Probability of the event knowing that I received no informations First I want to thank you if you pay attention to my post. I apologize if it seems elementary to you, note that I searched a lot an answer before posting.

I have a particular informational framework where there are 3 events denoted A, B, C.

A= This guy is cool.

$\overline{A}$= This guy is not cool

B= Someone is here

$\overline{B}$= No one is here

C= He tells me if he is cool

$\overline{C}$= He does not tell me if he is cool

P(A), P(B) and P(C) are the probabilities associated with these events.

I assume that if the person who can advise me is not here, she cannot tell me something.

What is the probability that this guy is cool given the fact that no one is telling me something?

I propose the following formula:

$P(A|\emptyset)=\frac{p(A\cap B \cap \overline{C})+p(A\cap\overline{B})}{P(\emptyset)}$

Where $P(\emptyset)=P(A\cap B \cap \overline{C})+P(\overline{A}\cap B \cap \overline{C})+p(\overline{B})$.

Knowing the Bayes rule the previous equation bothers me because it contains a kind of union between events on the numerator. Maybe this is because there is a kind of redundancy between event B and C. My question is, is it the correct formula to compute the conditional probability in this framework?

Thank you
The right denominator should be $P(\bar C)$, and the right numerator should be $P(A\cap \bar C)$ since $\bar C$ is the event of no receiving information, either because the guy is not there or the guy is there but gives no information. The denominator should not depend on the true state of his "coolness". 

An easy way to picture this problem is by doing a diagram. Think of a square and divide it in half, say vertically where the mass (or probability) in the left side represents $P(A)$ and complementary probability in the right. Then divide the square in another half, say horizontally, so that the mass in the top half represents $P(B)$. each of the resulting smaller squares contains the mass of each of the possible intersections between these two events. Finally, you can divide the top half (created in the second step) in half (say horizontally again) where that top part represents $P(C)$, and the bottom part (from this last line up to the bottom of the square) represents $P(\bar C)$. Clearly, $C$ does not intersect with $\bar B$ and $\bar C$ contains $\bar B$, as desired. And you can clearly visualize the probability you are trying to calculate.