Trying to solve a set problem given several sets

Sep 2016
10
0
San Diego, California
Hello,

I've been trying to figure out how to solve this question below:

Let A, B, and C be sets in a universal set U. We are given n(U) = 82, n(A) = 46, n(B) = 41, n(C) = 42, n(A [FONT=&quot]∩[/FONT] B) = 25, n(A [FONT=&quot]∩[/FONT] C) = 18, n(B [FONT=&quot]∩[/FONT] C) = 23, n(A [FONT=&quot]∩[/FONT] B [FONT=&quot]∩[/FONT] CC) = 13. Find the following values.(a) n(AC [FONT=&quot]∩[/FONT] B [FONT=&quot]∩[/FONT] C)


(b) n(A [FONT=&quot]∩[/FONT] BC [FONT=&quot]∩[/FONT] CC)


Unfortunately, I've had no luck in solving this. My math lecturer did not bother to go over this type of problem as it is more complicated than what we have been doing. I understand the basic theorem that is supposed to help solve this but I still do not know even where to start. If anyone could explain this problem or potentially solve it, that would be a huge help. Even seeing how to just set everything up would be helpful at this point.

Thank you!



 
Last edited:

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Looks like a candidate for a Venn Diagram! Draw three interlocking circles and label them "A", "B", and "C". The three circles divide into 7 non-overlapping regions. "\(\displaystyle A\cap B\)" is the section where circles "A" and "B" overlap. It include the little section where all three circles overlap but "\(\displaystyle A\cap B\cap C^c\)" is the \(\displaystyle A\cap B\) without that last part. Write the number "13" in that. We are also told that \(\displaystyle n(A\cap B)= 25\) so "\(\displaystyle n(A\cap B\cap C\)" is 25- 13= 12. Write "12" in that little section. That should get you back to the more common kind of problem.
 
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Sep 2016
10
0
San Diego, California
I'm sorry I am still not following. I just think I don't understand how to solve this we've been doing much more simple problems than this in class.
 

romsek

MHF Helper
Nov 2013
6,725
3,030
California
You should really draw yourself a diagram of 3 overlapping circles and start filling it bit by bit.

If you look carefully (or use basic set theory) you'll see that

$A \cap B = (A \cap B \cap C ) \cup (A \cap B \cap C^c)$

we know that $N(A \cap B)=25$ and $N(A \cap B \cap C^c)=13$

so $N(A \cap B \cap C) = 25-13 = 12$

we can further note that

$B \cap C = (A \cap B \cap C ) \cup (A^c \cap B \cap C)$

we know $N(B \cap C) = 23$ and now we know $N(A \cap B \cap C)=12$ so

$N(A^c \cap B \cap C)=23 -12 = 11$

that answers (a)

the next piece you should find is $N(A \cap B^c \cap C)$

this is easy, you are told $N(A \cap C)$ and you now know $N(A \cap B \cap C)$

then given that, since you know $N(A)$ you can find $N(A \cap B^c \cap C^c)$
 
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romsek

MHF Helper
Nov 2013
6,725
3,030
California
I should add to the previous post that the reason we can just add the $N()$ of each set is that the sets mentioned are disjoint.
 
Sep 2016
10
0
San Diego, California
I got 12 as the answer for the last part. Do you know if this is correct?
 
May 2009
612
334
I got 12 as the answer for the last part. Do you know if this is correct?
I didn't get that.

n(A) = n(A ∩ BC ∩ CC) + n(A ∩ B ∩ CC) + n(A ∩ BC ∩ C) + n(A ∩ B ∩ C)
46 = n(A ∩ BC ∩ CC) + 13 + n(A ∩ BC ∩ C) + 12

What did you get for n(A ∩ BC ∩ C)?


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