The addition rule of probability

Content updated: October 22, 2026

When calculating the probability of a union, check whether the cases overlap. Add directly for mutually exclusive cases; otherwise subtract the part counted twice.

Learning goals: partition into mutually exclusive cases, use complements, and correct double counting in the union of two events.

Prerequisites: event unions and intersections, classical probability, and combinations.

On this page: Definition · Method · Examples · Practice · Further learning

1. The rule for two events

If $A\cap B=\varnothing$, additivity of probability implies

For any two events, the general rule is

To prove it, partition $A\cup B$ into three pairwise mutually exclusive parts: $A\setminus B$, $A\cap B$, and $B\setminus A$. In $P(A)+P(B)$ the intersection occurs twice, whereas in the union probability it occurs once. Subtracting one copy establishes the formula. When the intersection is empty, it reduces to the first rule.

2. Several cases and complements

If $A_0,\ldots,A_n$ are pairwise mutually exclusive, repeated addition makes their union probability the sum of their probabilities. Check every pair; different event descriptions alone do not suffice. If they partition all of $\Omega$, their probabilities sum to 1.

Since $A$ and $\overline A$ are mutually exclusive and their union is $\Omega$, we have $P(A)+P(\overline A)=1$. Hence

Subtract a probability on the right, rather than an event itself. “At least one” often has the complement “none”; if the complement has fewer cases, this shortens the calculation while retaining the same experiment model.

3. Three worked examples

Example 1. Three balls of the same color

A box contains 6 red and 4 white balls. Choose 3 uniformly without replacement, ignoring order. Let $A_j$ mean exactly $j$ red balls, for $j=0,1,2,3$. There are $\binom{10}3=120$ groups in total. The favorable count for $A_j$ is $\binom6j\binom4{3-j}$, so

These events are pairwise mutually exclusive because one group cannot have two different red-ball counts. They exhaust all outcomes, and the four probabilities sum to 1. The event $M$ that all 3 have the same color is $A_0\cup A_3$, so

Example 2. At least one red ball, in two ways

In the same experiment, $N=A_1\cup A_2\cup A_3$. Adding three mutually exclusive parts gives $P(N)=(9+15+5)/30=29/30$. Alternatively, $\overline N=A_0$, since no red balls means all 3 are white. Thus $P(N)=1-P(A_0)=1-1/30=29/30$.

Both methods use the same partition. When many counts would otherwise be added, the complement “no red balls” has just one count and often requires less calculation.

Example 3. Two performance groups among 30 students

Choose one student uniformly from the specified group of 30. There are 15 singers, 10 dancers, and 6 students who do both. Let $A,B$ be the events of singing and dancing. Participating in at least one performance is their union, so

There are 19 students in at least one group and 11 in neither. The 6 students were included in both 15 and 10, so subtract them once. The denominator 30 is the specified uniformly sampled student group; do not substitute a different class of unknown size.

4. Practice

1. For a fair die, let $A=\{2,4,6\}$ and $B=\{4,5,6\}$. Calculate $P(A\cup B)$.
2. In the ball experiment, find the probability that the 3 balls are not all the same color.
3. If $P(A)=0.4$, $P(B)=0.5$, and $P(A\cap B)=0.2$, calculate the union probability. May 0.4 and 0.5 simply be added?

5. Common mistakes

Do not substitute independence for mutual exclusivity when adding probabilities. “Or” in a union includes both events occurring. Write the complement in the same sample space. A calculated union probability must not exceed 1.

6. Further learning

Random experiments and events (English lesson) · Probability of an event (English lesson) · Conditional probability (English lesson) · The multiplication rule of probability (English lesson).

Original exercise resource on Scribd.