set theory - doubt about reasoning

[qaopm]

hello.

 

I have a doubt about my thought process on this exercise.

 

we have 3 sets: A,B,C.

 

A intersection B intersection C = 10%

A intersection C = 20%

B intersection C = 20%

A intersection B = 40%

A = 70%

B = 55%

C = 30%

 

what is the complement (A union B union C)?

 

first I determined that:

 

U = 100%

 

then I cut the problem in order to calculate A union B first.

 

A union B = (70+55)-40 = 85 (adding A and B minus common elements, to find the true content of A union B)

(A union B) union C = (85+30)-20-20 = 75 (same procedure, removing the common A intersection C and B intersection C)

 

then I add A intersection B intersection C to the last calculations:

 

75 + 10 (A intersection B intersection C) - and this is the point that I'm not sure about.

 

I add the (A intersection B intersection C) because in that triple intersection lies part of C, that is not present in the intersection between (A union B) intersection C.

 

but I'm not sure that this is the right way to do it... the result is correct: complement (A union B union C) = U - (A union B union C) = 15% but I'm not satisfied wity my reasoning at all.

 

it sounds cheesy.

 

what do you think about it?

 

thanks in advance.

[daniton]

The reason why you have to add AnBnC is because you subtracted it two times not once one in AuB two in (AuB)uC since it is a part of both(try to imagine it in circle or something like that).So, you have to make correction so you did ,you add it.

in general :

AuBuC=A+B+C - AnB - BnC - AnC + AnBnC

[qaopm]

hello and thanks for your answer.

 

I have tried to demonstrate what you suggested, but still I have some doubt about my reasoning.

 

A union B union C = A+B+C - (A intersect B) - (B intersect C) - (A intersect C) + (A intersect B intersect C)

 

I don't know what the + sign means in the set theory and I couldn't find any reference to that sign in set theory online, so I implemented it in a way that seems to be working for the example.

 

consider

 

A = {a; b; c; d; e}

 

B = {d; e; f; g}

 

C = {d; f; h; i}

 

A intersect B = {d; e}

 

A intersect C = {d}

 

B intersect C = {d; f}

 

A intersect B intersect C = {d}

 

so

 

A+B+C = {a; b; c; d; e; d; e; f; g; d; f; h; i}

 

as you see I considered double entries for some elements, since the + operator probably acts differently from the union operator.

 

(A+B+C) - (A intersect B) = {a; b; c; d; e; f; g; d; f; h; i}

 

(A+B+C) - (A intersect B) - (B intersect C) = {a; b; c; d; e; f; g; h; i}

 

(A+B+C) - (A intersect B) - (B intersect C) - (A intersect C) = {a; b; c; e; f; g; h; i}

 

and finally

 

(A+B+C) - (A intersect B) - (B intersect C) - (A intersect C) + (A intersect B intersect C) = {a; b; c; e; f; g; h; i; d}

 

 

is this correct?

 

thanks in advance

[daniton]

you raise good example.You can see that you add AnBnC: d three times when you mix A,B,C than you subtract it three times on AnB,AnC,BnC which you can understand you removed AnBnC from the calculation at all. So, you have to bring it back by adding it specifically.

BTW you can proof it for general case just do this

AUBUC = (AUB)UC = AUB +C - (AuB)nC

 

(AuB)nC= (AnB)u(BnC)

=AnB+BnC-(AnBnBnC)

AnBnBnC=AnBnC

so

AUBUC=A+B+C- AnB - AnC- BnC+ AnBnC

I think you understand the sign change there.

Just give it some thought!!

[qaopm]

hello thanks for the answer.

 

the demonstration I have attempted of your example shows that there is an error:

 

(A union B) intersect C seems not to be equal to (A intersect B) union (B intersect C).

 

since

 

A union B = {a; b; c; d; e; f; g}

 

C = {d; f; h; i}

 

so (A union B) intersect C = {d; f}

 

but

 

(A intersect B) union (B intersect C) = {d; e; f}

 

since

 

(A intersect B) = {d; e}

 

and

 

(B intersect C) = {d; f}

[daniton]

Oh!! my bad correct it with (AnC)u(BnC)

don't you know about the distributive laws of sets.

for more see:

http://en.m.wikipedia.org/wiki/Algebra_of_sets

[qaopm]

dear daniton, thanks again for your answer.

 

but my brain seems to crash with more than two sets. I can't understand the dynamics of three sets operations. I try to look at the diagram but I only seem to make it worse.

 

for example, here is another problem that I completely couldn't understand.

 

"in an exam there are 65 candidates, there are 3 sections. 5 candidates passed the three sections; all candidates that passed the third section have passed also the other two. 20 candidates passed only the first two sections, 3 candidates didn't pass any, and 50 passed the first section. how many candidates have passed only the first, and how many only the second?"

 

so I extracted the data:

 

Universe = 65

 

S1 = 50

 

S1 intersect S2 intersect S3 = 5

 

S1 intersect S2 = 20

 

complementary of S1 union S2 union S3 = 3

 

so I must find

 

S1 - (S2 union S3)

 

and

 

S2 - S3

 

then I calculated

 

S1 union S2 union S3 = 65 - 3 = 62

 

then

 

S1 intersect S3 = 5

 

S1 union S3 = 55

 

and then I'm lost. I've made other calculations but I prefer not to post them here, since I don't understand them, even if the resulting numbers are correct.

[qaopm]

what I am a bit more sure is this calculation:

 

S1 union S2 union S3 - S1 union S3 = S2 - (S1 union S3) = 62-55 = 7

 

and then I have to add the S1 intersect S2 intersect S3 since I removed it one time:

 

7 + 5 = 12

 

and this is S2 - (S1 union S3).

 

but then I'm lost.

 

I can't implement algebra of sets for some reason, my mind refuses it.

Edited June 4, 2013 by qaopm

[daniton]

When you try to find "sth only" for example A only you should subtract any intersection of any thing with A from A but those relation should be them and them only ,those intersection must have no intersection.

if in system of A,B,C

relations

AnC

AnB

AnBnC

AnC only = AnC - AnBnC

AnB only = AnB - AnBnC

AnBnC only = AnBnC

when you subtract all these from A you get :

A only = A - AnC - AnB + AnBnC

try it fresh with this.

Edited June 4, 2013 by daniton

[qaopm]

OK solved!

 

"in an exam there are 65 candidates, there are 3 sections. 5 candidates
passed the three sections; all candidates that passed the third section
have passed also the other two. 20 candidates passed only the first two
sections, 3 candidates didn't pass any, and 50 passed the first section.
how many candidates have passed only the first, and how many only the
second?"

 

U = 65

 

C = 5

 

A intersect B - C = 20

 

complementary of A union B = 3

 

A = 50

 

A - B = ?

B - A = ?

 

A - B = A - (A intersect B) + C = 50 - 20 + 5 = 25

 

A union B = 65 - 3 = 62

 

A union B - A = B - A = 62 - 50 = 12

 

and that's it! =)

 

I was making the wrong diagram, since C is subset of A intersect B.

[daniton]

By the way you should be more careful on the question it is a little tricky. Any way you did great.

Can I ask what grade are you in?