# Find the final probability

Lets' assume we have 3 events that consist of pressing / not pressing buttons $$A$$, $$B$$, and $$C$$. Distribution of probablity for buttons is shown in the picture. In the case of dependend buttons pressing when $$A$$ is pressed, the conditional probability of pressing $$C$$ is $$P(C|A)= 0.5-0.4=0.1$$.

There is a logical function $$F = (\overline{x} \wedge \overline{y}) \vee (x \wedge y) \vee (y \vee \overline{z})$$. Where $$x=P(A)$$, $$y=P(B)$$, $$z=P(C)$$.

I should use Law of total probability. If we take event $$B$$ to calculate the probability for $$F$$, we get the following: $$P(F) = P(F|B) \cdot P(B) + P(F|\overline{B}) \cdot P(\overline{B})$$.

When button pressing is indepenent of each other i can solve it like this:

$$P(F|B) = 0 \vee x \vee \overline{z} = P(A \cup \overline{C})=P(A)+P(\overline{C})-P(A\cap \overline{C})$$

$$P(F|\overline B) = \overline x \vee 0 \vee 0 = P(\overline A)$$

$$P(F) =(P(A)+P(\overline{C})-P(A\cap \overline{C})) \cdot P(B) + P(\overline A) \cdot P(\overline B)$$

$$P(F) = (0.2 + 0.5 - 0.2 \cdot 0.5) \cdot 0.3 + 0.8 \cdot 0.7 = 0.74$$

And it's matches with Matlab result for indepentent events. But for depended events Matlab simulation shows that the result should be around $$0.6$$.

It seem's like in case of dependent events conditional probability of $$P(F|B)$$ will be different. $$P(F|B)=P(A \cup \overline C | B) = \frac{P((A \cup \overline C) \cap B))}{P(B)}$$
$$P(F|\overline B)=P(\overline A | \overline B) = \frac{P(\overline A \cap \overline B)}{P(\overline B)}$$