Conditional Probability

Context: FIT1058_MOC · the probability of $A$ given $B$ occurred · restricts the sample space to $B$ and rescales · the basis of Bayes’ Theorem

Quick Revision

• 🎯 Objective: $\mathrm{Pr}(A\mid B)=\mathrm{Pr}(A\cap B)/\mathrm{Pr}(B)$ ➔ treat $B$ as the new sample space.
• 📦 Core Components: restrict to $B$ ➔ rescale by $1/\mathrm{Pr}(B)$ ➔ multiplication rule.
• ⚡ Key Constraint: requires $\mathrm{Pr}(B)>0$; independence ⟺ $\mathrm{Pr}(A\mid B)=\mathrm{Pr}(A)$.

📝 Core

1. The Definition

• Formula ➔ $\mathrm{Pr}(A\mid B)=\frac{\mathrm{Pr}(A\cap B)}{\mathrm{Pr}(B)}$ ($\mathrm{Pr}(B)>0$).
• Restrict ➔ $B$ becomes the new sample space.
• Multiplication rule ➔ $\mathrm{Pr}(A\cap B)=\mathrm{Pr}(B)\mathrm{Pr}(A\mid B)=\mathrm{Pr}(A)\mathrm{Pr}(B\mid A)$.

2. Why Divide by $\mathrm{Pr}(B)$

• Rescale ➔ retained outcomes sum to $\mathrm{Pr}(B)\le 1$; divide by $\mathrm{Pr}(B)$ to renormalise.
• Only $A\cap B$ counts ➔ within $B$, $A$ is realised by $A\cap B$.

3. Special Cases

• $\mathrm{Pr}(B\mid B)=1$, $\mathrm{Pr}(\varnothing \mid B)=0$.
• Disjoint ➔ $\mathrm{Pr}(A\mid B)=0$.
• Independent ➔ $\mathrm{Pr}(A\mid B)=\mathrm{Pr}(A)$.

Key identities:

$\mathrm{Pr}(A\mid B)=\frac{\mathrm{Pr}(A\cap B)}{\mathrm{Pr}(B)},\mathrm{Pr}(A\cap B)=\mathrm{Pr}(B)\mathrm{Pr}(A\mid B)$

When It Flips: independence ⟺ $\mathrm{Pr}(A\mid B)=\mathrm{Pr}(A)$ ⟺ $\mathrm{Pr}(A\cap B)=\mathrm{Pr}(A)\mathrm{Pr}(B)$; the symmetric multiplication rule yields Bayes' Theorem.

📊 Exam Execution Trace

Manual Execution Trace

Scrabble: $\mathrm{Pr}(\text{vowel})=0.42$, $\mathrm{Pr}(\text{non-blank})=0.98$:

Step / State	Quantity	Value
0 (Init)	—	—
1	vowel ∩ non-blank	= vowel = 0.42
2	$\mathrm{Pr}(\text{non-blank})$	0.98
3	$\mathrm{Pr}(\text{vowel}\mid \text{non-blank})$	$\approx 0.43$

⚠️ Common Mistakes

• 💡 Numerator is $A\cap B$, not $A$ ➔ $\mathrm{Pr}(X)/\mathrm{Pr}(B)$ only when $X\subseteq B$; otherwise use $A\cap B$. Conditioning on an impossible event is undefined.

🧠 Active Recall

Define $\mathrm{Pr}(A\mid B)$ and justify the division by $\mathrm{Pr}(B)$.

• Hint: Renormalise on $B$.

Answer

• Short answer: $\mathrm{Pr}(A\mid B)=\mathrm{Pr}(A\cap B)/\mathrm{Pr}(B)$; $B$ is the new sample space.
• Why: Sum to 1 ➔ retained masses sum to $\mathrm{Pr}(B)$; dividing renormalises.

How does conditional probability characterise independence, and what is the multiplication rule?

• Hint: Unchanged conditional.

Answer

• Short answer: Independent ⟺ $\mathrm{Pr}(A\mid B)=\mathrm{Pr}(A)$ ⟺ $\mathrm{Pr}(A\cap B)=\mathrm{Pr}(A)\mathrm{Pr}(B)$.
• Why: Symmetric rule ➔ $\mathrm{Pr}(A\cap B)=\mathrm{Pr}(B)\mathrm{Pr}(A\mid B)=\mathrm{Pr}(A)\mathrm{Pr}(B\mid A)$ gives Bayes.