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Math Help - A problem come from "Introduction to Probability Models"

  1. #1
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    A problem come from "Introduction to Probability Models"

    " If the occurrence of B makes A more likely, does the occurrence of A make B more likely?

    I guess yes. But I could not figure out the reason. Thanks~~
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  2. #2
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    Hello, shiningstarpxx!

    An interesting question . . .


    If the occurrence of B makes A more likely,
    does the occurrence of A make B more likely?
    The answer is "Yes".
    We need Bayes' Theorem: . P(X\,|\,Y) \;=\;\frac{P(X \wedge\,Y)}{P(Y)}


    We are told that, if B happens, A is more likely to happen.
    . . That is, the probability of A, given B, is greater than the probability of A.

    In symbols: . P(A\,|\,B) \:> \:P(A)

    This means: . \frac{P(A \wedge B)}{P(B)} \:> \:P(A) \quad\Rightarrow\quad P(A\wedge B) \:>\:P(A)\!\cdot\!P(B)

    Hence, we have: . \frac{P(B \wedge A)}{P(A)} \:>\:P(B) \quad\Rightarrow\quad P(B\,|\,A) \:>\:P(B)


    Therefore, the probability of B, given A, is greater than the probability of B.
    . . The occurence of A makes B more likely.

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  3. #3
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    Quote Originally Posted by shiningstarpxx View Post
    " If the occurrence of B makes A more likely, does the occurrence of A make B more likely?

    I guess yes. But I could not figure out the reason. Thanks~~
    What you're asking is the following:

    Does Pr(A | B) > Pr(A) => Pr(B | A) > Pr(B) ?

    The answer is yes. Consider:

    \Pr(A | B) \, \Pr(B) = \Pr(B | A) \, \Pr(A) \Rightarrow \frac{\Pr(A | B)}{\Pr(A)} = \frac{\Pr(B | A)}{\Pr(B)}.

    But it's given that \Pr(A | B) > \Pr(A) \Rightarrow \frac{\Pr(A | B)}{\Pr(A)} > 1.

    Therefore \frac{\Pr(B | A)}{\Pr(B)} > 1 and the implication follows.


    Edit: Too fast for me this time, Soroban. But I like to think our replies ...... complement each other lol!
    Last edited by mr fantastic; June 2nd 2008 at 05:17 AM. Reason: See edit.
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