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04 Sep 2008, 21:27
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Can someone explain to me what the difference between the below formulae is?
I actually don't understand how the first formula could be correct: If you've subtracted AnB, AnC, BnC, you've taken out AnBnC three times. So you need to add it back in twice, right?
For 3 sets A, B, and C: P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)
No of persons in atleast one set:
P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)
Any ideas?
I actually don't understand how the first formula could be correct: If you've subtracted AnB, AnC, BnC, you've taken out AnBnC three times. So you need to add it back in twice, right?
For 3 sets A, B, and C: P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)
No of persons in atleast one set:
P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)
Any ideas?
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05 Sep 2008, 00:17
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I am not sure I understant your expressions, so I'll go with mine.
We have 3 sets: A, B, and C. These sets overlap, so that there are three subsets of the type AB and one subset ABC (say, students who study all three languages). The total number (of people), elimitnating overlaps, looks like this:
A + B + C -AB -AC -BC -2*ABC
or:
Aonly + Bonly + Conly +AB + AC + BC + ABC
Math wizards, please correct me if the above is wrong.
We have 3 sets: A, B, and C. These sets overlap, so that there are three subsets of the type AB and one subset ABC (say, students who study all three languages). The total number (of people), elimitnating overlaps, looks like this:
A + B + C -AB -AC -BC -2*ABC
or:
Aonly + Bonly + Conly +AB + AC + BC + ABC
Math wizards, please correct me if the above is wrong.
05 Sep 2008, 04:47
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P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC) is correct.
You can understand it by drawing 3 circles intersecting each other.
when you add P(A) + P(B) + P(C) , you have included each of P(AnB) , P(AnC) and P(BnC) twice.
for example P(AnB) is a part of P(A) as well as P(B) ; so you deduct once.
Now when you deduct -P(AnB) – P(AnC) – P(BnC) , you are deducting P(AnBnC) three times and also P(A) + P(B) + P(C) has added P(AnBnC) 3 times. So you need to add 1 P(AnBnC) back to the expression.
You can understand it by drawing 3 circles intersecting each other.
when you add P(A) + P(B) + P(C) , you have included each of P(AnB) , P(AnC) and P(BnC) twice.
for example P(AnB) is a part of P(A) as well as P(B) ; so you deduct once.
Now when you deduct -P(AnB) – P(AnC) – P(BnC) , you are deducting P(AnBnC) three times and also P(A) + P(B) + P(C) has added P(AnBnC) 3 times. So you need to add 1 P(AnBnC) back to the expression.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
