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  1. #1
    Occasional Member Evan's Avatar
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    Default Help with probability

    I know I'm probably missing something really easy, but I can't figure this out right now and I'm burnt out from hours of programming.

    Anyone?

    (The easy answers seem too easy, but they might be right; I can't even think)
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  2. #2
    mountain surfing nomadic's Avatar
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    i think since...

    (0,0) = .08
    (1,0) = .01

    so (1,1) + (0,1) should = .91

    im guessing (1,1) * 8 = (0,1)

    (1,1) = .101 and (0,1) = .808

    can't help but wonder if bayes method applies more tho... too hung over to look it up for you...

  3. #3
    Occasional Member Evan's Avatar
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    oops i just realized this is super easy. sorry y'all.

  4. #4

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    Quote Originally Posted by dissonance View Post
    oops i just realized this is super easy. sorry y'all.
    Really? It seems like you are missing information.

    If you are assuming Y is independent of X because Joe's grading is not influenced by the professors writing, then P(Y=1|X=1)=P(Y=1)=1-P(Y=0)=1-[P(X=0,Y=0)+P(X=1,Y=0)]=1-[0.08+0.01]=0.91.

    However, if we believe that Joe cannot grade a problem set unless the professor is writing...P(Y=1|X=0)=0 which means P(Y=1,X=1)=0.91, which means P(Y=1|X=1)=P(Y=1,X=1)/P(X=1)=0.91/(0.01+0.91)=0.989 approximately.

    I'm wondering how you solved the conundrum. Independence seems like a more plausible interpretation.

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