# Prove the Probability of Two Events

I'm taking my first course in Probability and one of my homework problems is to prove that for any two sets: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Note that the function $P$ is the probability of an event happening. From the third axiom, we are given that for any two disjoint sets, that is $A \cap B = \emptyset$:

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

My thought process follows from this third axiom, but I run into a problem near the end:

Since $A$ and $B$ are disjoint sets, $P(A)=P(A-B)$ and $P(B)=P(B-A)$, so

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

From this, it appears that we are essentially taking the symmetric difference of A and B. This, to me, seems to be equivalent to:

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

Hence, by my logic:

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

Which, in essence, doesn't care whether two sets are disjoint or not as we are subtracting the intersection of them. If the sets are disjoint, then the intersection is merely null. If they intersect, then we are removing the "extra pieces".

I feel like this is inadequate proof despite my ability to convince myself that it is true.. Where, if anywhere, did I go wrong?

Thanks!

For any sets $A$ and $B$, we have the disjoint union

$$A \cup B = (A- B) \cup (A \cap B) \cup (B-A).$$

Then, by the axiom,

$$P (A \cup B ) = P(A- B) + P(A \cap B) + P(B - A).$$

Since $P (A-B)= P(A) - P(A \cap B)$ and $P (B-A)= P(B)- P(A \cap B)$, the result follows.

Here's how I'd do it:

Note that $$(*)\quad P(A\bigcup B) = P(A \bigcup A^{c}B) = P(A)+P(A^{c}B)$$

Then since $B = AB \bigcup A^{c}B$ we get that

$$P(B) = P(AB) + P(A^{c}B)$$

or equivalently $$P(A^{c}B) = P(B) - P(AB)$$ Plugging this into $*$ gives the desired result and thus completing the proof. Hope that helps!

A rigorous proof of the statement $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ is performed using not only the axioms of probability theory but also the identities of set theory and the axioms of real numbers.

PROVE: For any arbitrary probability experiment consisting of a finite sample space, $$S$$, and events $$A$$ and $$B$$ s.t. $$A \subseteq S$$ and $$B \subseteq S$$, the following is true: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$.

$$1$$. Let $$S$$ be the finite sample space of some arbitrary probability experiment with events $$A$$ and $$B$$ s.t. $$A \subseteq S$$ and $$B \subseteq S$$. First, we show $$P(A \cup B) = P(A \cup (B \cap A^C))$$.

$$A \cup B = (A \cup B) \cap S$$ by the identity law, where $$S$$, the sample space, is our universal set

$$= (A \cup B) \cap (A \cup A^C)$$ by the negation law

$$= A \cup (B \cap A^C)$$ by the distributive law

Hence, $$A \cup B = A \cup (B \cap A^C)$$; thus, we know

(1) $$P(A \cup B) = P(A \cup (B \cap A^C))$$

$$2$$. Second, we show the sets $$A$$ and $$B \cap A^C$$ are mutually exclusive or "disjoint" - or in other words, that $$A \cap (B \cap A^C)=\emptyset$$.

$$A \cap (B \cap A^C)=A \cap (A^C \cap B)$$ by the commutative law

$$=(A \cap A^C) \cap B$$ by the associative law

$$=\emptyset \cap B$$ by the negation law

$$=\emptyset$$ by the domination law

Hence, $$A \cap (B \cap A^C)=\emptyset$$, implying that sets $$A$$ and $$B \cap A^C$$ are disjoint.

$$3$$. Since $$A$$ and $$B \cap A^C$$ are disjoint, then by the finite additvity axiom (axiom iii of modern probability theory) we may conclude $$P(A \cup (B \cap A^C))=P(A) + P(B \cap A^C)$$. Then, by substitution with equation (1), we have

(2) $$P(A \cup B) = P(A) + P(B \cap A^C)$$.

$$4$$. Next, we show $$P(B)=P((B \cap A) \cup (B \cap A^C))$$.

$$B=B \cap S$$ by the identity law

$$=B \cap (A \cup A^C)$$ by the negation law

$$=(B \cap A) \cup (B \cap A^C)$$ by the distribution law

Hence, $$B=(B \cap A) \cup (B \cap A^C)$$; thus, we know

(3) $$P(B)=P((B \cap A) \cup (B \cap A^C))$$

$$5$$. Now we show that sets $$B \cap A$$ and $$B \cap A^C$$ are disjoint - or in other words, that $$(B \cap A) \cap (B \cap A^C)=\emptyset$$.

$$(B \cap A) \cap (B \cap A^C)=B \cap (A \cap A^C)$$ by the distributive law

$$=B \cap \emptyset$$ by the negation law

$$=\emptyset$$ by the domination law

Hence, $$(B \cap A) \cap (B \cap A^C)=\emptyset$$, implying that sets $$B \cap A$$ and $$B \cap A^C$$ are disjoint.

$$6$$. Since $$B \cap A$$ and $$B \cap A^C$$ are disjoint, then by the finite additvity axiom we may conclude $$P((B \cap A) \cup (B \cap A^C))=P(B \cap A) + P(B \cap A^C)$$. Then, by substitution with equation (3), we have $$P(B)=P(B \cap A) + P(B \cap A^C)$$.

$$\Rightarrow P(B)-P(B \cap A)=P(B \cap A^C)$$ by the addition property of equality, additive inverse law, and additive identity law. Then, by the reflexive property of equality, we have

(4) $$P(B \cap A^C)=P(B)-P(B \cap A)$$

$$7$$. By equation (2) we know $$P(A \cup B) = P(A) + P(B \cap A^C)$$, and by equation (4) we know $$P(B \cap A^C)=P(B)-P(B \cap A)$$. By substitution with equations (2) and (4), we have $$P(A \cup B) = P(A) + P(B)-P(B \cap A)$$.

$$8$$. Thus, if events $$A$$ and $$B$$ are subsets of a finite sample space of some arbitrary probability experiment, then $$P(A \cup B) = P(A) + P(B)-P(B \cap A)$$.

$$\therefore$$ For any arbitrary probability experiment consisting of a finite sample space, $$S$$, and events $$A$$ and $$B$$ s.t. $$A \subseteq S$$ and $$B \subseteq S$$, the following is true: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$.

Q.E.D.

Use these three facts to get started. If $C\subset D$ are events, $P(D - C) = P(D) - P(C).$ If $C$ and $D$ are any events $C\cup D = C \cup (D - C)$ is a disjoint union, and $D - C = D - (C\cap D)$.
Fool with this and you should be able to get it.

One way to prove this is to define indicator random variables $I_A, I_B, I_{A \cap B}$ and $I_{A \cup B}$, where $I_E = 1$ if the event $E$ occurs and $I_E = 0$ if $E$ does not occur.

Now suppose that $I_{A \cup B} = 1$, which means $A$ or $B$ has happened, and consider the three ways in which this can occur. If $A$ occurs but not $B$, then $I_A = 1$, $I_B = 0$ and $I_{A \cap B} = 0$. If $B$ occurs but not $A$ then similarly $I_A = 0$, $I_B = 1$ and $I_{A \cap B} = 0$. Finally if both $A$ and $B$ occur, then $I_A = 1$, $I_B = 1$ and $I_{A \cap B} = 1$.

If on the other hand $I_{A \cup B} = 0$, then neither $A$ nor $B$ has occurred and $I_A = 0$, $I_B = 0$ and $I_{A \cap B} = 0$

Notice we're just shown that $I_{A \cup B} = I_A + I_B - I_{A \cap B}$ and hence $\mathbb{E}(I_{A \cup B}) = \mathbb{E}(I_A) + \mathbb{E}(I_B) - \mathbb{E}(I_{A \cap B})$, or $\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B)$. (The basic idea here is that the inclusion-exclusion formula in probability is nothing but simple counting.)