Results 1 to 8 of 8

Math Help - independence

  1. #1
    Newbie
    Joined
    Aug 2009
    From
    Sydney
    Posts
    15

    independence

    Hi, im just confirming if this logic is right, for my assignment. Well i know that events are mutually exclusive if  P (A \cap B) = 0 . And then this is not independent.

    so if i wanted to determine is something is independent then is it just if  P (A \cap B) is not equal to 0?

    coz in this subject we havent covered stuff like chi-squares so im not allowed to do that. thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Katina88 View Post
    Hi, im just confirming if this logic is right, for my assignment. Well i know that events are mutually exclusive if  P (A \cap B) = 0 . And then this is not independent.

    so if i wanted to determine is something is independent then is it just if  P (A \cap B) is not equal to 0?

    coz in this subject we havent covered stuff like chi-squares so im not allowed to do that. thanks
    The test for the independence of two events is clearly stated here: http://en.wikipedia.org/wiki/Indepen...ability_theory)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2009
    From
    Sydney
    Posts
    15
    ok then is this right:
    if
     P(A|B) = P(A|B') = P(A) then it is independent?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Katina88 View Post
    ok then is this right:
    if
     P(A|B) = P(A|B') = P(A) then it is independent?
    Yes. But the condition \Pr(A \cap B) = \Pr(A) \cdot \Pr(B) might be the better one to apply.
    Last edited by mr fantastic; September 5th 2009 at 10:38 PM. Reason: Fixed a typo - thanks baldeagle.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Quote Originally Posted by mr fantastic View Post
    Yes. But the condition \Pr(A \cup B) = \Pr(A) \cdot \Pr(B) might be the better one to apply.

    OMG that's a BAD typo
    you mean intersection and not union here.
    Another clarification...
    Mutually exclusive or disjoint means that the intersection of two sets is empty,
    THAT implies that P(AB)=0.
    BUT P(AB)=0 does not mean they are disjoint.

    Also If two sets are independent then..........

    P(A|B)=P(A)=P(A|B')
    and
    P(B|A)=P(B)=P(B|A')
    and
    P(AB)=P(A)P(B)
    and
    P(AB')=P(A)P(B')
    and
    P(A'B)=P(A')P(B)
    and
    P(A'B')=P(A')P(B')


    IF any one hold, ALL HOLD.
    And if one fails, they all do.



    NOW to the question at hand, which I did in class 2 weeks ago.
    IF A and B are disjoint (and they have positive probability) then they are dependent.

    That's because P(AB)=0 while P(A)P(B)>0, so

    P(AB)\ne P(A)P(B)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    Quote Originally Posted by Katina88 View Post
    ok then is this right:
    if
     P(A|B) = P(A|B') = P(A) then it is independent?
    If  P(A|B) = P(A) they are independent

    It then follows that  P(A|B') = P(A) too.

    You only need show that 2 of these 3 are equal
     P(A|B) = P(A|B') = P(A)
    the other must also be equal to the first two.

    These are the more intuitive definitions.
    IF A and B are independent then knowing one does not change the probability of the other.
     P(A|B)=P(A)
    Last edited by matheagle; September 6th 2009 at 12:44 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Aug 2009
    From
    Sydney
    Posts
    15
    Quote Originally Posted by matheagle View Post
    OMG that's a BAD typo
    lol i was thinking that i was doing something wrong. hehe thanks
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    night
    Last edited by matheagle; September 6th 2009 at 09:33 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: November 15th 2010, 01:24 PM
  2. Independence
    Posted in the Statistics Forum
    Replies: 0
    Last Post: October 26th 2010, 04:20 PM
  3. Independence
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: February 6th 2010, 07:41 PM
  4. independence
    Posted in the Advanced Statistics Forum
    Replies: 8
    Last Post: January 27th 2010, 10:29 PM
  5. Independence with mod
    Posted in the Number Theory Forum
    Replies: 5
    Last Post: September 18th 2009, 10:53 PM

Search Tags


/mathhelpforum @mathhelpforum