How can I derive this equation?

Question

This equation is from this paper and I can't understand how the right-hand can be derived from the left-hand using Bayes' rule: $$\frac{p(O_{fg} \mid I, I_t)}{p(O_{bg} \mid I, I_t)} = \frac{p(I \mid I_t, O_{fg}) p(O_{fg})}{p(I \mid I_t, O_{bg}) p(O_{bg})}$$

Answer 1

From Bayes' rule:

\begin{align} \frac{p(O_{fg} \mid I, I_t)}{p(O_{bg} \mid I, I_t)}&=\frac{p(I \mid O_{fg}, I_t)p(O_{fg} \mid I_t)}{p(I\mid I_t)} \frac{1}{\frac{p(I \mid O_{bg}, I_t) p(O_{bg} \mid I_t)}{p(I\mid I_t)}}\\ &= \frac{p(I \mid O_{fg}, I_t)p(O_{fg}\mid I_t)}{p(I\mid I_t)} \frac{p(I\mid I_t)}{p(I \mid O_{bg}, I_t) p(O_{bg} \mid I_t)}\\ &= \frac{p(I \mid O_{fg}, I_t)p(O_{fg}\mid I_t)}{p(I \mid O_{bg}, I_t) p(O_{bg} \mid I_t)} \end{align} Now I am not too sure about this topic exactly, but I think that $O_{fg}$ and $O_{bg}$ are independent from $I_t$, in which case $p(O_{fg} \mid I_t)=p(O_{fg})$ and similarly for $p(O_{bg}\mid I_t)=p(O_{bg})$, hence we get $$ \frac{p(O_{fg} \mid I, I_t)}{p(O_{bg} \mid I, I_t)}= \frac{p(I \mid O_{fg}, I_t)p(O_{fg})}{p(I \mid O_{bg}, I_t) p(O_{bg})} $$ as wanted. But please correct me if I'm wrong.