Results 1 to 4 of 4
Like Tree2Thanks
  • 1 Post By emakarov
  • 1 Post By emakarov

Math Help - Conditional Probability Proof

  1. #1
    Newbie
    Joined
    Mar 2012
    From
    Netherlands
    Posts
    5

    Conditional Probability Proof

    Hello,

    I am new here! Nice forum by the way.

    So i have a weird proof to solve there that kind of remind this one but its seems that its quite different, after all.

    B, C are mutually exclusive.

    And i have to proove this:

    P(A | B ∪ C) = [ P(B) * P(A | B) / P(B) + P(C) ] + [ P(C) * P(A | C) / P(B) + P(C) ]

    I have tried load of things, but mostly i am doing cycles around. I am definatelly missing some keypoint ?

    Shoud i try proove it by taking the long part and reach P(A | B ∪ C) , or should i take small part reach the big one?

    Honestly i have tryed both way and nothing absolutely nothing.


    On my best attempt i reached this point (by taking the long part trying to reach the 1st part):

    P(A∩B) + P(A∩C) / P(B) + P(C) and i cant continue from this point. i mean that if "+" was like "or" i could probably say that

    P(A∩B) + P(A∩C) / P(B) + P(C) = P(A∩(B∪C)) / P(B) +P(C) and finish it here (but its wrong)...

    Any ideas there ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,530
    Thanks
    774

    Re: Conditional Probability Proof

    Quote Originally Posted by Yonas View Post
    And i have to proove this:

    P(A | B ∪ C) = [ P(B) * P(A | B) / P(B) + P(C) ] + [ P(C) * P(A | C) / P(B) + P(C) ]
    You do know that in the standard order or operations division comes before addition, right?

    Quote Originally Posted by Yonas View Post
    On my best attempt i reached this point (by taking the long part trying to reach the 1st part):

    (P(A∩B) + P(A∩C)) / (P(B) + P(C)) and i cant continue from this point. i mean that if "+" was like "or" i could probably say that

    (P(A∩B) + P(A∩C)) / (P(B) + P(C)) = P(A∩(B∪C)) / (P(B) +P(C)) and finish it here (but its wrong)...
    It's not wrong. First, since B and C are mutually exclusive (i.e., are disjoint subsets of the sample space), we have P(B) + P(C) = P(B ∪ C). Second, for the same reason A ∩ B and A ∩ C are also disjoint, so P(A ∩ B) + P(A ∩ C) = P((A ∩ B) ∪ (A ∩ C)) = P(A ∩ (B ∪ C)).
    Thanks from Yonas
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2012
    From
    Netherlands
    Posts
    5

    Re: Conditional Probability Proof

    Quote Originally Posted by emakarov View Post
    You do know that in the standard order or operations division comes before addition, right?
    Yes mate, i know that, and its quite trashy from me not using parentheses, excuse me.

    Quote Originally Posted by emakarov View Post
    It's not wrong. First, since B and C are mutually exclusive (i.e., are disjoint subsets of the sample space), we have P(B) + P(C) = P(B ∪ C). Second, for the same reason A ∩ B and A ∩ C are also disjoint, so P(A ∩ B) + P(A ∩ C) = P((A ∩ B) ∪ (A ∩ C)) = P(A ∩ (B ∪ C)).
    Oh, i see! Quite enlighting answer pal. Thank you a lot.

    So, do you think my way to proove going from [P(B) * P(A | B) / ( P(B) + P(C) ) ] + [ P(C) * P(A | C) / ( P(B) + P(C) ) ] to P(A | B ∪ C) is right, just like i described it later? I mean, do i really proove it like this, it seems that i really do, but i cant be sure like 100% , i am not very familiar with this field. Can you call this "legitimate" proof ?

    Thanks
    Last edited by Yonas; March 24th 2012 at 05:43 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,530
    Thanks
    774

    Re: Conditional Probability Proof

    Ultimately, it does not matter whether you prove equality by converting the left-hand side into the right-hand side or vice versa. It may be easier to come up with a transformation in one direction, but once the chain of equalities is written, it can be read both ways.

    If you'd like, post a complete proof for a feedback.
    Thanks from Yonas
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. conditional probability proof
    Posted in the Advanced Statistics Forum
    Replies: 6
    Last Post: October 15th 2010, 06:21 PM
  2. conditional probability proof...
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: September 23rd 2010, 03:53 PM
  3. Conditional probability proof
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: July 11th 2010, 12:02 PM
  4. Conditional Probability Proof
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: September 22nd 2009, 10:37 AM
  5. conditional probability proof
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: September 8th 2009, 01:52 PM

Search Tags


/mathhelpforum @mathhelpforum