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.
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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$)$
$P(.|C)$ 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
$$P(A)=P(A,B)+P(A,B^c)$$