How to calculate intersection and union of probabilities?
The Product Rule applies to events which are independent. Only then is the probability of the intersection equal to the product of the probabilities of the events.
$\mathsf P(A\cap B) ~=~ \mathsf P(A)~\mathsf P(B)~$ only when events $A$ and $B$ are independent.
(When dealing with more than two events we require mutual independence.)
Otherwise conditional probability must be used: $\mathsf P(A\cap B)~=~\mathsf P(A)~\mathsf P(B\mid A)\\\qquad\qquad~=~\mathsf P(A\mid B)~\mathsf P(B)$
The Addition Rule applies only when the events are mutually exclusive (also known as disjoint). Only then is the probability of the union equal to the sum of probabilities of the event.
$\mathsf P(A\cup B)~=~\mathsf P(A)+\mathsf P(B)$
Otherwise if the events are not disjoint (ie they have common outcomes) then we would be over measuring and must exclude the measure of the intersection.
$\mathsf P(A\cup B)~=~\mathsf P(A)+\mathsf P(B) - \mathsf P(A\cap B)$
When dealing with more than two events, the principle of inclusion and exclusion is required
$\begin{align}\mathsf P(A\cup B\cup C)~=~&\mathsf P(A)+\mathsf P(B)+\mathsf P(C) - \mathsf P(A\cap B)-\mathsf P(A\cap C)-\mathsf P(B\cap C)+\mathsf P(A\cap B\cap C)\end{align}$
... and so on.
$\Box$
This is the inclusion-exclusion principle. We have that
$$
\mathbb{P}(A_1\cup A_2\cup A_3)=\mathbb{P}(A_1)+\mathbb{P}(A_2)+\mathbb{P}(A_3)-\mathbb{P}(A_1\cap A_2)-\mathbb{P}(A_1\cap A_3)-\mathbb{P}(A_2\cap A_3)+\mathbb{P}(A_1\cap A_2\cap A_3)
$$
for any three events $A_1$, $A_2$ and $A_3$. The following picture illustrates the idea behind this formula:
