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rao
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SET theory ... BIG CONFUSION Updated on: 17 Jul 2008, 20:13
Originally posted by rao on 17 Jul 2008, 18:43.
Last edited by rao on 17 Jul 2008, 20:13, edited 1 time in total.
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Hi All,

I was going thru some set theory questiona dthen got confused. Checked some we pages and onw of 'em had following info:

Formulas for three-component set problems:
u = union
n = intersection

1. For 3 sets A, B, and C: P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)

2. No of persons in exactly one set:
P(A) + P(B) + P(C) – 2P(AnB) – 2P(AnC) – 2P(BnC) + 3P(AnBnC)

3. No of persons in exactly two of the sets: P(AnB) + P(AnC) + P(BnC) – 3P(AnBnC)

4. No of persons in exactly three of the sets: P(AnBnC)

5. No of persons in two or more sets: P(AnB) + P(AnC) + P(BnC) – 2P(AnBnC)

6. No of persons in atleast one set:
P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)



Can somebody please explain what is the difference between 1 and 6?


Thanks much in advance!



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SET theory ... BIG CONFUSION 17 Jul 2008, 19:12
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Another problem that I have is in uderstanding AuBuC

I always get this .. please let me know where I am wrong?

P(AuBuC)=P(A)+P(B)+P(C) - P(AnB) - P(AnC) + P(AnBnC) - P(BnC) + P(AnBnC)

since when I first subtracted P(AnB) that was also including P(AnBnC). So next time when I am substracting P(AnC) and P(AnB), I am adding P(AnBnC) to avoind any duplicaiton
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SET theory ... BIG CONFUSION 17 Jul 2008, 20:17
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rao_1857
Another problem that I have is in uderstanding AuBuC

I always get this .. please let me know where I am wrong?

P(AuBuC)=P(A)+P(B)+P(C) - P(AnB) - P(AnC) + P(AnBnC) - P(BnC) + P(AnBnC)

since when I first subtracted P(AnB) that was also including P(AnBnC). So next time when I am substracting P(AnC) and P(AnB), I am adding P(AnBnC) to avoind any duplicaiton
Show more

let me try.... AuBuC should include everying that is in A, B and C and common areas should be counted only once...

AnB is a part of A
AnB is a part of B

so when we write A+B+C we are actully counting AnB twice .... so we have to substratc AnB once to get the final answer ... similar logic for AnC and BnC ....

AnBnC is a part of A, B, C, AnB, AnC and BnC....
so when we add A+B+C and substract - AnB-AnC-BnC we are adding AnBnC 3times and substraticg it 3 times .... so we have to add in once to get the final answer....

HTH
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durgesh79
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SET theory ... BIG CONFUSION 17 Jul 2008, 20:21
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For set theory questions ... i'd suggets to always go by Venn diagrams...

these formula can be confusing ... which one to apply where ....
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alpha_plus_gamma
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SET theory ... BIG CONFUSION 17 Jul 2008, 20:33
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durgesh79
For set theory questions ... i'd suggets to always go by Venn diagrams...

these formula can be confusing ... which one to apply where ....
Show more

Completely agree with you on this.

Most of the times venn diagram is the fastest approach. It gives a clear picture that formulae.
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SET theory ... BIG CONFUSION 18 Jul 2008, 13:12
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alpha_plus_gamma
durgesh79
For set theory questions ... i'd suggets to always go by Venn diagrams...

these formula can be confusing ... which one to apply where ....

Completely agree with you on this.

Most of the times venn diagram is the fastest approach. It gives a clear picture that formulae.
Show more

Sure ... I always do it with venn diagrams but htis time I got confused with deriving formulae for this one.
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rao
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SET theory ... BIG CONFUSION 18 Jul 2008, 13:14
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durgesh79
rao_1857
Another problem that I have is in uderstanding AuBuC

I always get this .. please let me know where I am wrong?

P(AuBuC)=P(A)+P(B)+P(C) - P(AnB) - P(AnC) + P(AnBnC) - P(BnC) + P(AnBnC)

since when I first subtracted P(AnB) that was also including P(AnBnC). So next time when I am substracting P(AnC) and P(AnB), I am adding P(AnBnC) to avoind any duplicaiton

let me try.... AuBuC should include everying that is in A, B and C and common areas should be counted only once...

AnB is a part of A
AnB is a part of B

so when we write A+B+C we are actully counting AnB twice .... so we have to substratc AnB once to get the final answer ... similar logic for AnC and BnC ....

AnBnC is a part of A, B, C, AnB, AnC and BnC....
so when we add A+B+C and substract - AnB-AnC-BnC we are adding AnBnC 3times and substraticg it 3 times .... so we have to add in once to get the final answer....

HTH
Show more

Make sense so far. Hope I dont go too deep into this. My problem is that I go very deep and eventually ned up in confusing myself. So far your explanation makes perfect sense.

Thanks

PS: +1
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SET theory ... BIG CONFUSION 24 Mar 2010, 08:51
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Guys, great work...have been following this for a long time...so whats with the original question?


1. For 3 sets A, B, and C: P(AuBuC) = P(A) + P(B) + P(C) – P(AnB) – P(AnC) – P(BnC) + P(AnBnC)

6. No of persons in atleast one set:
P(A) + P(B) + P(C) - P(AnB) - P(AnC) - P(BnC) + 2 P(AnBnC)



"Can somebody please explain what is the difference between 1 and 6?"

I dont think 6 makes sense...Any takers??



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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
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Thank you for understanding, and happy exploring!