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In this answer, we are given the solution for calculating the probability of at least of of two events occurring. How can we generalize that for 3 or more events?

For example, what is the probability of at least one of A or B or C occurring? Can I just induct the answer recursively by computing at least one of (A or B) and then using that answer (as shown in the above link) to compute the probability of (A or B) or C? Will this give the correct result and thus provide a pattern for generalizing for any number of options? Can the solution be written more cleanly as some kind of summation or the like with indexed events A sub 1, A sub 2, ... A sub n?

Community
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WilliamKF
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2 Answers2

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Generally, if we're looking at the probability of at least one of a group of events occurring, $$p(A_1\cup A_2\cup\ldots\cup A_n) $$ We use the fact that the probability of an event $E$ and the probability of its complement $E^c$ add to 1, hence, the above probability is equivalent with $$1-p(A_1\cup A_2\cup\ldots\cup A_n)^c,$$ $$=1-p(A_1^c\cap A_2^c\cap\ldots\cap A_n^c),$$ by De Morgan's Law. That way, instead of having to calculate all possible ways of at least one event occurring, we can find the single probability of no event occurring and subtract that from 1.

Brent
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Yes, you could just repeatedly apply the approach in the other answer; if you have $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ then it follows from your though that $$P(A\cup B\cup C)=P(A\cup B) + P(C) - P((A\cup B)\cap C)$$ and as $(A\cup B)\cap C=(A\cap C)\cup (B\cap C)$ we can simplify to $$P(A\cup B\cup C)=P(A\cup B) + P(C) - P((A\cap C)\cup (B\cap C))$$ then, applying the approach to the two unions on the left: $$P(A\cup B\cup C)=(P(A)+P(B)-P(A\cup B)) + P(C) - (P(A\cap C) + P(B\cap C) + P(A\cap C \cap B \cap C))$$ or more cleanly written $$P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C) +P(A\cap B\cap C).$$

You could certainly continue to apply this process, however it will become tedious before too long, as each addition term of the union about doubles the number of terms in the expansion. However, a simple generalization does exist, called the inclusion-exclusion principle. Essentially, what it says is that to calculate $P(A_1\cup A_2 \cup A_3 \cup \ldots)$, you sum up the individual probabilities $P(A_i)$ for each $A$, then subtract the probabilities $P(A_i\cap A_j)$ for distinct $i$ and $j$ (the probabilities of any two events happening), then add all $P(A_i\cap A_j \cap A_k)$ into the sum, subtract $P(A_i\cap A_j\cap A_k \cap A_s)$ and so on. The Wikipedia page illustrates it fairly well (though it uses cardinality of sets - but it applies more widely to any measure, of which probability and cardinality are examples)

Milo Brandt
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