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I have a problem that I can't work out

I've two conditional independent A,B such as

$P(A,B|C) = P(A|C)P(B|C)$

Now I've to find posterior formula for:

$P(C | A,B)$, now what I got was pretty straigthforward application of bayes:

$\frac{P(B|C)P(A|C)P(A)}{P(A\cap B)}$

With few variants (e.g. get an intersection on numerator)

but I can't get the lecturer solution that is:

$\frac{P(B|C)P(C|A)}{P(B|A)}$

Any help?

(note is not homework, but self studying on some pdfs)

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Note that from the assumption you get: $\mathbb P(B|A,C)=\mathbb P(B|C)$. Therefore: $$ \mathbb P(C|A,B)\mathbb P(B|A)=\mathbb P(B,C|A)=\mathbb P(C|A)\mathbb P(B|C,A)=\mathbb P(C|A)\mathbb P(B|C). $$

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