# 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 do not seem to know anything about the interaction between $I$ and $F$ – Henry Jan 1 '15 at 18:41
• @Henry Does that mean the problem isn't solvable with just what zedoul gave us? – Zubin Mukerjee Jan 1 '15 at 18:43
• @Zubin Mukerjee: I suspect not – Henry Jan 1 '15 at 18:47

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 $0$.

• Thank you for your detailed solution. Let me ask one more thing. In order to calculate $P(¬H∩I∩¬F)$, first I can divide them into $P(¬H∩I∩¬F)=p(¬F|I∩¬H)p(I|¬H)p(¬H)$ However, how can I calculate $p(¬F|I∩¬H)$ in this setting? – zedoul Jan 1 '15 at 19:10
• I do not think you can. The best you may be able to do is look at possibilities. – Henry Jan 1 '15 at 19:12
• The value for $P(¬H|I,¬F)$ should be $0.3$ just with those setting. Anyway thank you for your answer! – zedoul Jan 1 '15 at 19:19
• To get $0.3$ (which is within the range I gave) you have probably assumed that $I$ and $F$ are conditionally independent give $H$ or given $\neg H$. But the question does not suggest that you can assume this. – Henry Jan 1 '15 at 19:27