Two identities that connect union, intersection and conditional probability.
6 min read2 worked examples
The addition formula
Adding $\mathrm{P}(A)$ and $\mathrm{P}(B)$ double-counts the overlap, so it has to be subtracted back off. This gives the addition formula, which holds for any two events $A$ and $B$:
Addition formula
$$\mathrm{P}(A\cup B) = \mathrm{P}(A) + \mathrm{P}(B) - \mathrm{P}(A\cap B)$$
Where it comes from
Let $a=\mathrm{P}(A)$, $b=\mathrm{P}(B)$ and $i=\mathrm{P}(A\cap B)$. The three regions of $A\cup B$ have probabilities $a-i$, $i$ and $b-i$.
DerivationSumming the three regions of A ∪ B
1
Let $a=\mathrm{P}(A)$, $b=\mathrm{P}(B)$, $i=\mathrm{P}(A\cap B)$.
2
$A\cup B$ is made of three non-overlapping regions: $A$ only $(a-i)$, the intersection $(i)$, and $B$ only $(b-i)$.
3
$\mathrm{P}(A\cup B) = (a-i)+i+(b-i) = a+b-i$.
4
So $\mathrm{P}(A\cup B)=\mathrm{P}(A)+\mathrm{P}(B)-\mathrm{P}(A\cap B)$.
▪ QED
Check: mutually exclusive events
If $A$ and $B$ are mutually exclusive, $\mathrm{P}(A\cap B)=0$, and the formula reduces to $\mathrm{P}(A\cup B)=\mathrm{P}(A)+\mathrm{P}(B)$ — exactly the rule from Set Notation.
Worked example
Rearranging the addition formula
$\mathrm{P}(A)=0.5$, $\mathrm{P}(B)=0.65$ and $\mathrm{P}(A\cup B)=0.85$. Find $\mathrm{P}(A\cap B)$.
The completed diagram once $\mathrm{P}(A\cap B)=0.3$ has been found: $A$ only is $0.5-0.3=0.2$, $B$ only is $0.65-0.3=0.35$, and outside both is $1-0.85=0.15$.
The multiplication formula
Rearranging the definition of conditional probability, $\mathrm{P}(A\mid B)=\dfrac{\mathrm{P}(A\cap B)}{\mathrm{P}(B)}$, for $\mathrm{P}(A\cap B)$ gives a second essential identity:
Multiplication formula
$$\mathrm{P}(A\cap B) = \mathrm{P}(A\mid B)\times \mathrm{P}(B) = \mathrm{P}(B\mid A)\times \mathrm{P}(A)$$
This is the formula that justifies multiplying along the branches of a tree diagram to find the probability of a combined outcome — the next lesson relies on it directly.
Using the multiplication formula
$\mathrm{P}(A)=0.4$ and $\mathrm{P}(B\mid A)=0.25$. Find $\mathrm{P}(A\cap B)$.
$\mathrm{P}(A\cap B) = \mathrm{P}(B\mid A)\times\mathrm{P}(A) = 0.25\times 0.4 = 0.1$.
Check yourself
$\mathrm{P}(A)=0.6$, $\mathrm{P}(B)=0.25$ and $\mathrm{P}(A\cap B)=0.1$. Find $\mathrm{P}(A\cup B)$, then find $\mathrm{P}(A\mid B)$.