Inclusion-Exclusion Principle

Study guides: Exam P-1, Exam MAS-I

The Inclusion-Exclusion Principle is a counting rule for the probability (or size) of a union of overlapping events: add the individual probabilities, subtract the pairwise intersections, add back the triple intersections, and so on, so that no outcome is counted more than once.

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

$\begin{aligned} P(A \cup B \cup C) = {}& P(A) + P(B) + P(C) \\ &- P(A \cap B) - P(A \cap C) - P(B \cap C) \\ &+ P(A \cap B \cap C) \end{aligned}$

The signs alternate by intersection size: add single events, subtract pairwise intersections, add triple intersections, and so on.
For disjoint events the intersection terms are $0$ , so the rule reduces to $P(A \cup B) = P(A) + P(B)$ .
It is the general form of the addition rule of Probability and relies on Set Operations for the union and intersection.

Probability of At Least One Event Occurring Among a group of insurance claims, $P(A) = 0.5$ , $P(B) = 0.4$ , and $P(A \cap B) = 0.2$ . What is the probability that at least one of $A$ or $B$ occurs?
Answer "At least one of $A$ or $B$ " is the event $A \cup B$ :
 $\begin{align*} P(A \cup B) &= P(A) + P(B) - P(A \cap B) \\ &= 0.5 + 0.4 - 0.2 \\ &= 0.7 \end{align*}$ 

Three Overlapping Coverages A policyholder may file an auto ( $A$ ), home ( $B$ ), or liability ( $C$ ) claim. Given $P(A) = P(B) = P(C) = 0.3$ , each pairwise intersection $= 0.1$ , and $P(A \cap B \cap C) = 0.05$ , find the probability of at least one claim.
Answer
 $\begin{align*} P(A \cup B \cup C) &= 0.3 + 0.3 + 0.3 \\ &\quad - 0.1 - 0.1 - 0.1 \\ &\quad + 0.05 \\ &= 0.65 \end{align*}$ 
There is a 65% chance the policyholder files at least one of the three claims.