(C) P(AnB) P(AnC) P(BnC) + P(AnBnC)


    Number of people in exactly one set =
    P(A) + P(B) + P(C) 2P(AnB) 2P(AnC) 2P(BnC) + 3P(AnBnC)


    Number of people in exactly two of the sets =
    P(AnB) + P(AnC) + P(BnC) 3P(AnBnC)


    Number of people in exactly three of the sets =
    P(AnBnC)


    Number of people in two or more sets =
    P(AnB) + P(AnC) + P(BnC) 2P(AnBnC)
    Last edited by TwinnSplitter; 04-30-2005 at 05:23 AM. Reason: typo

  5. #15
    ish
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    Re: sets

    [QUOTE=If there are three sets A, B, and C, then
    P(AnBnC) = P(A) + P(B) + P(C) P(AnB) P(AnC) P(BnC) + P(AnBnC)


    =/QUOTE]

    shouldnt this be a UNION of sets???

  6. #16
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    Re: sets

    sorry about that ish, I just fixed it

  7. #17
    ish
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    Re: sets

    these formulas are really good.Thanks for the trouble.

  8. #18
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    Re: sets

    Let X be P(AnB) + P(AnC) + P(BnC).
    Using the formulas above, we have

    85 = 50+30+20-X+5

    X = 20

    Number of people having two or more = X - 2(AnBnC) = 20 - 10 = 10

    Answer is 10

  9. #19
    hem
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    Re: sets

    To add to TwinSplitter's forumla list:

    No of people in atleast 1 set =
    P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)

    This formula can also be derived from:
    No of people in atleast 1 set =
    no of people in exactly 1 set + no of people in exactly 2 set + no of people in exactly 3 set



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  10. #20
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    Quote Originally Posted by hem
    To add to TwinSplitter's forumla list:

    No of people in atleast 1 set =
    P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)

    This formula can also be derived from:
    No of people in atleast 1 set =
    no of people in exactly 1 set + no of people in exactly 2 set + no of people in exactly 3 set
    Shouldnt this be

    P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + P(AnBnC)
    When I add no of people in exactly 1 set with no of people in exactly 2 set I see that 3P(AnBnC) gets cancelled and then I add P(AnBnc) for exactly 3 set.

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