venn diagram

[logistic_guy]

here is the question

Find the regions of each event.

(a) \(\displaystyle A \cup C\)
(b) \(\displaystyle B' \cap A\)
(c) \(\displaystyle A \cap B \cap C\)
(d) \(\displaystyle (A \cup B) \cap C'\)



my attemb
(a) and (b) and (c) is very easy
(a) \(\displaystyle A \cup C = 1,2,3,4,5,7\)
(b) \(\displaystyle B' \cap A = 4\)
(c) \(\displaystyle A \cap B \cap C = 1\)
(d) \(\displaystyle (A \cup B) \cap C'\) this i don't know

 

[Dr.Peterson]

here is the question

Find the regions of each event.

(a) \(\displaystyle A \cup C\)
(b) \(\displaystyle B' \cap A\)
(c) \(\displaystyle A \cap B \cap C\)
(d) \(\displaystyle (A \cup B) \cap C'\)

View attachment 38986

my attemb
(a) and (b) and (c) is very easy
(a) \(\displaystyle A \cup C = 1,2,3,4,5,7\)
(b) \(\displaystyle B' \cap A = 4\)
(c) \(\displaystyle A \cap B \cap C = 1\)
(d) \(\displaystyle (A \cup B) \cap C'\) this i don't know

Your answer to (b) is wrong! How did you decide that?

For (d), what is \(\displaystyle (A \cup B)\)? What is \(\displaystyle C'\)? What is their intersection?

 

[logistic_guy]

Your answer to (b) is wrong! How did you decide that?

why wrong?
i want something not in \(\displaystyle B\) but in \(\displaystyle A\)
there's three options
1. \(\displaystyle 4\)
2. \(\displaystyle 7\)
3. \(\displaystyle 4,7\)

first and second option i think equivalent, but i chose \(\displaystyle 4\) because it have the privilige to be in \(\displaystyle C\) as well

For (d), what is \(\displaystyle (A \cup B)\)? What is \(\displaystyle C'\)? What is their intersection?

\(\displaystyle A \cup B = 1,2,3,4,6,7\)
\(\displaystyle C' = 2,6,7\)

i'm not sure about their intersection but i think \(\displaystyle (A \cup B) \cap C' = 2,6,7\). is it correct?

 

[Dr.Peterson]

why wrong?
i want something not in \(\displaystyle B\) but in \(\displaystyle A\)
there's three options
1. \(\displaystyle 4\)
2. \(\displaystyle 7\)
3. \(\displaystyle 4,7\)

first and second option i think equivalent, but i chose \(\displaystyle 4\) because it have the privilige to be in \(\displaystyle C\) as well

But [imath]B'\cap A[/imath] is the set of all elements that are in A but not in B, not just whichever one you happen to like.

\(\displaystyle A \cup B = 1,2,3,4,6,7\)
\(\displaystyle C' = 2,6,7\)

i'm not sure about their intersection but i think \(\displaystyle (A \cup B) \cap C' = 2,6,7\). is it correct?

Again, the intersection is the set of all elements that are in both sets. So, yes, this is correct.

They could have included an element that is not in any of the sets; then [imath]C'[/imath] would be different from [imath](A\cup B)\cap C'[/imath].

 

[logistic_guy]

But [imath]B'\cap A[/imath] is the set of all elements that are in A but not in B, not just whichever one you happen to like.

i miss understand the concept as any point

so \(\displaystyle B' \cap A = 4,7\)

Again, the intersection is the set of all elements that are in both sets. So, yes, this is correct.

i get the idea now. it's all elements

They could have included an element that is not in any of the sets; then [imath]C'[/imath] would be different from [imath](A\cup B)\cap C'[/imath].

you mean like if there is number \(\displaystyle 8\) in region \(\displaystyle S\)

\(\displaystyle A \cup B = 1,2,3,4,6,7\)
\(\displaystyle C' = 2,6,7,8\)

i think the intersection is still the sameso what you mean exactly by different

 

[Dr.Peterson]

you mean like if there is number \(\displaystyle 8\) in region \(\displaystyle S\)

The name S applies to the entire rectangle (the universe), but yes, that outside region where they wrote S is what I mean.

\(\displaystyle A \cup B = 1,2,3,4,6,7\)
\(\displaystyle C' = 2,6,7,8\)

i think the intersection is still the sameso what you mean exactly by different

I meant what I said:

... then [imath]C'[/imath] would be different from [imath](A\cup B)\cap C'[/imath].

In this case, [imath]C'[/imath] would be {2, 6, 7, 8}, while [imath](A\cup B)\cap C'[/imath] would still be {2, 6, 7}, because 8 would not be in [imath](A\cup B)[/imath]. Those are different.

 

[logistic_guy]

The name S applies to the entire rectangle (the universe), but yes, that outside region where they wrote S is what I mean.

I meant what I said:

In this case, [imath]C'[/imath] would be {2, 6, 7, 8}, while [imath](A\cup B)\cap C'[/imath] would still be {2, 6, 7}, because 8 would not be in [imath](A\cup B)[/imath]. Those are different.

i think i get what you mean

in the op it just happen \(\displaystyle C' = (A \cup B) \cap C'\)
but in the new scenario when we include number \(\displaystyle 8\) in the white area of the rectangle
\(\displaystyle C' \neq (A \cup B) \cap C'\)