The axioms gave the addition rule for two events: add, then subtract the overlap that was counted twice. With three events the overlaps overlap too, so one more correction is needed. The result, inclusion–exclusion, is the standard way to find the chance that at least one of several events happens.
Key idea
$$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C).$$ Add the singles, subtract the pairs, add back the triple.
Why it works
Look at one outcome that lies in exactly $k$ of the three events, and count how many times the right side counts it. The singles count it $k$ times, the pairs $\binom{k}{2}$ times, the triple $\binom{k}{3}$ times. For $k = 1$: $1 - 0 + 0 = 1$. For $k = 2$: $2 - 1 + 0 = 1$. For $k = 3$: $3 - 3 + 1 = 1$. Every outcome in the union is counted exactly once, and outcomes outside the union are never counted, so the sum is $P(A \cup B \cup C)$. The same count works for $n$ events, because $$\sum_{j \ge 1} (-1)^{j+1}\binom{k}{j} = 1 - (1 - 1)^{k} = 1$$
Worked examples
Worked example: Three activities
In a gym, $P(\text{swim}) = 0.35$, $P(\text{run}) = 0.26$, $P(\text{cycle}) = 0.22$, the pairwise overlaps are $P(S \cap R) = 0.09$, $P(S \cap C) = 0.10$, $P(R \cap C) = 0.07$, and $P(S \cap R \cap C) = 0.03$. Find the probability that a member does at least one.
- Singles: $0.35 + 0.26 + 0.22 = 0.83$.
- Pairs: $0.09 + 0.10 + 0.07 = 0.26$.
- $$P(S \cup R \cup C) = 0.83 - 0.26 + 0.03 = 0.60$$
Answer: $0.6$
Worked example: Divisible by 2, 3 or 5
A whole number is chosen at random from 1 to 60. Find the probability that it is divisible by 2, 3 or 5.
- Let A, B, C be "divisible by 2", "by 3", "by 5". Divisible by both 2 and 3 means divisible by 6, and so on, because 2, 3 and 5 share no factors.
- Counts: $30 + 20 + 12$ singles; $10 + 6 + 4$ pairs (multiples of 6, 10, 15); $2$ multiples of 30.
- $$30 + 20 + 12 - 10 - 6 - 4 + 2 = 44$$ numbers, so the probability is $\frac{44}{60} = \frac{11}{15}$.
Answer: $\frac{11}{15}$
Worked example: Exactly one of the three
With the gym numbers above, find the probability that a member does exactly one of the three activities.
- The sum of the singles counts each "exactly two" region twice and the triple region three times; the sum of the pairs counts each "exactly two" region once and the triple region three times.
- So $$P(\text{exactly one}) = \Sigma\,\text{singles} - 2\,\Sigma\,\text{pairs} + 3P(\text{triple})$$
- $$= 0.83 - 2(0.26) + 3(0.03) = 0.40$$
- In the same way, $$P(\text{exactly two}) = \Sigma\,\text{pairs} - 3P(\text{triple}) = 0.26 - 0.09 = 0.17$$ each pairwise overlap contains the triple region once.
Answer: $0.4$ (and $0.17$ for exactly two)
The formula can also be run backward. If the union is known and one term is missing, such as $P(A \cap B \cap C)$, write out inclusion–exclusion and solve for it. And "none of the three" is the complement of the union.
Common mistake
Stopping after subtracting the pairs. The triple region was added three times (once in each single) and subtracted three times (once in each pair), so without the last term it is not counted at all. Check: your answer must be at least as large as the largest single probability, and at most their sum.
For $n$ events, add all singles, subtract all pairs, add all triples, subtract all quadruples, and so on. One famous use: if $n$ people grab coats at random, the chance that nobody gets their own coat is $1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \cdots \pm \frac{1}{n!}$, which is close to $\frac{1}{e} \approx 0.368$ for any $n$ beyond 5.