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Thread: Conditinal probability theorem

  1. #1
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    Conditinal probability theorem

    I have to either prove or disprove the following:

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

    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 $\displaystyle P(A) \leq P(B) $

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

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

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

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

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

    Thank you for your help.
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  2. #2
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    Quote Originally Posted by Jame View Post
    I have to either prove or disprove the following:
    If $\displaystyle P(A) \leq P(B)$, then $\displaystyle P(A|C) \leq P(B|C)$ (Assuming $\displaystyle P(C) \geq 0$)
    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 $\displaystyle P(A) \leq P(B) $
    i.e. $\displaystyle \frac{1}{6} \leq \frac{3}{6}$
    but I believe that $\displaystyle P(B|C) = 0$

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

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

    So this contradicts the hypothesis, but I am not sure if I calculated the probabilities of $\displaystyle P(A \cap C)$ and $\displaystyle P(B \cap C)$ right.
    That is correct.
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  3. #3
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    Woot excellent. Thank you for your help!
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