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.
