Question: Let's say we have a problem of predicting whether a storm is coming or not. We have a model `P(storm is coming | how many clouds are outside)`, and have another model `P(storm is coming | how scared the dogs are)`.

My question in general is how to combine these two models in a sensible way. It seems to make sense if I just multiply the two together - however, I don't know what this expression evaluates to:

`P(storm is coming | how many clouds are outside) * P(storm is coming | how scared the dogs are) = ?`

Or, with simpler notation:

`P(A|B)P(A|C) = ?` (where B and C are independent)

~~I have a feeling the answer is `P(A|B)P(A|C) = P(A|B,C)` but I can't prove it.~~ (As many pointed out, this is wrong)

Edit: Sorry for the confusion, the question here is "what is `P(A|B)P(A|C) = ?`"

Answer: 
Not really a simplification (if that is possible at all) but, if you wish a relation between $P(a \vert b)P(a \vert c)$ and $P(a \vert b,c)$, you could say that 

$P(a \vert b)P(a \vert c) = P(a \vert b,c)^2 \frac{P(c \vert b)}{P(c \vert a,b)} \frac{P(b \vert c)}{P(b \vert a,c)}$ 

I think that this relation is more theoretic and fun than useful. In practical cases you would use the simpler

$P(a \vert b) = P(a \vert b,c) \frac{P(c \vert b)}{P(c \vert a,b)}$ 

$P(a \vert c) = P(a \vert b,c) \frac{P(b \vert c)}{P(b \vert a,c)}$

one use is maybe that this tells how you can't give a usefull interpretation for $P(a \vert b)P(a \vert c)$. Say $P(\text{cancer} \vert \text{smoking})=0.1$ and $P(\text{cancer} \vert \text{sports})=0.001$ then how these relate to $P(\text{cancer} \vert \text{smoking and sports})^2$ is dependent on the terms in the fractions and it can be much different from $0.0001$ either larger or smaller and depends on the case.
As JonMark Perry has already mentioned, Bayes' theorem prohibits that your initial suspicion is true. The rules for conditional probabilities specifically allow for multiplication of probabilities only if conditioning under the same event (either B or C or both simultaneously). To show you a visualistion of the 2 probabilities take a look at [this image of a tree diagram.](http://maths.nayland.school.nz/Year_13_Maths/3.13_Probability/Images_3.3_Probability/Probab8.jpg). Within a branch of the tree, you may multiply the probabilities which is interpreted as both events having to happen simultaneously P(A and B) = P(A|B) \* P(B) or P(A and C) = P(A|C) \* P(C) . In order to combine to two different branches (the directions would be B and C) the probabilities have to be added. P(A|B) \* P(B)+ P(A|C) \* P(C) is then the probability that the storm comes up, independent of the clouds coming up or how scared the dogs are.