Express $P(A|B, C)$ in terms of $P(A), P(B|A)$ and $P(C|A) $ if $B$ and $C$ are independent. Given 3 events $A, B, C$ where $B$ and $C$ are independent, express $P(A|B, C)$ in terms of $P(A), P(B|A)$ and $P(C|A) $.
My attempt: By definition of conditional probability, we have
$P(A|B, C)=P(A, B, C) /P(B, C). $
I got stuck here.
I don't know how independence can simplify the calculation.
Any hint is appreciated. 
 A: Okay... so upon some thought, it seems to me that this is only true if $B$ and $C$ are also assumed conditionally independent given $A$.
Indeed, by two applications of Bayes' rule and the independence of $B$ and $C$, we have
$$
P(A|B,C)=P(B|A,C) \frac{P(B|C)}{P(A|C)}=P(B|A,C) \frac{P(B|C)}{P(C|A)}\frac{P(A)}{P(C)}=P(B|A,C) \frac{P(B)}{P(C|A)}\frac{P(A)}{P(C)}
$$
By symmetry,
$$
P(A|B,C)=P(C|A,B)\frac{P(C)}{P(B|A)}\frac{P(A)}{P(B)}
$$
Hence, multiplying these two expressions together, we get
$$
P(A|B,C)^2=\frac{P(A)^2}{P(B|A)P(C|A)} P(B|A,C)P(C|A,B)=\frac{P(A)^2}{P(B|A)P(C|A)} \frac{P(B,C|A)}{P(C|A)} \frac{P(B,C|A)}{P(B|A)}
$$
Collecting terms, we get that
$$
P(A|B,C)^2=\frac{P(A)^2}{P(B|A)^2P(C|A)^2}\cdot P(B,C|A)^2
$$
Now, the fraction here is a function of $P(A),P(C|A),P(B|A)$, but the latter factor is not.
 Indeed, if it were, then taking $B$ and $C$ to be conditionally indendent given $C$, we would get $P(B,C|A)=P(B|A)P(C|A)$. We wish to demonstrate that this is not always the case.
Indeed, let our probability space be $\{0,1\}^2$ with the uniform measure and let $A=\{(0,0),(1,1)\}$, $B=\{(0,0),(1,0)\}$ and $C=\{(0,0),(0,1)\}$. Then, $P(B|A)=P(C|A)=P(B,C|A)=\frac{1}{2}$ so that the formula $P(B,C|A)=P(B|A)P(C|A)$ (which as argued is the only possible formula) does not hold in general for independent $B$ and $C$. Hence, the above is demonstrably not a function $P(A)$, $P(B|A)$ and $P(C|A)$ unless we add the additional hypothesis that $B$ and $C$ remain independent conditional on $A$.
