Worked examples — Mutually exclusive events — addition rule

Every symbol used here is defined in the parent note. Quick refresher of the two workhorses:

---

The scenario matrix

Every addition-rule problem falls into exactly one of these case classes. The table lists them all, and the examples below are labelled with the cell they hit.

Cell	Case class	What makes it tricky	Example
C1	Disjoint events, just add	nothing — the friendly case	Ex 1
C2	Overlapping events, must subtract	the double-count trap	Ex 2
C3	Three+ pairwise-disjoint events	the sum $\sum P(A_i)$	Ex 3
C4	Complement route ($P(A^{\prime })=1-P(A)$)	"not" phrasing	Ex 4
C5	Degenerate: an impossible ($P=0$) or certain ($P=1$) event	zero / one edge inputs	Ex 5
C6	Impossibility check: $P(A)+P(B)>1$	can they even be disjoint?	Ex 6
C7	Real-world word problem	translating words → sets	Ex 7
C8	Exam twist: solve backwards for an unknown overlap	rearranging the general rule	Ex 8

We use Venn diagrams throughout — two overlapping ovals inside a rectangle $S$. See Venn diagrams in probability.

---

Example 1 — Disjoint, just add (Cell C1)

Step 1 — Write the events as sets. $A=\{1\}$, $B=\{6\}$. Why this step? Turning words into subsets of the sample space $S=\{1,2,3,4,5,6\}$ lets us see whether they overlap. See Probability — basic definitions & sample space.

Step 2 — Check the overlap. $A\cap B=\varnothing$ — no face is both a 1 and a 6. Why this step? This is the question the parent note begs us to ask: can both happen at once? No → the subtraction term is zero.

Step 3 — Add. Each face is equally likely, so $P(A)=\frac{1}{6}$, $P(B)=\frac{1}{6}$, and $P(A\cup B)=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}.$ Why this step? With zero overlap the special rule applies — plain addition.

---

Example 2 — Overlapping, must subtract (Cell C2)

Step 1 — Name the events. $A=\{\text{hearts}\}$, $B=\{\text{face cards}\}$. Why this step? We need the sizes and, crucially, the overlap.

Step 2 — Find the overlap. Cards that are both a heart and a face card: the Jack, Queen, King of hearts — that's $3$ cards. So $P(A\cap B)=\frac{3}{52}$. Why this step? These three cards live in both ovals of the Venn diagram (Fig 1). Adding $P(A)+P(B)$ counts each of them twice.

Step 3 — Apply the general rule. $P(A\cup B)=\frac{13}{52}+\frac{12}{52}-\frac{3}{52}=\frac{22}{52}=\frac{11}{26}.$ Why this step? We add the pieces, then remove the double-counted overlap once — inclusion–exclusion. See General addition rule & inclusion–exclusion.

---

Example 3 — Three pairwise-disjoint events (Cell C3)

Step 1 — Confirm pairwise disjoint. A single spin gives exactly one colour, so no two colour-events can occur together: every pairwise intersection is $\varnothing$. Why this step? The $n$-event rule $P(A_1\cup \cdots \cup A_n)=\sum P(A_i)$ only applies when the events are pairwise mutually exclusive.

Step 2 — Sum the three probabilities. $P=0.4+0.25+0.2=\mathrm{0.85.}$ Why this step? With every overlap zero, every subtraction term vanishes — the union collapses to a plain sum.

---

Example 4 — The complement route (Cell C4)

Step 1 — Identify the event and its complement. Let $A=\{\text{picks correctly}\}$, so $A^{\prime }=\{\text{does not pick correctly}\}$. Why this step? "Not" phrasing screams complement. $A$ and $A^{\prime }$ are mutually exclusive and together fill $S$ (exhaustive). See Complementary events.

Step 2 — Use the complement corollary. Since $P(A)+P(A^{\prime })=1$, $P(A^{\prime })=1-P(A)=1-\frac{1}{5}=\frac{4}{5}.$ Why this step? Computing the four wrong options directly also works, but the complement is one clean subtraction — far less error-prone.

---

Example 5 — Degenerate inputs: zero and certain (Cell C5)

Step 1 — Assign the edge probabilities. No face shows 7, so $A=\varnothing$ and $P(A)=0$. Every face is $\le 6$, so $B=S$ and $P(B)=1$. Why this step? These are the boundary values probability allows: $0\le P\le 1$. They test whether the rule still behaves.

Step 2 — Union. Since $A=\varnothing$, "$A$ or $B$" is just $B$: $P(A\cup B)=P(\varnothing \cup S)=P(S)=1.$ Check via the rule: $A\cap B=\varnothing \cap S=\varnothing$, so $P(A\cup B)=0+1-0=1.$ ✓ Why this step? Adding an impossible event changes nothing — its probability contributes $0$.

Step 3 — Intersection. $A\cap B=\varnothing \cap S=\varnothing$, so $P(A\cap B)=0$. Why this step? An impossible event shares no outcome with anything, so $A$ and $B$ here are (trivially) mutually exclusive.

---

Example 6 — Can they be disjoint at all? (Cell C6)

Step 1 — Test the disjoint assumption. If mutually exclusive, $P(\text{rain}\cup \text{wind})=0.7+0.5=1.2$. Why this step? Any probability, including a union, must satisfy $P\le 1$. A value of $1.2$ is forbidden.

Step 2 — Conclude they must overlap. Because $1.2>1$, they cannot be disjoint. Use the general rule with the constraint $P(\text{rain}\cup \text{wind})\le 1$: $P(\text{rain}\cap \text{wind})=P(\text{rain})+P(\text{wind})-P(\text{rain}\cup \text{wind})\ge 0.7+0.5-1=\mathrm{0.2.}$ Why this step? Rearranging the general rule turns the "$\le 1$" ceiling into a floor on the overlap.

---

Example 7 — Real-world word problem (Cell C7)

Step 1 — Translate to probabilities. With $n(S)=30$: $P(F)=\frac{18}{30},P(Sp)=\frac{12}{30},P(F\cap Sp)=\frac{5}{30}.$ Why this step? Equally-likely students mean probability = (count)/(30).

Step 2 — Apply the general rule (they overlap). $P(F\cup Sp)=\frac{18}{30}+\frac{12}{30}-\frac{5}{30}=\frac{25}{30}=\frac{5}{6}.$ Why this step? The 5 both-language students sit in the overlap of the Venn diagram; adding without subtracting would double-count them.

Step 3 — "Neither" via complement. $P(\text{neither})=1-P(F\cup Sp)=1-\frac{25}{30}=\frac{5}{30}=\frac{1}{6}.$ Why this step? "Studies at least one" and "studies none" are complementary.

---

Example 8 — Exam twist: solve backwards (Cell C8)

Step 1 — Rearrange the general rule for the overlap. $P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.55+0.4-0.8=\mathrm{0.15.}$ Why this step? The general rule has four quantities; given any three, algebra hands you the fourth. This is the exam's favourite move.

Step 2 — Mutually exclusive? No: $P(A\cap B)=0.15\ne 0$, so there is an overlap. Why this step? Mutual exclusivity is exactly the statement $P(A\cap B)=0$; a nonzero overlap rules it out.

Step 3 — Independent? For independence we would need $P(A\cap B)=P(A)P(B)$. Check: $0.55\times 0.4=0.22\ne 0.15$. So they are not independent either. See Independent events — multiplication rule and Conditional probability. Why this step? Independence is a multiplication test on the intersection — a completely different condition from exclusivity (which is an addition-side condition).

---

Coverage check — did we hit every cell?

Recall Matrix cell → example

C1 disjoint add ::: Example 1 (die, 1 or 6) C2 overlapping subtract ::: Example 2 (heart or face card) C3 three pairwise-disjoint ::: Example 3 (spinner colours) C4 complement route ::: Example 4 (wrong guess) C5 degenerate 0 / 1 inputs ::: Example 5 (roll a 7 / roll $\le 6$) C6 sum > 1 impossibility ::: Example 6 (rain & wind) C7 real-world word problem ::: Example 7 (French / Spanish) C8 exam twist, solve backwards ::: Example 8 (find the overlap)

---

Connections

• Mutually exclusive events — addition rule (parent)
• General addition rule & inclusion–exclusion
• Complementary events
• Independent events — multiplication rule
• Conditional probability
• Venn diagrams in probability
• Probability — basic definitions & sample space