Conditional probability for three events I want to calculate 
$P(\neg H\mid I, \neg F)$, 
where 
$p(H) = p(\neg H) = 0.5$, 
$p(I\mid H) = 0.8$, $p(\neg I\mid H) = 0.2$,
$p(I\mid\neg H) = 0.4$, $p(\neg I\mid\neg H) = 0.6$,
$p(F\mid H) = 0.3$, $p(\neg F\mid H) = 0.7$,
$p(F\mid\neg H) = 0.4$, $p(\neg F\mid\neg H) = 0.6$,
and $p(H\mid I) = 2/3$
However, I feel hard to make a formula with bayes rule.
It would be very helpful if somebody can derive it.
Answer:Thanks to Henry, I can understand how it derives. Let me add short answer to the post for whom who would face this problem like me.
$P(\neg H\mid I \cap \neg F) = \dfrac{P(\neg H \cap I \cap \neg F)}{P( I \cap \neg F)}$
$= \dfrac{P(\neg H \cap I) \times P(\neg H \cap \neg F)}{P(\neg H \cap I) \times P(\neg H \cap \neg F) + P(H \cap I) \times P(H \cap \neg F)}$ since $F \perp I$
$= 0.3$

You can calculate 


*

*$P(I \cap H)=0.4$

*$P(\neg I \cap H)=0.1$

*$P(I \cap \neg H)=0.2$

*$P(\neg I \cap \neg H)=0.3$

*$P(F \cap H)=0.15$

*$P(\neg F \cap H)=0.35$

*$P(F \cap \neg H)=0.2$

*$P(\neg F \cap \neg H)=0.3$


and things based on this such as $P(I)=0.6$ and $P(\neg F)=0.65$. 
But what you want to know is $ \dfrac{P(\neg H \cap I \cap \neg F)}{P( I \cap \neg F)}$ and the best you can do is put wide bounds on it.
For example, I think you can have $P(\neg H \cap I \cap \neg F)=0.2$ and $P(H \cap I \cap \neg F)=0.25$ to make your conditional probability $\frac49$.
I also think you can have $P(\neg H \cap I \cap \neg F)=0$ and $P(H \cap I \cap \neg F)=0.35$ to make your conditional probability $0$.
