2005 April 30th, 05:18 AM
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#13 (permalink) |
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TestMagic Guru-in-Training
 
Join Date: Dec 2004
Posts: 566
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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)) |
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2005 April 30th, 06:17 AM
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#14 (permalink) |
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Within my grasp!
Join Date: Dec 2004
Location: Chicago
Posts: 421
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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 : 2005 April 30th at 06:23 AM. Reason: typo |
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2005 April 30th, 06:21 AM
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#15 (permalink) |
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Dumbledore's Army
Join Date: Jan 2005
Posts: 2,334
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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??? 
_ _ _ _ SIG _ _ _ _
If you can't change your mind, are you sure you have one? Be weird whenever you have the chance Don't lurk! Please participate  OF COURSE I DON'T LOOK BUSY...I DID IT RIGHT THE FIRST TIME. |
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2005 April 30th, 06:25 AM
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#17 (permalink) |
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Dumbledore's Army
Join Date: Jan 2005
Posts: 2,334
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Re: sets
these formulas are really good.Thanks for the trouble.
_ _ _ _ SIG _ _ _ _
If you can't change your mind, are you sure you have one? Be weird whenever you have the chance Don't lurk! Please participate OF COURSE I DON'T LOOK BUSY...I DID IT RIGHT THE FIRST TIME. |
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2005 May 28th, 05:22 PM
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#19 (permalink) |
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I JUST got here.
 
Join Date: Apr 2005
Posts: 28
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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 ~~~~~~~~ More you look, more you see! |
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2005 November 23rd, 05:16 AM
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#20 (permalink) | |
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Within my grasp!
Join Date: Aug 2005
Posts: 495
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Quote:
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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