Results 1 to 8 of 8

Math Help - Probability proof?

  1. #1
    Junior Member
    Joined
    Feb 2009
    Posts
    57

    Probability proof?

    This is really confusing to me... how could I prove this?

    If A1, A2, and A3 are three events and the union of A1 and A2 != 0 and the union of A1 and A3 !=0 but the union of A2 and A3 = 0 show that

    P(at least one Ai) = P(A1) + P(A2) + P(A3) - 2P(A1 n A2)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by vexiked View Post
    This is really confusing to me... how could I prove this?

    If A1, A2, and A3 are three events and the union of A1 and A2 != 0 and the union of A1 and A3 !=0 but the union of A2 and A3 = 0 show that

    P(at least one Ai) = P(A1) + P(A2) + P(A3) - 2P(A1 n A2)
    This can be found here : Inclusion-exclusion principle - Wikipedia, the free encyclopedia


    For a simple proof... :
    P(at least one Ai)= \mathbb{P}(A_1\cup A_2\cup A_3)

    Now we know that \mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)

    Apply this to (A_1\cup A_2) and A_3

    Then try to go from here.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2009
    Posts
    57
    I think I got it with what you have shown but not sure....

    P(A1 n A2 n A3) = P(A1) + P(A2) + P(A3) - P(A1 n A2) - P(A1 n A2) + P(A1 n A2 n A3)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2009
    Posts
    57
    Can someone help me out with this one? I'm not sure if I am following the above posters directions correctly.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,956
    Thanks
    1780
    Awards
    1
    Quote Originally Posted by vexiked View Post
    Can someone help me out with this one? I'm not sure if I am following the above posters directions correctly.
    Look at your question.

    Quote Originally Posted by vexiked View Post
    If A1, A2, and A3 are three events and the union of A1 and A2 != 0 and the union of A1 and A3 !=0 but the union of A2 and A3 = 0 show that
    P(at least one Ai) = P(A1) + P(A2) + P(A3) - 2P(A1 n A2)
    .
    There very little there that makes any sense what so ever.
    The notation is incorrect; the vocabulary is wrong.
    Example: “union of A1 and A2 != 0” has no meaning whatsoever.
    Neither does “the union of A2 and A3 = 0”.

    We cannot possibly answer a posting that cannot be read.
    You post is not readable.
    I suspect that you don’t know union from intersection.

    The inclusion/exclusion principle applied to probability function is:
     P\left( {A \cup B \cup C} \right) = P\left( A \right) + P\left( B \right) + P\left( C \right) - P\left( {A \cap B} \right) - P\left( {A \cap C} \right) - P\left( {C \cap B} \right) + P\left( {A \cap B \cap C} \right)
    Last edited by mr fantastic; October 1st 2009 at 02:48 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Feb 2009
    Posts
    57
    Oh I am sorry. I definately meant to use intersection.. here it goes with some lateX. Thanks again.

    If A1, A2, and A3 are three events. The events A2 and A3 are mutually exclusive.

    Show that
    <br />
P\left( at least one Ai \right) = P\left( A1 \right) + P\left( A2\right) + P\left( A3 \right) - 2P\left( {A1 \cap A2} \right)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,956
    Thanks
    1780
    Awards
    1
    Quote Originally Posted by vexiked View Post
    Oh I am sorry. I definately meant to use intersection.. here it goes with some lateX. Thanks again.

    If A1, A2, and A3 are three events. The events A2 and A3 are mutually exclusive.

    Show that
    <br />
P\left( at least one Ai \right) = P\left( A_1 \right) + P\left( A_2\right) + P\left( A_3 \right) - \color{red}{2P\left( {A_1 \cap A_2} \right)}
    Well that is false unless you are given that:
    P\left( A_1 \cap A_2 \right)=P\left( {A_1 \cap A_3} \right) and A_2\cap A_3=\emptyset

    BTW: Learn to make subscripts.
    [tex]A_2\cap A_3=\emptyset[/tex] gives A_2\cap A_3=\emptyset.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Feb 2009
    Posts
    57
    Thanks for your help. The question should read

    If A_1, A_2, A_3 are three events and P\left(A_1 \cap A_2 \right) = P\left(A_1 \cap A_3 \right) \neq 0 but P\left(A_2 \cap A_3\right) = 0, show that,
    <br />
P\left( at least one Ai \right) = P\left( A1 \right) + P\left( A2\right) + P\left( A3 \right) - 2P\left( {A1 \cap A2} \right)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Probability proof
    Posted in the Statistics Forum
    Replies: 3
    Last Post: November 4th 2011, 08:12 AM
  2. probability proof
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: March 15th 2011, 04:52 AM
  3. help with probability proof
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: March 12th 2009, 10:08 AM
  4. probability proof help
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: September 17th 2008, 08:39 AM
  5. Probability Proof
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: September 15th 2008, 04:27 PM

Search Tags


/mathhelpforum @mathhelpforum