Independent Events

Context: FIT1058_MOC · the occurrence of one does not affect the other · defined by $\mathrm{Pr}(A\cap B)=\mathrm{Pr}(A)\mathrm{Pr}(B)$ · multiplicative, the dual of disjoint additivity

Quick Revision

• 🎯 Objective: both occur ⟹ product ➔ $\mathrm{Pr}(A\cap B)=\mathrm{Pr}(A)\mathrm{Pr}(B)$.
• 📦 Core Components: numerical test ➔ multiplicative over intersections ➔ series/parallel reliability.
• ⚡ Key Constraint: independent ≠ mutually exclusive (disjoint positive-probability events are dependent).

📝 Core

1. The Definition

• Test ➔ $\mathrm{Pr}(A\cap B)=\mathrm{Pr}(A)\cdot \mathrm{Pr}(B)$ — purely numerical.
• Not mechanism ➔ decided by the numbers, not intuition.
• Multiplicative ➔ probability factors over independent intersections (dual of disjoint additivity).

2. Test, Don’t Guess

• Looks linked, is independent ➔ possible when the product equation happens to hold.
• Looks independent, isn’t ➔ always verify $\mathrm{Pr}(A\cap B)=\mathrm{Pr}(A)\mathrm{Pr}(B)$.

3. vs Mutually Exclusive

• Disjoint ➔ $\mathrm{Pr}(A\cap B)=0\ne \mathrm{Pr}(A)\mathrm{Pr}(B)$ for positive-probability events ⟹ dependent.
• Exclusivity ➔ one prevents the other, the opposite of independence.

Key identities:

$\mathrm{Pr}(A\cap B)=\mathrm{Pr}(A)\mathrm{Pr}(B)$ $\text{series: }p\cdot p=p^2;\text{parallel: }1-(1-p{)}^2=2p-p^2$

⚖️ Core Decision Matrix

Structure	Survival	Method
series (AND)	$p^2$	multiply
parallel (OR)	$2p-p^2$	complement $1-(1-p{)}^2$
independent	$\mathrm{Pr}(A)\mathrm{Pr}(B)$	product
mutually exclusive	dependent	—

When It Flips: probability is multiplicative over independent intersections (as it is additive over disjoint unions) — the tool for decomposing a complex event into independent pieces; the complement route is quickest for "at least one".

📊 Exam Execution Trace

Applied Exercise

Problem: Two independent parallel links (each survives w.p. $p$) — survival probability, two ways. Derivation Proof / Hand-Calculation Walkthrough:

$$\begin{array}{rl}\text{union}& :p+p-p\cdot p=2p-p^2\\ \text{complement}& :1-(1-p{)}^2=2p-p^2\end{array}$$

Final Extracted Output: $2p-p^2$; the complement route is quicker.

⚠️ Common Mistakes

• 💡 Independent ≠ mutually exclusive ➔ disjoint positive-probability events are dependent ($0\ne \mathrm{Pr}(A)\mathrm{Pr}(B)$); “separate on a Venn diagram” is not independence.

🧠 Active Recall

Define independence and why mutually exclusive events (positive probability) are never independent.

• Hint: Product vs zero.

Answer

• Short answer: Independent ⟺ $\mathrm{Pr}(A\cap B)=\mathrm{Pr}(A)\mathrm{Pr}(B)$; disjoint gives $\mathrm{Pr}(A\cap B)=0$.
• Why: $>0\ne 0$ ➔ exclusivity means $A$ prevents $B$ — strongest dependence.

For two independent links each surviving w.p. $p$, give series and parallel survival.

• Hint: AND multiplies, OR via complement.

Answer

• Short answer: Series $p^2$; parallel $1-(1-p{)}^2=2p-p^2$.
• Why: Complement simplest ➔ “at least one” = $1-$“both fail”.