The Rules of Probability: Complement, Addition & Inclusion–Exclusion

Every event $A$ and its complement $A^c$ are disjoint and together fill the whole sample space $\Omega$, so additivity gives the complement rule $\mathbb{P}(A)+\mathbb{P}(A^c)=1,\qquad\text{equivalently}\qquad \mathbb{P}(A^c)=1-\mathbb{P}(A).$ Its value is tactical: whenever $A$ is awkward to count but its complement is simple, compute $\mathbb{P}(A^c)$ and subtract. The headline case is the complement trick for 'at least one' events, $\mathbb{P}(\text{at least one})=1-\mathbb{P}(\text{none}),$ since 'none' is usually one clean scenario while 'at least one' hides many. Example 10.4 is the template: a fair die is rolled four times and we want $\mathbb{P}(\text{some value appears more than once})$. Counting the ways a repeat happens (a value twice, thrice, four times, or two values twice each) invites overcounting; instead the complement 'all four rolls different' is a single ordered selection. There are $6\cdot5\cdot4\cdot3=360$ ways to get four distinct values out of $\#\Omega=6^4=1296$ equally likely outcomes, so $\mathbb{P}(A^c)=\tfrac{360}{1296}=\tfrac{5}{18}$ and $\mathbb{P}(A)=1-\tfrac{5}{18}=\tfrac{13}{18}$.

Two events are mutually exclusive (or disjoint) when they cannot both occur, i.e. $A\cap B=\varnothing$. Probability axiom 3 (additivity) says that for pairwise disjoint events $A_1,A_2,A_3,\dots$ with union $A$, $\mathbb{P}(A)=\mathbb{P}(A_1)+\mathbb{P}(A_2)+\mathbb{P}(A_3)+\cdots\qquad(\textbf{Disjoint Sum Rule}).$ This is how a complicated event is tamed: break it into simpler, non-overlapping pieces and add. Example 10.3 does exactly this \u2014 drawing two balls from an urn of $30$ red, $20$ green and $10$ yellow, the event 'exactly one red or exactly one yellow' splits into three disjoint cases (red+green, yellow+green, red+yellow) whose counts $600+200+300=1100$, out of $\binom{60}{2}=1770$, give $\tfrac{1100}{1770}=\tfrac{110}{177}$. A warning that motivates the rest of the lesson: the clean identity $\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)$ holds only for disjoint events. The moment $A$ and $B$ can occur together, adding $\mathbb{P}(A)$ and $\mathbb{P}(B)$ counts the overlap twice and overshoots.

When $A$ and $B$ overlap, the General Sum Rule repairs the double-count by subtracting the intersection once: $\mathbb{P}(A\cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)\qquad(\textbf{addition rule}).$ The reason is a disjoint decomposition read off a Venn diagram: $A\cup B$ is the disjoint union of $A\cap B^c$, $A\cap B$, and $A^c\cap B$, so $\mathbb{P}(A\cup B)=\mathbb{P}(AB^c)+\mathbb{P}(AB)+\mathbb{P}(A^cB)$. Since $\mathbb{P}(A)=\mathbb{P}(AB^c)+\mathbb{P}(AB)$ and $\mathbb{P}(B)=\mathbb{P}(A^cB)+\mathbb{P}(AB)$, the sum $\mathbb{P}(A)+\mathbb{P}(B)$ includes $\mathbb{P}(AB)$ twice \u2014 hence the single subtraction. When $A\cap B=\varnothing$ the correction term is $0$ and we recover the Disjoint Sum Rule, so the addition rule is the genuinely general statement. The same decomposition underlies the Law of Total Probability $\mathbb{P}(B)=\mathbb{P}(A\cap B)+\mathbb{P}(A^c\cap B)$, which splits any event $B$ along $A$ and $A^c$, and the Monotonicity Rule: if $A\subseteq B$ then $\mathbb{P}(A)\le\mathbb{P}(B)$.

Remark 10.3 draws two consequences. First, rearranging the addition rule expresses the intersection through the other three quantities: $\mathbb{P}(A\cap B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cup B).$ This is often the only route to $\mathbb{P}(A\cap B)$ when the union is what you are given. Second, the principle extends to any finite number of events by inclusion\u2013exclusion: add the singles, subtract the pairwise intersections, add back the triple. For three events, $\mathbb{P}(A\cup B\cup C)=\mathbb{P}(A)+\mathbb{P}(B)+\mathbb{P}(C)-\mathbb{P}(A\cap B)-\mathbb{P}(A\cap C)-\mathbb{P}(B\cap C)+\mathbb{P}(A\cap B\cap C).$ The alternating signs correct the overcount region by region: the triple overlap is added three times by the singles and removed three times by the pairs, so the final $+\mathbb{P}(A\cap B\cap C)$ restores it exactly once. Example 10.5 revisits the urn of Example 10.3 with $A=\{\text{exactly one red}\}$ and $B=\{\text{exactly one yellow}\}$, so $A\cap B=\{\text{one red and one yellow}\}$, and $\mathbb{P}(A\cup B)=\tfrac{900}{1770}+\tfrac{500}{1770}-\tfrac{300}{1770}=\tfrac{1100}{1770}=\tfrac{110}{177}$ \u2014 the same answer the disjoint method gave, now via inclusion\u2013exclusion.