# Probability Question

1. Jan 27, 2013

### GreenPrint

1. The problem statement, all variables and given/known data

Let A and B be two events in a sample space. Under what condition(s) is A$\bigcap (A \bigcup B)^{c}$ empty?

2. Relevant equations

De'Morgan's law

$(A \bigcup B)^{c} = (A^{c} \bigcap B^{c})$

3. The attempt at a solution

A$\bigcap (A \bigcup B)^{c}$

I use De'Morgan's Law

$A \bigcap (A^{c} \bigcap B^{c})$

I don't know if I can do this or not but I think it's what I'm supposed to do.

$(A \bigcap A^{c}) \bigcap (A \bigcap B^{c})$

If I'm not mistaken

$A \bigcap A^{c}$ = 1

so

$1 \bigcap (A \bigcap B^{c})$

If I'm not mistaken this can be simplified some more to

$A \bigcap B^{c}$

So I guess the answer is when

$A \bigcap B^{c} = ∅$

Does this look right?

Thanks for any help!

2. Jan 27, 2013

### Dick

1? What kind of a set is 1? Isn't $A \bigcap A^{c} = \phi$, the empty set?

3. Jan 27, 2013

### haruspex

But unfortunately you are mistaken. Is it what you intended to write... that (in the language of set theory) the intersection of a set with its complement is the universal set?

4. Jan 28, 2013

### GreenPrint

What do you mean by universal set? Now that I think of it I think perhaps empty set is the answer to that. Isn't it empty under all conditions since

∅ = A$\bigcap A^{c}$

5. Jan 28, 2013

### Dick

Yes, it is. It's empty regardless of what B is.

6. Jan 29, 2013

### HallsofIvy

Staff Emeritus
Given a set, A, Ac is the set of all objects that are NOT in A. But what is meant by "all objects"? In order that we not have to include nonsensical things like "Jupiter's fourth moon" or "the fairies that live down by the creek", we have to have a specific "domain of discourse"- all those things that we are talking about. The set containing all things that in set A or B or, in fact, all the things we allow to be in the sets we are talking about is the "universal set". You cannot talk about the "complement of a set" without having a "universal set" so I suspect you just know it by a different name.

Last edited: Jan 29, 2013
7. Jan 29, 2013

### rollingstein

Isn't that always empty?

I used a Venn Diagram

8. Jan 29, 2013

### HallsofIvy

Staff Emeritus
Isn't what always empty? If you are referring to the original post, $A\cap (A \cup B)^c$, yes, that is empty no matter what A and B are. In order to be in $A\cap (A\cup B)^c$, x must be in both A and $(A\cup B)^c$. In order to be in $(A\cup B)^c$, x must not be in $A\cup B$. But $A\cup B$ includes all members of A so that x cannot be in A. There is NO x in $A\cap (A\cup B)^c$.

Last edited: Jan 29, 2013
9. Jan 29, 2013

Yes. Thanks.