Compliments/Universe

[Caldus]

OK, if I have sets such as:

|A| = 7
|B| = 10
|A intersection B| = 5
(And the universe equaled 19)

Then how do you find the compliment of |A intersection B|? And what would the answer be in this case? Thanks.

[matt grime]

Do you really want to take the complement of a cardinality?

[ahrkron]

You may find useful to make a drawing of the situation.

[Caldus]

OK, here is the problem:

If Universe = 19 and:
A = People who dislike NDP
B = People who dislike Liberals
C = People who dislike Conservatives
D = People who dislike Canadian Alliance
|A| = 7
|B| = 10
|C| = 11
|D| = 6
|B intersection A| = 5
|A intersection C| = 5
|B intersection C| = 6
|A intersection D| = 3
|B intersection D| = 4
|C intersection D| = 5
|C intersection B intersection A| = 3
|B intersection A intersection D| = 2
|C intersection A intersection D| = 3
|C intersection B intersection D| = 4
|A intersection B intersection C intersection D| = 2

Are all given, then how many like all 4 parties (not dislike)? Can someone point me in the right direction for this? Thank you.

[Zurtex]

I've never done this before so this may be a really stupid comment, but wouldn't it make it just that little easier if:

|A intersection B intersection C intersection D| = 2

Then you can delete this line and take 2 away from all these 4 lines:

|A| = 7
|B| = 10
|C| = 11
|D| = 6

And the universe.

[Caldus]

Not sure...

(Not really sure where to start with this myself...)

[matt grime]

You want A^c\cap B^c\cap C^c \cap D^C call this set E

Let U denote the set of all people asked (the universe)

By definition E = (U\A)n(U\B)n(U\C)n(U\D)

can you work with the rules of sets to simplify that?

Or can you think of a better way of doing it? Such as: (AuB)^c = (A^c)n(B^c)?