• mofoss@alien.topB
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    1 year ago

    P(K=1) = 1/2

    P(a=1|K=1) = P(a=1,K=1)/P(K=1) = (1/4)/(1/2)=1/2

    P(b=1|K=1) = P(b=1,K=1)/P(K=1) = (1/8)/(1/2)=1/4

    P(c=0|K=1) = P(c=0, K=1)/P(K=1) = (1/4)/(1/2)=1/2

    P(a=1, b=1, c=0, K=1) = 0

    P(a=1, b=1, c=0, K=0) = 1/8

    [0.5 * 0.25 * 0.5] / (0 + 1/8) = (1/16) / (1/8) = 1/2

    For conditionals, convert it into joints and priors first and THEN use the table to count instances out of N samples.

    P(X|Y) = P(X,Y)/P(Y)

    :)

    • Terrible_Button_1763@alien.topB
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      1 year ago

      At the very least your calculation does not agree with your formula of P(X|Y) = P(X,Y)/P(Y).

      How is the numerator a calculation of P(X,Y)? [0.5 * 0.25 * 0.5] is P(a = 1 | K =1 ) * P(b = 1 | K = 1) * P(c = 0 | K = 1) which is (in Naive Bayes) P(X|Y) by Naive Bayes and not P(X, Y).

      • mofoss@alien.topB
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        1 year ago

        Uh not sure what Fubini’s theorem is, I just use the equivalence of P(X|Y)P(Y) = P(Y|X)P(X) = P(X,Y)

        • Terrible_Button_1763@alien.topB
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          1 year ago

          That’s not what the questions is asking. And that’s not Bayes’ rule. The denominator is not even calculating P(Y) under Naive Bayes.

          Hmm, maybe machine learning is not just import tensorflow/pytorch/llm.

          • mofoss@alien.topB
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            1 year ago

            Features are independent when conditioned on the dependent is pretty much what I know about Naive Bayes, I personally don’t care for the semantics.

            Also the last time I was using naive bayes was grad school 7 years ago so things are fuzzy, sorry

    • Kruki37@alien.topB
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      1 year ago

      Seems like you dropped one of the 1/2s from the numerator. Maybe I’m missing something but the answer looks like 1/4 to me as your workings show