1. The Addition Side — Union Rules
Before we can talk about independence and multiplication, we need the addition rules firmly in place. These tell us how to compute the probability of at least one of several events occurring — i.e., the probability of a union.
1a. Addition Rule for Mutually Exclusive Events
Two events $A$ and $B$ are mutually exclusive (also called incompatible or disjoint) when they cannot both happen at the same time: $A \cap B = \varnothing$.
For mutually exclusive events:
Why it works: Since $A$ and $B$ share no outcomes, there is nothing counted twice when we add.
Quick example: Roll a fair die. Let $A = \{1, 2\}$ and $B = \{5, 6\}$. These are disjoint, so $P(A \cup B) = \tfrac{2}{6} + \tfrac{2}{6} = \tfrac{4}{6} = \tfrac{2}{3}$.
1b. General Addition Rule — Inclusion-Exclusion for Two Events
When events can overlap (they are compatible), simply adding their probabilities double-counts the outcomes in the intersection. We must subtract the overlap:
Counted example — resolving the double-count:
Roll a fair die. Let $A = \{2, 4, 6\}$ (even) and $B = \{3, 6\}$ (greater than 2 and divisible by 3).
- $P(A) = 3/6$, $P(B) = 2/6$
- $A \cap B = \{6\}$, so $P(A \cap B) = 1/6$
- Naïve sum: $3/6 + 2/6 = 5/6$ — but outcome 6 was counted twice.
- Correct: $P(A \cup B) = 3/6 + 2/6 - 1/6 = 4/6 = 2/3$.
Check by listing: $A \cup B = \{2, 3, 4, 6\}$, which has 4 outcomes out of 6. ✓
1c. Inclusion-Exclusion for Three Events
The same double-counting hazard extends. With three events we over-subtract, so we must add back the triple intersection: