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Hey everyone I have a question about deconstructing this conditional probability. After using Bayes Rule and then attempting to use $P(A | B,C) = P(A,B | B,C)$ and $P(A|B,C) = \frac{P(A,B | C)}{P(B|C)}$ I can not for the life of me figure out how to make both sides equal.

See this Photo

Thank You

P.S. First time poster so hopefully everything is ok with question, tried searching prior to asking. $($Already asked this on wrong forum $-1$ for me$)$

1 Answers 1

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$P(.|C)$ is a Probability function itself. So

$$P(A|B,C)=\frac{P(A,B|C)}{P(B|C)}$$ Just like $$P(A|B)=\frac{P(A,B)}{P(B)}$$

For the second question, Analogously, it's only the Total Probability Theorem. Just Like

$$P(A)=P(A,B)+P(A,B^c)$$

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    Ah I see I didn't think you could expand the Total Probability Theorem in that way. Thank you so much for the edit and explanation. Mersi – 2017-02-13
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    @user415853 Your welcome... Mersi with "S"?! It's somehow weird! – 2017-02-13
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    Merci* it's been a long math-filled night aha thanks again – 2017-02-13