Probabity of Joint The is a well known fact, that
$\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(AB)$
I found an interesting proof, that start from the stating the fact
$A \cup B = (A\bar{B}) \cup (AB) \cup (\bar{A}B)$, these event are disjoint.
The problem is I don't understand why above equality is true.
 A: Starting from the fact
$$A \cup B = (A\bar{B}) \cup (AB) \cup (\bar{A}B)$$
This statement is stating that things that belong to the union of set $A$ and set $B$ must either belong to: 
Set $A$ and not set $B$
Set $B$ and not set $A$
Both set $A$ and set $B$
This should be obvious if you think about it but it may be a little clearer if you plot a logic truth table as shown below.
$$ \begin{array}{cc|ccc|cl} A & B & (A\bar{B}) & (AB) & (\bar{A}B) & A \cup B & \\ 
\hline
0 & 0 & 0 & 0 & 0 & 0 \\ 
0 & 1 & 0 & 0 & 1 & 1 \\ 
1 & 0 & 1 & 0 & 0 & 1 \\
1 & 1 & 0 & 1 & 0 & 1 \\
\end{array} $$
It's important as we are trying to to calculate probabilities that these sets are disjoint otherwise we would count one or more possibilities twice   
The formula you have for the probability of event A or B recognises that both A and can occur and ensures its not counted twice. 
$$\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(AB)$$
A: Assume that $x\in A$. Then 


*

*either $x\in B$, hence $x\in AB\subset$ RHS,

*of $x\in \bar B$: in this case, $x\in A\bar B\subset$ RHS.


Hence $A\cup B\subset A\bar B\cup \bar AB\cup AB$.
Conversely, $AB\subset A\cup B$, $A\bar B\subset A\cup B$ and $A\bar B\subset A\cup B$, which shows the  reverse inclusion.
