Additivity for Disjoint Events

Why Disjointness Makes Addition Work

When several events cannot occur together, the probability that at least one occurs is the sum of their individual probabilities. This is useful whenever outcomes are grouped into distinct categories: a result may fall into exactly one diagnostic category, a selected item has exactly one color, or an experiment follows exactly one of several possible cases. The essential condition is disjointness. If an outcome belongs to more than one event, adding the probabilities counts that outcome more than once.

The two-event version of this rule follows from the probability axioms. The purpose here is to extend it carefully to any finite number of events, then use the result for partitions and case-based calculations. No assumption that the events are equally likely is needed.

Definition: Events $A_1,\ldots,A_n$ are pairwise disjoint if $A_i\cap A_j=\varnothing$ whenever $i\ne j$. Their union $A_1\cup\cdots\cup A_n$ is the event that at least one of them occurs. For pairwise disjoint events, no outcome in this union is counted in more than one event.

For this tutorial, a probability model consists of a sample space $S$, a collection of events that includes the events being considered and their finite unions, and a probability function $P$. It satisfies $P(A)\ge 0$, $P(S)=1$, and the disjoint-additivity axiom: if events $A$ and $B$ are disjoint, then $P(A\cup B)=P(A)+P(B)$. The finite-union result below uses this axiom repeatedly.

Finite Additivity

Theorem: If $A_1,\ldots,A_n$ are pairwise disjoint events, where $n$ is a positive integer, then $$ P\!\left(\bigcup_{i=1}^{n} A_i\right)=\sum_{i=1}^{n}P(A_i). $$

The result is a finite extension of the two-event axiom. The proof must check that at each step the union of events already combined remains disjoint from the next event.

Start with one event.
For $n=1$, the union is just $A_1$, and $P(A_1)=P(A_1)$.

Assume the rule holds for $n-1$ events.
Suppose the formula is true for every collection of $n-1$ pairwise disjoint events.

Separate the last event.
Pairwise disjointness implies $A_n\cap(\bigcup_{i=1}^{n-1}A_i)=\varnothing$. Apply the two-event axiom to this union and $A_n$, then use the induction assumption.

In detail, the base case $n=1$ holds as stated. For the induction step, let $B=\bigcup_{i=1}^{n-1}A_i$. If an outcome belonged to $A_n\cap B$, it would belong to $A_n$ and to some $A_i$ with $i<n$, contradicting pairwise disjointness. Thus $B\cap A_n=\varnothing$. The two-event axiom and the induction assumption give

P\!\left(\bigcup_{i=1}^{n}A_i\right) =P(B\cup A_n) =P(B)+P(A_n) =\sum_{i=1}^{n-1}P(A_i)+P(A_n) =\sum_{i=1}^{n}P(A_i).

This completes the induction and proves finite additivity. $\square$

A particularly important application is a partition. A collection $C_1,\ldots,C_m$ is a partition of $S$ if its members are pairwise disjoint and their union is the whole sample space. Because $P(S)=1$, finite additivity immediately gives a normalization identity.

Result for a partition: If $C_1,\ldots,C_m$ partition $S$, then $$ \sum_{j=1}^{m}P(C_j)=P(S)=1. $$ For any event $A$, the events $A\cap C_1,\ldots,A\cap C_m$ are pairwise disjoint and partition $A$. Therefore $$ P(A)=\sum_{j=1}^{m}P(A\cap C_j). $$

To verify the second statement, first note that if an outcome belonged to both $A\cap C_i$ and $A\cap C_j$, it would belong to both $C_i$ and $C_j$, which is impossible when $i\ne j$. Also, since the $C_j$ cover $S$, every outcome in $A$ belongs to some $C_j$, and hence to some $A\cap C_j$. Conversely, every outcome in $A\cap C_j$ is in $A$. Thus these intersections are disjoint and their union is $A$, so finite additivity applies. $\square$

Worked Applications

Worked Example: Summing Mutually Exclusive Outcomes

A fair six-sided die is rolled once. Let $A_1=\{1\}$, $A_2=\{2\}$, and $A_3=\{3\}$, where an event is written as the set of die outcomes that produce it. Find the probability of rolling one of the numbers 1, 2, or 3.

The three singleton events are pairwise disjoint: a single roll cannot equal two different numbers. For a fair die, each face has probability $1/6$, so

P(A_1\cup A_2\cup A_3) =P(A_1)+P(A_2)+P(A_3) =\frac{1}{6}+\frac{1}{6}+\frac{1}{6} =\frac{3}{6} =\frac{1}{2}.

The union is exactly $\{1,2,3\}$, which has three of the six equally likely outcomes. The answer $1/2$ therefore agrees with direct counting.

Worked Example: Finding a Probability From a Partition

A laboratory classifies each test result as negative, mildly positive, moderately positive, or strongly positive. The four categories are exhaustive and mutually exclusive. Their probabilities are $0.50$, $0.25$, $0.17$, and $0.08$, respectively. Find the probability that a result is positive at any level.

The event “positive at any level” is the union of the three positive categories. Because a result belongs to exactly one category, these three events are disjoint. Finite additivity gives

P(\text{positive}) =0.25+0.17+0.08 =0.50.

As a check, the four category probabilities sum to $0.50+0.25+0.17+0.08=1.00$, as they must for an exhaustive partition of the possible results. The negative category is excluded from the positive event; the remaining three categories contribute exactly $0.50$.

Worked Example: Adding Across Cases

A randomly selected appointment is either routine, urgent, or follow-up. These cases form a partition, with probabilities $0.50$, $0.30$, and $0.20$. The probabilities of a no-show within these cases are $0.10$, $0.40$, and $0.80$, respectively. Find the overall probability of a no-show.

Let $N$ be the event of a no-show, and let $C_1,C_2,C_3$ denote the routine, urgent, and follow-up cases. The events $N\cap C_1$, $N\cap C_2$, and $N\cap C_3$ are disjoint and together make up $N$. By the definition $P(N\mid C_j)=P(N\cap C_j)/P(C_j)$, when $P(C_j)>0$, we have $P(N\cap C_j)=P(C_j)P(N\mid C_j)$. Therefore,

\begin{aligned} P(N) &=P(N\cap C_1)+P(N\cap C_2)+P(N\cap C_3)\\ &=(0.50)(0.10)+(0.30)(0.40)+(0.20)(0.80)\\ &=0.05+0.12+0.16\\ &=0.33. \end{aligned}

Thus the overall no-show probability is $0.33$. Each case contributes its own joint probability of being in that case and having a no-show. The conditional percentages alone should not be added: the cases have different probabilities of occurring.

What Fails When Events Overlap?

The disjointness condition is not a technical detail that can be ignored. For example, on a fair six-sided die let $A=\{1,2,3\}$ and $B=\{3,4\}$. Then $P(A)=3/6$ and $P(B)=2/6$. Adding gives $5/6$, but $A\cup B=\{1,2,3,4\}$, so its probability is $4/6$. The outcome 3 lies in both events and is counted twice in the sum.

One way to make an addition rule applicable is to separate the overlap: $A\cup B$ is the disjoint union of $A$ and $B\setminus A$. In this example, $B\setminus A=\{4\}$, so

P(A\cup B)=P(A)+P(B\setminus A)=\frac{3}{6}+\frac{1}{6}=\frac{4}{6}.

In general, when events overlap, do not apply finite additivity directly to them. Instead, identify disjoint pieces or use the two-event union rule, $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, which corrects for the double-counted intersection.

Common mistake: “The probability of $A$ or $B$ is always $P(A)+P(B)$.” This is wrong unless $A$ and $B$ are disjoint. Before adding probabilities, check whether an outcome can belong to more than one event. For a partition, this check is built into the definition; for other collections, it must be done explicitly.

Why the Rule Matters

Finite additivity turns complicated events into manageable case calculations. First split the possible outcomes into a finite collection of mutually exclusive cases; then calculate the contribution from each case and add. This approach is reliable only when the cases are exhaustive for the event being calculated and do not overlap. If a case split omits an outcome, the sum is too small; if cases overlap, the sum may count some probability more than once.

A useful check is to ask two questions before summing: Can two of these cases happen together? and Do the cases cover the event whose probability I want? For a partition of the whole sample space, the probabilities should sum to one. For a partition of an event $A$, the probabilities of its pieces should sum to $P(A)$.

Check Your Understanding

Use the finite additivity rule and the disjointness condition to answer these questions.

Suppose $A_1,A_2,A_3,A_4$ are pairwise disjoint and have probabilities $0.12, 0.23, 0.31,$ and $0.09$. Find the probability of their union.
What two properties must events $C_1,\ldots,C_m$ satisfy to form a partition of the sample space?
A sample space is partitioned into events with probabilities $0.4$, $0.35$, and $q$. Find $q$.
On a fair die, let $A=\{1,2,3\}$ and $B=\{3,4,5\}$. Explain why $P(A)+P(B)$ is not $P(A\cup B)$, and find $P(A\cup B)$.
An event $E$ is split into three pairwise disjoint cases $E\cap C_1,E\cap C_2,E\cap C_3$. State the finite-additivity equation for $P(E)$.

What You'll Learn