Mutually Exclusive Events

Context: FIT1058_MOC · disjoint events · probability is additive over disjoint unions · the probabilistic Addition Principle

Quick Revision

• 🎯 Objective: disjoint events ($A\cap B=\varnothing$) ⟹ probability adds ➔ $\mathrm{Pr}(A\bigsqcup B)=\mathrm{Pr}(A)+\mathrm{Pr}(B)$.
• 📦 Core Components: pairwise disjoint ➔ additive ➔ decomposition tactic.
• ⚡ Key Constraint: add only when disjoint; mutually exclusive ≠ independent.

📝 Core

1. The Definition

• Disjoint ➔ $A\cap B=\varnothing$; one occurring prevents the other.
• Additive ➔ $\mathrm{Pr}(A\bigsqcup B)=\mathrm{Pr}(A)+\mathrm{Pr}(B)$ (union defined only when disjoint).

2. Additivity over $n$ Events

• Pairwise disjoint ➔ $\mathrm{Pr}({⨆}_i{A}_i)={\sum }_i\mathrm{Pr}(A_i)$.
• Foundational ➔ $\mathrm{Pr}(A)={\sum }_{x\in A}\mathrm{Pr}(x)$ is this rule on singletons.

3. Tactics & Contrasts

• Decompose ➔ split into disjoint simpler events, add.
• Only if disjoint ➔ overlap needs $-\mathrm{Pr}(A\cap B)$.
• ≠ independent ➔ disjoint positive-probability events are dependent.

Key identities:

$A\cap B=\varnothing \Rightarrow \mathrm{Pr}(A\bigsqcup B)=\mathrm{Pr}(A)+\mathrm{Pr}(B)$ $\mathrm{Pr}(A_1\bigsqcup \cdots \bigsqcup A_n)={\sum }_i\mathrm{Pr}(A_i)$

⚖️ Core Decision Matrix

Relation	Joint	Rule
mutually exclusive	$\mathrm{Pr}(A\cap B)=0$	add
overlapping	$>0$	inclusion–exclusion
independent	$\mathrm{Pr}(A)\mathrm{Pr}(B)$	multiply
disjoint + positive	dependent	—

When It Flips: this is the Addition Principle for measures — disjoint "or" adds, mirroring $|A\bigsqcup B|=|A|+|B|$. "Separate on a Venn diagram" ≠ independent; disjoint positive-probability events are in fact dependent.

📊 Exam Execution Trace

Manual Execution Trace

Two dice, total $\le 5$ or $\ge 10$:

Step / State	Event	$\mathrm{Pr}$	Disjoint?
0 (Init)	—	—	—
1	total $\le 5$	$\frac{10}{36}$	yes
2	total $\ge 10$	$\frac{6}{36}$	yes
3	union	$\frac{16}{36}=\frac{4}{9}$	—

⚠️ Common Mistakes

• 💡 Add only when disjoint ➔ $\mathrm{Pr}(A\cup B)=\mathrm{Pr}(A)+\mathrm{Pr}(B)$ iff $A\cap B=\varnothing$; else subtract $\mathrm{Pr}(A\cap B)$.

🧠 Active Recall

When is $\mathrm{Pr}(A\cup B)=\mathrm{Pr}(A)+\mathrm{Pr}(B)$, and how does it define probability itself?

• Hint: Disjoint additivity.

Answer

• Short answer: Exactly when $A\cap B=\varnothing$; generalises to pairwise-disjoint families.
• Why: Singletons ➔ $\mathrm{Pr}(A)={\sum }_{x\in A}\mathrm{Pr}(x)$ is additivity over disjoint $\{x\}$.

Why are two mutually exclusive events (both positive) necessarily not independent?

• Hint: Zero vs product.

Answer

• Short answer: $\mathrm{Pr}(A\cap B)=0$ but independence needs $\mathrm{Pr}(A)\mathrm{Pr}(B)>0$.
• Why: Prevention ➔ $A$ occurring rules out $B$ — the opposite of no influence.