Q:

Conditional probability given two events

Is the following probability correct given that $W$ and $H$ depend on $G$?
$$P(G\mid w,h) = \frac{P(w\mid G)\cdot P(h\mid G)\cdot P(G)}{\left(P(w\mid G) P(G)+P(w\mid \overline{G}) P(\overline{G})\right) \cdot \left(P(h\mid G) P(G)+P(h\mid \overline{G}) P(\overline{G})\right)} $$

A:

More information on the dependancies of W and H is needed.
Generally:
$$\mathsf P(G\mid W,H) ~=~ \dfrac{\mathsf P(W, H\mid G)~\mathsf P(G)}{\mathsf P(W, H)}$$
Now, if events $W$ and $H$ depend on event $G$ but are conditionally independent given $G$, then the numerator you want occurs:
$$\mathsf P(G\mid W,H) ~=~ \dfrac{\color{navy}{\mathsf P(W\mid G)~\mathsf P(H\mid G)~\mathsf P(G)}}{\mathsf P(W\mid G)~\mathsf P(H\mid G)~\mathsf P(G)+\mathsf P(W\mid\overline G)~\mathsf P(H\mid\overline G)~\mathsf P(\overline G)}$$
On the other hand if events $W$ and $H$ depend on event $G$ but are pairwise independent, then the denominator you want occurs:
$$\begin{align}
\mathsf P(G\mid W,H) ~=~& \dfrac{\mathsf P(W, H\mid G)~\mathsf P(G)}{\mathsf P(W)~\mathsf P(H)}
\\ = ~& \dfrac{\mathsf P(W, H\mid G)~\mathsf P(G)}{\color{navy}{\Big(\mathsf P(W\mid G)~\mathsf P(G)+\mathsf P(W\mid\overline G)~\mathsf P(\overline G)\Big)~\Big(\mathsf P(H\mid G)~\mathsf P(G)+\mathsf P(H\mid\overline G)~\mathsf P(\overline G)\Big)}}
\end{align}$$
To mix and match, both scenarios would need to be so.

