Finite Additivity Across Disjoint Events

Why Disjointness Matters

When possible outcomes are grouped into categories that cannot occur together, the probability of a union of categories is the sum of their probabilities. This rule is useful whenever a question asks for the probability of an event described by several distinct cases: for example, a claim falling into one of several severity bands, or a reportable event arising in one of several regions.

The essential condition is pairwise disjointness. If two events can occur at the same time, adding their probabilities counts their shared outcomes more than once. Finite additivity is therefore not an unrestricted rule for combining events; it applies to a finite collection whose members are mutually exclusive.

Definition: Events $A_1,\ldots,A_n$ are pairwise disjoint if $A_i\cap A_j=\varnothing$ whenever $i\ne j$. Their union, $\bigcup_{i=1}^{n} A_i$, is the event that at least one of them occurs. A partition of a sample space $\Omega$ is a finite collection of pairwise disjoint events whose union is $\Omega$.

Finite Additivity

In the probability axioms, additivity applies to disjoint events: if $A$ and $B$ are disjoint, then $P(A\cup B)=P(A)+P(B)$. Finite additivity extends this rule to any finite number of pairwise disjoint events. The result does not require outcomes to be equally likely.

Theorem: If $A_1,\ldots,A_n$ are pairwise disjoint events, then

P\!\left(\bigcup_{i=1}^{n}A_i\right)=\sum_{i=1}^{n}P(A_i).

The theorem is a direct consequence of the two-event additivity axiom. Here is a proof by induction, which shows why the pairwise-disjoint condition is sufficient.

Base case.
For $n=1$, the union contains only $A_1$, so $P(\bigcup_{i=1}^{1}A_i)=P(A_1)$, as required.

Inductive assumption.
Assume the result holds for $n-1$ pairwise disjoint events. Write $U=\bigcup_{i=1}^{n-1}A_i$. By the assumption, $P(U)=\sum_{i=1}^{n-1}P(A_i)$.

Add the last event.
Because $A_n$ is disjoint from every earlier event, it is disjoint from their union $U$. Two-event additivity gives $P(U\cup A_n)=P(U)+P(A_n)$.

Substitute and conclude.
Substituting the inductive assumption yields $P(\bigcup_{i=1}^{n}A_i)=\sum_{i=1}^{n-1}P(A_i)+P(A_n)=\sum_{i=1}^{n}P(A_i)$. Thus the result holds for $n$, completing the induction.

For a finite sample space with probability assigned to each outcome, the same result can be seen by grouping outcomes. Disjoint events contain no common outcomes, so adding the probabilities within each event and then across events counts each outcome in the union exactly once. This is the practical reason that disjoint categories can be safely aggregated.

Using a Partition to Break Up an Event

A partition is especially useful when the event of interest is not itself one of the listed categories. Suppose $B_1,\ldots,B_n$ partition $\Omega$, and let $A$ be any event. The pieces $A\cap B_1,\ldots,A\cap B_n$ are pairwise disjoint: two different partition cells cannot contain the same outcome. Their union is all of $A$, because every outcome in $A$ belongs to exactly one cell of the partition. Therefore finite additivity gives:

A=\bigcup_{i=1}^{n}(A\cap B_i) \qquad\text{and}\qquad P(A)=\sum_{i=1}^{n}P(A\cap B_i).

This decomposition is valid whether or not $A$ is disjoint from the partition cells themselves. It is the pieces formed by intersecting $A$ with different cells that are disjoint. This distinction is important in applications involving several possible sources, periods, or categories.

Worked Example: Aggregating Claim Severity Bands

For a hypothetical portfolio, consider the finite probability model in which one policy is classified by its payment during a specified period. The mutually exclusive and exhaustive categories are: no payment, a payment greater than $0 and at most $1,000, a payment greater than $1,000 and at most $5,000, and a payment greater than $5,000. Their probabilities are $0.74$, $0.16$, $0.07$, and $0.03$, respectively. Find the probability of a payment greater than $1,000 and the probability of any positive payment.

Method. Treat the four categories as a partition and add the probabilities of the cells that make up each event.

The partition check is $0.74+0.16+0.07+0.03=1.00$. Thus the listed categories account for the full probability.

A payment greater than $1,000 occurs in either of the last two categories. These events are disjoint, so $P(\text{payment greater than } \$1{,}000)=0.07+0.03=0.10$.

A positive payment occurs in any of the three payment categories: $P(\text{positive payment})=0.16+0.07+0.03=0.26$.

Check. The positive-payment probability is also $1-0.74=0.26$, consistent with the category sum. The common mistake to avoid is to include the no-payment category when calculating a positive-payment event.

Worked Example: Summing Event Pieces Across Regions

A hypothetical insurer classifies each inspection by exactly one of three regions, $R_1,R_2,R_3$, with probabilities $0.45$, $0.35$, and $0.20$. Let $D$ be the event that an inspection finds a reportable defect. The probabilities of a defect and each region occurring together are $P(D\cap R_1)=0.009$, $P(D\cap R_2)=0.014$, and $P(D\cap R_3)=0.006$. Find $P(D)$, and then find the probability of a reportable defect in region $R_2$ or $R_3$.

Method. Intersect the event of interest with the partition cells, then add the resulting disjoint pieces.

The regional probabilities sum to $0.45+0.35+0.20=1.00$, so the regions form a partition. The three events $D\cap R_1$, $D\cap R_2$, and $D\cap R_3$ are disjoint and together make up $D$. Hence $P(D)=0.009+0.014+0.006=0.029$.

For a defect in region $R_2$ or $R_3$, add just those two pieces: $P((D\cap R_2)\cup(D\cap R_3))=0.014+0.006=0.020$.

Check. Each joint probability is no greater than the probability of its region, and the defect probability $0.029$ lies between 0 and 1. A common mistake is to add the region probabilities $0.35+0.20$; that gives the chance of being in either region, not the chance of both being in one of those regions and having a defect.

Worked Example: Decomposing an Event by Time Category

For a hypothetical equipment-repair plan, one repair outcome belongs to exactly one of three duration categories: at most one year, more than one and at most three years, or more than three years. Their probabilities are $0.50$, $0.30$, and $0.20$. Let $E$ be the event that the plan requires reimbursement. The probabilities that reimbursement is required in the three categories, in the same order, are $0.035$, $0.024$, and $0.006$. Find the probability that reimbursement is required within three years, and find the probability that it is required at any duration.

Method. Use the intersections of the event with the partition cells. Since the duration categories are disjoint, the relevant intersections are disjoint as well.

The probability of reimbursement within three years includes the first two duration categories: $0.035+0.024=0.059$.

The event of reimbursement at any duration consists of all three pieces: $P(E)=0.035+0.024+0.006=0.065$.

Check. The probability of reimbursement within three years, $0.059$, is no greater than the probability at any duration, $0.065$. The category probabilities sum to $1.00$, while the three reimbursement probabilities sum to $0.065$; these are different sums because the first covers all outcomes and the second covers only outcomes in $E$.

Common Trap: Adding Events That Overlap

Finite additivity cannot be applied merely because a problem lists several events. Consider a hypothetical outcome space divided into four disjoint regions: $A$ only has probability $0.22$, both $A$ and $B$ has probability $0.08$, $B$ only has probability $0.12$, and neither has probability $0.58$. These four probabilities sum to $1.00$.

The incorrect calculation is $P(A)+P(B)=(0.22+0.08)+(0.12+0.08)=0.50$, reported as $P(A\cup B)$. The shared region of probability $0.08$ has been counted twice. To correct the calculation, describe the union using its disjoint regions: $A$ only, both $A$ and $B$, and $B$ only. Then $P(A\cup B)=0.22+0.08+0.12=0.42$. The lesson is not to add overlapping events directly; first identify disjoint pieces that cover the event of interest.

Key takeaway: Before adding probabilities, verify that the events are pairwise disjoint and that their union is exactly the event being asked about. If a partition is available, intersect the target event with each partition cell and add those disjoint pieces.

Check Your Understanding

For each problem, identify the relevant disjoint events before calculating.

In a hypothetical policy model, the probabilities of no payment, payment up to $2,000, payment greater than $2,000 and at most $8,000, and payment greater than $8,000 are $0.61$, $0.24$, $0.11$, and $0.04$. Find the probability of a payment greater than $2,000. (Answer: $0.15$)
A sample space is partitioned into $B_1,B_2,B_3$, and $P(A\cap B_1)=0.015$, $P(A\cap B_2)=0.012$, $P(A\cap B_3)=0.003$. Find $P(A)$. (Answer: $0.030$)
A reportable event can arise in exactly one of three regions. The probabilities of a reportable event in regions 1, 2, and 3 are $0.008$, $0.017$, and $0.005$. Find the probability of a reportable event in region 1 or region 3. (Answer: $0.013$)
Four disjoint outcome regions have probabilities $0.19$ for $A$ only, $0.07$ for both $A$ and $B$, $0.11$ for $B$ only, and $0.63$ for neither. Find $P(A\cup B)$, and explain why adding $P(A)$ and $P(B)$ directly would be wrong. (Answer: $0.37$; direct addition counts the shared $0.07$ twice.)
A partition has four cells. The probabilities of the target event occurring in those cells are $0.040$, $0.030$, $0.015$, and $0.005$. Find the target-event probability, and find the probability that it occurs in one of the first two cells. (Answer: $0.090$ overall; $0.070$ in the first two cells.)

What You'll Learn