We usually speak of the union of two events in terms of either/or disjunctive language. That is, what is the probability that the outcome wil be either E1 or E2? For example, if there is a chow chow and a German shepherd, what is the possibility that Felipe will choose either a German shepherd or a chow chow? We can express this as:
P(E1 ∪ E2) = |E1 ∪ E2| / |S|
In order to combine this, we cannot just add E1 and E2 together. But why? Because the outcome might include both events. For example, if there are 4 chow chows and 5 German shepherds, and one event has Felipe choosing 2 chow chows and a German shepherd, but another has him choosing 3 chow chows only, combining them through simple addition would entail counting some of the chow chows twice, which would skew the results.
Instead, “when computing the number of outocmes in the union of E1 and E2, we need to add the outcomces in E1 to the outcomes in E2, and to subtract the outcomes in both E1 and E2″(Steinhart, p. 112). Thus, as Steinhart notes, we would formally represent |E1 ∪ E2| as:
|E1 ∪ E2| = |E1| + |E2| – |E1 ∩ E2|
P(E1 ∪ E2) = (|E1| + |E2| – |E1 ∩ E2|) / |S|
Steinhart notes that “we can divide each term in the numerator by |S| to get a nicer formula expressing the probability of the union only in temrs of other probabilities”:
P(E1 ∪ E2) = |E1| / |S| + |E2| / |S| – |E1 ∩ E2| / |S|;
P(E1 ∪ E2) = P(E1) + P(E2) – P(E1 ∩ E2)
Steinhart illustrates this using the example of picking either a heart or a face card from a deck of cards:
P(Heart ∪ Face) = P(Heart) + P(Face) – P(Heart ∩ Face);
P(Heart ∪ Face) = 13/52 + 12/52 – 3/52 = 22/52.
What about when two groups are mutually exclusive? This means that they have an empty intersection because they have nothing in common. The probability, therefore, of any given outcome being in both E1 and E2 is zero. We can represent E1 and E2 as mutually exclusive as following:
P(E1 ∪ E2) = P(E1) + P(E2)
Steinhart, Eric. “More Precisely: The Mathematics You Need To Do Philosophy.” Broadview Press, 2009.