1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability Question

  1. Jan 27, 2013 #1
    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[itex]\bigcap (A \bigcup B)^{c}[/itex] empty?


    2. Relevant equations

    De'Morgan's law

    [itex](A \bigcup B)^{c} = (A^{c} \bigcap B^{c})[/itex]

    3. The attempt at a solution

    A[itex]\bigcap (A \bigcup B)^{c}[/itex]

    I use De'Morgan's Law

    [itex]A \bigcap (A^{c} \bigcap B^{c})[/itex]

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

    [itex](A \bigcap A^{c}) \bigcap (A \bigcap B^{c})[/itex]

    If I'm not mistaken

    [itex]A \bigcap A^{c}[/itex] = 1

    so

    [itex] 1 \bigcap (A \bigcap B^{c})[/itex]

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

    [itex]A \bigcap B^{c}[/itex]

    So I guess the answer is when

    [itex] A \bigcap B^{c} = ∅ [/itex]

    Does this look right?

    Thanks for any help!
     
  2. jcsd
  3. Jan 27, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    1? What kind of a set is 1? Isn't [itex]A \bigcap A^{c} = \phi[/itex], the empty set?
     
  4. Jan 27, 2013 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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?
     
  5. Jan 28, 2013 #4
    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[itex]\bigcap A^{c}[/itex]
     
  6. Jan 28, 2013 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, it is. It's empty regardless of what B is.
     
  7. Jan 29, 2013 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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
  8. Jan 29, 2013 #7

    rollingstein

    User Avatar
    Gold Member

    Isn't that always empty?

    I used a Venn Diagram
     
  9. Jan 29, 2013 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Isn't what always empty? If you are referring to the original post, [itex]A\cap (A \cup B)^c[/itex], yes, that is empty no matter what A and B are. In order to be in [itex]A\cap (A\cup B)^c[/itex], x must be in both A and [itex](A\cup B)^c[/itex]. In order to be in [itex](A\cup B)^c[/itex], x must not be in [itex]A\cup B[/itex]. But [itex]A\cup B[/itex] includes all members of A so that x cannot be in A. There is NO x in [itex]A\cap (A\cup B)^c[/itex].
     
    Last edited: Jan 29, 2013
  10. Jan 29, 2013 #9

    rollingstein

    User Avatar
    Gold Member

    Yes. Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Probability Question
  1. A probability question (Replies: 3)

  2. Probability Question (Replies: 2)

  3. Probability Question (Replies: 12)

Loading...