$P(A\cap B)=P(A)P(B)$

In the event that the events are mutually exclusive, if A does not occur, B and if B does not occur, A.

In this case the probability that both occur simultaneously is zero. Thus

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The probability of A or B occurring corresponds to each outcome that one or the other produces. This corresponds to the union of both A \cup B events and is calculated by adding both probabilities.

$P(A\cup B)=P(A)+P(B)$

If the events are NOT mutually exclusive, the sets can have points in common, that is, their intersection is NOT empty

equation

If you want to calculate the probability that A or B will occur as the sum, you will have the problem that the area of the intersection will be counted twice. Therefore, it is necessary to subtract once the area of the intersection in order to have the total number of events in a unique way.

When the events A and B are NOT mutually exclusive, the probability is calculated as the sum of the P(A) probabilities that will occur A and P(B) occur B, there being the problem that the set A \cap B in which they can coincide, would be adding twice. Therefore the probability is the sum minus the probability that they coincide:

equation

The sum never exceeds unity since both sets do not intercept and the sum cannot be greater than all possible cases.

$P(A\mid B)=\displaystyle\frac{P(A\cap B)}{P(B)}$