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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.

Graph of the probablity distribution

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For dependent events:

$P(F|B)=P(A \cup \overline C | B) = \frac{P((A \cup \overline C) \cap B))}{P(B)}$

$P(F|\overline B)=P(\overline A | \overline B) = \frac{P(\overline A \cap \overline B)}{P(\overline B)}$

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