Probability Rules for Events

Context: FIT1058_MOC · set operations on events give probability rules · complement, difference, general union · mirror the set-size laws

Quick Revision

• 🎯 Objective: set operations on events ⟹ probability identities ➔ complement / difference / union.
• 📦 Core Components: $\mathrm{Pr}(\overline{A})=1-\mathrm{Pr}(A)$ ➔ difference ➔ two-event inclusion–exclusion.
• ⚡ Key Constraint: these mirror the set-size laws divided by $|U|$; complement is the workhorse shortcut.

📝 Core

1. The Three Rules

• Complement ➔ $\mathrm{Pr}(\overline{A})=1-\mathrm{Pr}(A)$.
• Difference ➔ $\mathrm{Pr}(B\setminus A)=\mathrm{Pr}(B)-\mathrm{Pr}(A\cap B)$.
• Union ➔ $\mathrm{Pr}(A\cup B)=\mathrm{Pr}(A)+\mathrm{Pr}(B)-\mathrm{Pr}(A\cap B)$.

2. Consequences

• Complement tactic ➔ “at least one” $=1-$“none”.
• Monotonicity ➔ $A\subseteq B\Rightarrow \mathrm{Pr}(A)\le \mathrm{Pr}(B)$.
• Union↔intersection ➔ knowing one gives the other.

3. Partition Trick

• Distributive ➔ $B={⨆}_i{B}_i\Rightarrow \mathrm{Pr}(A\cap B)={\sum }_i\mathrm{Pr}(A\cap B_i)$.
• Seed of ➔ Law of Total Probability.

Key identities:

$\mathrm{Pr}(\overline{A})=1-\mathrm{Pr}(A),\mathrm{Pr}(B\setminus A)=\mathrm{Pr}(B)-\mathrm{Pr}(A\cap B)$ $\mathrm{Pr}(A\cup B)=\mathrm{Pr}(A)+\mathrm{Pr}(B)-\mathrm{Pr}(A\cap B)$

⚖️ Core Decision Matrix

Target	Set form	Rule
$\overline{A}$	$U\setminus A$	$1-\mathrm{Pr}(A)$
$B\setminus A$	difference	$\mathrm{Pr}(B)-\mathrm{Pr}(A\cap B)$
$A\cup B$	union	$\mathrm{Pr}A+\mathrm{Pr}B-\mathrm{Pr}(A\cap B)$
$A\cap B$	intersection	$\mathrm{Pr}A+\mathrm{Pr}B-\mathrm{Pr}(A\cup B)$

When It Flips: these mirror the set-size laws over $|U|$ — dividing $|A\cup B|=|A|+|B|-|A\cap B|$ by $|U|$ gives the probability rule in the uniform case; the general-union rule is two-event inclusion–exclusion.

📊 Exam Execution Trace

Applied Exercise

Problem: Find $\mathrm{Pr}(\text{not blank})$ and $\mathrm{Pr}(\text{consonant})$. Derivation Proof / Hand-Calculation Walkthrough:

$$\begin{array}{rl}\mathrm{Pr}(\text{not blank})& =1-0.02=0.98\\ \mathrm{Pr}(\text{consonant})& =\mathrm{Pr}(\text{non-blank})-\mathrm{Pr}(\text{vowel})=0.98-0.42=0.56\end{array}$$

Final Extracted Output: 0.98 (complement); 0.56 (difference, since vowels $\subseteq$ non-blanks).

⚠️ Common Mistakes

• 💡 Difference needs $A\cap B$, not $A$ ➔ $\mathrm{Pr}(B)-\mathrm{Pr}(A)$ only holds when $A\subseteq B$; in general subtract $\mathrm{Pr}(A\cap B)$.

🧠 Active Recall

Derive $\mathrm{Pr}(A)=1-\mathrm{Pr}(\overline{A})$ and $\mathrm{Pr}(A\cup B)=\mathrm{Pr}(A)+\mathrm{Pr}(B)-\mathrm{Pr}(A\cap B)$.

• Hint: Disjoint decomposition.

Answer

• Short answer: $A\bigsqcup \overline{A}=U$ ⟹ complement; $A\cup B=(A\setminus B)\bigsqcup (A\cap B)\bigsqcup (B\setminus A)$ ⟹ union.
• Why: Double-count fix ➔ adding $\mathrm{Pr}(A)+\mathrm{Pr}(B)$ counts $A\cap B$ twice.

Why does $A\subseteq B$ imply $\mathrm{Pr}(A)\le \mathrm{Pr}(B)$, and what replaces $\mathrm{Pr}(B)-\mathrm{Pr}(A)$ otherwise?

• Hint: Monotonicity.

Answer

• Short answer: $B=A\bigsqcup (B\setminus A)$ with nonnegative parts ⟹ $\mathrm{Pr}(B)\ge \mathrm{Pr}(A)$; general form uses $\mathrm{Pr}(A\cap B)$.
• Why: Special case ➔ $A\subseteq B$ makes $A=A\cap B$.