# Proposition on Probability

Proposition : $$P(E \cup F) = P(E) + P(F) - P(EF)$$
I have the following laws and axioms available to me:

### Laws from Set Theory

Commumutative Laws: $$E \cup F = F \cup E$$; $$EF=FE$$
Associative Laws: $$(E \cup F) \cup G = E\cup(F \cup G)$$; $$(EF)G = E(FG)$$
Distributive Laws: $$(E \cup F)G=EG\cup FG$$; $$EF \cup G = (E \cup G)(F \cup G)$$

### DeMorgan's Laws

$$(E \cup F)^c = E^cF^c$$
$$(EF)^c = E^c \cup F^c$$

### Axioms of Probability

Axiom 1: $$0 \leq P(E) \leq 1$$
Axiom 2: $$P(S) = 1$$
Axiom 3: $$P\left(\bigcup_{i=1}^{n}E_i\right) = \sum_{i=1}^{n}P(E_i)$$ For $$E_i$$ mutually exclusive

### My attempt

I tried to use Axiom 3 by constructing mutually exclusive sets: Sets $$(E-F), EF, (F-E)$$ are mutually exclusive and we can write $$P(E \cup F) = P(E-F)+P(EF)+P(F-E)$$ However, this requires another proposition that $$P(X-Y) = P(X) - P(XY)$$. The proof requires a Venn diagram. The first proposition has also been proved using a Venn diagram in the book. I am wondering if the proposition can be proved using the laws and axioms alone.

• Well, $X = X \cup (Y - Y) = XY \cup (X - Y),$ right? And $XY \cap (X - Y) = X \cap (Y - Y) = \emptyset.$ Jun 28, 2022 at 6:53
• @StephenDonovan didn't understand why $X \cup (Y-Y) = XY \cup (X-Y)$, which law are you using here? Jun 30, 2022 at 2:19
• My apologies, I didn't write it out before commenting and it seems I confused myself somehow: what I should've done was $X = X \cap S = X \cap (Y \cup Y^c) = (X \cap Y) \cup (X \cap Y^c) = XY \cup (X - Y).$ It's just using the distributive property Jun 30, 2022 at 2:28
• @StephenDonovan thank you, and it is easy to show that $XY \cap (X-Y) = \phi$ from set theory (hopefully this translates nicely to probability). Then we use the third axiom to prove the proposition. This is what I understood. If you want you can write the answer and I will accept it. Jun 30, 2022 at 3:03