Results 1 to 3 of 3

Math Help - Conditinal probability theorem

  1. #1
    Member
    Joined
    Feb 2011
    Posts
    83
    Thanks
    2

    Conditinal probability theorem

    I have to either prove or disprove the following:

    If P(A) \leq P(B), then  P(A|C) \leq  P(B|C) (Assuming P(C) \geq 0)

    After trying to prove this generally a million times and failing, I decided to find a counter example, I'm not sure if this works though.

    All the following events have to do with rolling one fair dice.
    A is getting a "2"
    B is getting an odd number
    C is getting an even number

    Now P(A) \leq P(B)

    i.e. \frac{1}{6} \leq \frac{3}{6}

    but I believe that P(B|C) = 0

    i.e P( B \cap C) = 0

    and P(A|C) = \frac{1/6}{3/6} = \frac{1}{3} (i.e P(A \cap C) = \frac{1}{6}

    So this contradicts the hypothesis, but I am not sure if I calculated the probabilities of P(A \cap C) and P(B \cap C) right.

    Thank you for your help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,792
    Thanks
    1687
    Awards
    1
    Quote Originally Posted by Jame View Post
    I have to either prove or disprove the following:
    If P(A) \leq P(B), then  P(A|C) \leq  P(B|C) (Assuming P(C) \geq 0)
    All the following events have to do with rolling one fair dice.
    A is getting a "2"
    B is getting an odd number
    C is getting an even number

    Now P(A) \leq P(B)
    i.e. \frac{1}{6} \leq \frac{3}{6}
    but I believe that P(B|C) = 0

    i.e P( B \cap C) = 0

    and P(A|C) = \frac{1/6}{3/6} = \frac{1}{3} (i.e P(A \cap C) = \frac{1}{6}

    So this contradicts the hypothesis, but I am not sure if I calculated the probabilities of P(A \cap C) and P(B \cap C) right.
    That is correct.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2011
    Posts
    83
    Thanks
    2
    Woot excellent. Thank you for your help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Conditinal Probability with numbers selected at random
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: October 6th 2010, 08:35 PM
  2. Probability-Bayes Theorem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: May 5th 2010, 11:13 PM
  3. Proof of total probability theorem
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: December 23rd 2009, 12:25 AM
  4. Replies: 2
    Last Post: August 24th 2008, 01:12 PM
  5. probability bayes theorem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 16th 2007, 07:55 PM

Search Tags


/mathhelpforum @mathhelpforum