# An axiomatic proof of the probability formula of union
## The CHOUAKRI Sid Ahmed Method $P(E 1 \cup E 2) = P(E 1) + P(E 2) - P(E 1 \cap E 2)$

Sid Ahmed Chouakri

► To cite this version:

Sid Ahmed Chouakri. An axiomatic proof of the probability formula of union The CHOUAKRI Sid Ahmed Method $P(E 1 \cup E 2) = P(E 1) + P(E 2) - P(E 1 \cap E 2)$. 2025. hal-05108372

HAL Id: hal-05108372

https://hal.science/hal-05108372v1

Submitted on 11 Jun 2025

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# An axiomatic proof of the probability formula of union

## The CHOUAKRI Sid Ahmed Method

$$P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$$

CHOUAKRI Sid Ahmed

11 juin 2025

## Declaration

I, the undersigned, **CHOUAKRI Sid Ahmed**, certify that the demonstration presented below is a personal and original construction. Any similarity to any other demonstration would be pure coincidence.

## Axioms of probability used

1. $P(E) \ge 0$

2. $P(\Omega) = 1$

3. If $E_1$ et $E_2$ are mutually exclusive : $P(E_1 \cup E_2) = P(E_1) + P(E_2)$

4. $P(E_1 \cap E_2) = P(E_1) \cdot P(E_2 | E_1)$ (conditional probability)

## Demonstration

We want to demonstrate : :

$$P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$$

We know that :

$$E_1 \cup E_2 = E_1 \cup (E_2 \cap \overline{E_1})$$

And as :

$$E_1 \cap (E_2 \cap \overline{E_1}) = \emptyset$$

So, by applying Axiom 3 :

$$P(E_1 \cup E_2) = P(E_1) + P(E_2 \cap \overline{E_1})$$

Remains to verify :

$$P(E_2 \cap \overline{E_1}) = P(E_2) - P(E_1 \cap E_2)$$
Apply Axiom 4 :

$$P(E_2 \cap \overline{E_1}) = P(E_2) \cdot P(\overline{E_1} | E_2)$$

Well-knowing :

$$P(\overline{E_1} | E_2) = 1 - P(E_1 | E_2)$$

Then :

$$P(E_2 \cap \overline{E_1}) = P(E_2) \cdot (1 - P(E_1 | E_2)) = P(E_2) - P(E_2) \cdot P(E_1 | E_2)$$

And so :

$$P(E_2 \cap \overline{E_1}) = P(E_2) - P(E_1 \cap E_2)$$

Finally, we get :

$$P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \blacksquare$$

This proof is a personal proposal of the author CHOUAKRI Sid Ahmed, in accordance with the classical axioms of probability theory.