# Probability of $A$ union with not $B$

Given:

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

$P\left(A\right) + P\left( \overline{A} \right) = 1$

$P\left( A|B\right) = \frac{P\left( A \cap B\right) }{P\left( B\right) }$

Is it possible to find:

$P\left( A \cap \overline{B}\right) = P\left( A\right) - P\left( A \cap B\right)$

Or the latter didn't derive from the first ones?

The more verbose the answer the better.

• I don't know what your question means, too. – Takahiro Waki Jan 8 '17 at 6:42

## Proof

It is generally true that $A=\left( A \cap B \right) \cup \left( A \cap \overline{B} \right)$.

In fact, since $\left( A \cap B \right) \cap \left( A \cap \overline{B} \right) = \emptyset$ then $$A=\left( A \cap B \right) \sqcup \left( A \cap \overline{B} \right)$$ where $\sqcup$ denotes a disjoint union.

Since $\sigma$-additivity implies finite additivity (can you prove this?), we then have $$P\left(A\right)=P\left(\left(A\cap B\right)\sqcup\left(A\cap\overline{B}\right)\right)=P\left(A\cap B\right)+P\left(A\cap\overline{B}\right)$$ By subtracting $P\left( A\cap B \right)$ from both sides we get the required result.

## Discussion

We proved the identity (almost) directly from axioms. I believe the first two identities you provided are usually also derived directly from finite additivity, and the third one is simply a definition.

So to answer your question, we did not need the former claims to prove the latter (at least in my proof).

• Question edited to make it clear (removing "came from the other way means"), is it? – KcFnMi Jan 7 '17 at 20:51
• Did you mean to ask "Or the latter cannot be derived without knowing the first ones?" – 8l2s Jan 7 '17 at 20:58
• Question edited (again). – KcFnMi Jan 8 '17 at 1:38
• Do I now answer your question? – 8l2s Jan 8 '17 at 9:14

First, $A\cap B$ and $A \cap \bar B$ are disjoint. This means that $(A\cap B)\cap(A \cap B) = \emptyset.$ In other words the two events contain no common outcomes; they can never occur at the same time. To see this, note that every outcome in $A\cap B$ is in $B$ and every outcome in $A\cap \bar B$ is in $\bar B$. Since $B$ and $\bar B$ have no common elements by definition, nor do $A\cap B$ and $A \cap \bar B.$

Second, $(A\cap B) \cup (A\cap\bar B) = A.$ This is because $A\cap B$ has all elements of $A$ that are also in $B$, while $A\cap \bar B$ has all the elements of $A$ that are also in $\bar B.$ Together (their union), the contain all elements of $A$ since all outcomes are either in $B$ or $\bar B.$

If two events $C,D$ are disjoint (which means they can't happen at the same time) then the probability of their union (either C or D happens) must be $P(C\cup D) = P(C) + P(D).$ You can either get this from your formula $P(C\cup D) = P(C) + P(D) - P(C\cap D)$, or reason that since only one of them can happen, the probability of either of them happening must just be the sum. (You can also derive the whole formula intuitively by drawing a Venn diagram).

So since they're disjoint, $$P(A) = P((A\cap B) \cup (A\cap\bar B)) = P(A\cap B) + P(A\cap\bar B)$$

The facts used in the derivation were $$P(A\cup B) = P(A) + P(B) -P(A\cap B)$$ and $$P(A) = \sum_{i-1}^nP(A\cap D_i)$$ where $D_i$ are disjoint events with $\sum_i P(D_i) = 1.$ And that $B$ and $\bar B$ are disjoint P(B) + B(\bar B) = 1.$• Would you mind to expand the explanation in the first line? – KcFnMi Jan 7 '17 at 20:28 • of course. Will edit with more detail. (FWIW the accepted answer is essentially the same, I will try to complement it by being more intuitive and less axiomatic) – spaceisdarkgreen Jan 7 '17 at 20:54 • Introduction of$C$and$D\$ adds complexity, perhaps too much (at least for me). – KcFnMi Jan 8 '17 at 0:15