Conditional Probability and Independence

🎯 Learning Objectives

• Compute conditional probability $P(A|B)$ for any two events
• Understand when and why conditional probability differs from joint probability
• Determine whether events are independent using the formal definition
• Apply the multiplication rule for dependent events
• Use probability trees to visualize conditional probabilities

📋 Prerequisites

• Probability Intro (11-probability-intro) — sample spaces, events, basic probability
• Probability Rules (12-probability-rules) — addition and multiplication rules
• Basic algebra

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📖 Core Content

14.1 Intuition: How New Information Changes Probability

Everyday analogy: Before a cricket match, you estimate India's chance of winning at 60%. But then you learn the pitch is a spin-friendly one (favoring India), and you update your estimate to 75%. That's conditional probability — updating based on new information. 🔑 Key Insight: Almost all real-world probability is conditional. When you check the weather forecast, you're looking at $P(\text{rain} | \text{current weather conditions})$.

14.2 Formal Definition

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

• $P(B) > 0$. If $P(B) = 0$, $P(A|B)$ is undefined (you can't condition on an impossible event).

14.3 Intuitive Interpretation

14.4 The Multiplication Rule (General Form)

$$P(A \cap B) = P(A) \times P(B|A) = P(B) \times P(A|B)$$ $$P(\text{H}_2 | \text{H}_1) = \frac{12}{51}$$ $$P(\text{H}_1 \cap \text{H}_2) = \frac{13}{52} \times \frac{12}{51} = \frac{156}{2652} = \frac{1}{17}$$

14.5 Independence Revisited

$$P(A|B) = P(A)$$

Disjoint vs. Independent

CRITICAL: Disjoint events are NEVER independent (unless one has probability 0). Proof: If A and B are disjoint and $P(A) > 0$, $P(B) > 0$, then $P(A|B) = 0 \neq P(A)$. Knowing B happened tells you A definitely didn't happen — that's dependence.

14.6 Conditional Probability from Contingency Tables

$$P(\text{Coffee} | \text{Male}) = \frac{P(\text{Coffee} \cap \text{Male})}{P(\text{Male})} = \frac{40/150}{70/150} = \frac{40}{70} = \frac{4}{7}$$

14.7 Law of Total Probability

$$P(A) = P(A|B_1)P(B_1) + P(A|B_2)P(B_2) + \cdots + P(A|B_k)P(B_k)$$ $$P(\text{defective}) = P(D|M_1)P(M_1) + P(D|M_2)P(M_2)$$ $$= 0.05 \times 0.60 + 0.08 \times 0.40 = 0.030 + 0.032 = 0.062$$

14.8 Worked Examples

Example 1: Basic Conditional Probability (Easy)

$$P(6|\text{even}) = \frac{1/6}{1/2} = \frac{1}{3}$$

Example 2: From Contingency Table (Medium)

Example 3: Testing Independence (Harder)

14.9 Edge Cases & Gotchas

Conditioning on a Zero-Probability Event

Symmetry of Conditional Probability

14.10 Why This Matters

• Bayes' theorem (next topic): Updating beliefs with evidence
• Statistical inference: All conclusions are conditional on the data
• Machine learning: Naive Bayes classifier, Bayesian networks
• Medical testing: Positive predictive value = $P(\text{disease} | \text{positive test})$

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📐 Key Formulas / Concepts

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⚠️ Common Pitfalls

Pitfall 1: Confusing $P(A|B)$ with $P(B|A)$

Pitfall 2: Assuming Independence Without Checking

Pitfall 3: Forgetting That Conditional Probabilities Are Probabilities

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📝 Practice Questions

Q1: Basic Conditional Probability

</strong>

Two fair dice are rolled. Given that the sum is 7, what's the probability that one die shows a 6?

<details> <strong>Solution</strong>

Step 1: Outcomes with sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes Step 2: Among these, those with a 6: (1,6), (6,1) → 2 outcomes Step 3: $P = 2/6 = 1/3$

$\boxed{1/3}$

</details> > **Q2: From Table** > > </strong> > >

Find $P(\text{Smartphone} | \text{Age < 30})$.

<details> <strong>Solution</strong>

$P(\text{Smartphone} | <30) = \frac{45}{60} = \frac{3}{4} = 0.75$

$\boxed{0.75}$

</details> > **Q3: Multiplication Rule** > > </strong> > > A bag has 5 red and 7 blue chips. Draw 2 without replacement. Probability both are the same color? > > <details> <strong>Solution</strong> > > $P(\text{both red}) = \frac{5}{12} \times \frac{4}{11} = \frac{20}{132}$ > > $P(\text{both blue}) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132}$ > > $P(\text{same color}) = \frac{20}{132} + \frac{42}{132} = \frac{62}{132} = \frac{31}{66}$ > > $\boxed{31/66 \approx 0.470}$ </details> > **Q4: Independence Check** > > </strong> > > $P(A) = 0.3$, $P(B) = 0.6$, $P(A \cap B) = 0.18$. Are A and B independent? > > <details> <strong>Solution</strong> > > $P(A) \times P(B) = 0.3 \times 0.6 = 0.18$ $P(A \cap B) = 0.18$ > > Since $0.18 = 0.18$, they **are independent**. > > Alternatively: $P(A|B) = 0.18/0.6 = 0.3 = P(A)$ ✅ </details> > **Q5: Law of Total Probability** > > </strong> > > A company has 3 divisions. Division A has 20% of employees (5% managers), Division B has 50% (10% managers), Division C has 30% (3% managers). What percent of all employees are managers? > > <details> <strong>Solution</strong> > > $P(\text{manager}) = P(M|A)P(A) + P(M|B)P(B) + P(M|C)P(C)$ $= 0.05(0.20) + 0.10(0.50) + 0.03(0.30)$ $= 0.01 + 0.05 + 0.009 = 0.069$ > > So 6.9% of all employees are managers. > > $\boxed{0.069 \text{ or } 6.9\%}$ </details> > **Q6: Two-Stage Drawing** > > </strong> > > A bag has 4 red, 3 blue, 5 green marbles. Draw 2 without replacement. $P(\text{second is blue})$? > > <details> <strong>Solution</strong> > > Using the law of total probability: > > $P(\text{second blue}) = P(\text{second blue} | \text{first red})P(\text{first red}) + P(\text{second blue} | \text{first blue})P(\text{first blue}) + P(\text{second blue} | \text{first green})P(\text{first green})$ > > $P(\text{second blue}) = \frac{3}{11} \times \frac{4}{12} + \frac{2}{11} \times \frac{3}{12} + \frac{3}{11} \times \frac{5}{12}$ > > $= \frac{12}{132} + \frac{6}{132} + \frac{15}{132} = \frac{33}{132} = \frac{1}{4}$ > > **Notice:** $P(\text{second blue}) = 3/12 = 1/4 = P(\text{first blue})$! The marginal probability of drawing a blue on the second draw is the same as the first. This is always true for random draws without replacement. > > $\boxed{1/4}$ </details> > **Q7: Conditional from Joint** > > </strong> > > $P(A) = 0.4$, $P(B) = 0.5$, $P(A \cup B) = 0.7$. Find $P(A|B)$. > > <details> <strong>Solution</strong> > > **Step 1:** Find $P(A \cap B)$: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $0.7 = 0.4 + 0.5 - P(A \cap B)$ $P(A \cap B) = 0.9 - 0.7 = 0.2$ > > **Step 2:** $P(A|B) = \frac{0.2}{0.5} = 0.4$ > > $\boxed{0.4}$ </details> > **Q8: Application — Weather** > > </strong> > > In a certain city: $P(\text{rain}) = 0.3$, $P(\text{cloudy}) = 0.5$, $P(\text{rain} \cap \text{cloudy}) = 0.25$. > > a) Find $P(\text{rain} | \text{cloudy})$. b) Find $P(\text{cloudy} | \text{rain})$. c) Are rain and cloudy independent? > > <details> <strong>Solution</strong> > > a) $P(\text{rain} | \text{cloudy}) = \frac{0.25}{0.5} = 0.5$ > > b) $P(\text{cloudy} | \text{rain}) = \frac{0.25}{0.3} \approx 0.833$ > > c) Check independence: $P(\text{rain}) \times P(\text{cloudy}) = 0.3 \times 0.5 = 0.15$ $P(\text{rain} \cap \text{cloudy}) = 0.25$ Since $0.15 \neq 0.25$, they're **dependent** (as expected). > > $\boxed{0.5,\ 0.833,\ \text{Dependent}}$ </details> > **Q9: Sensitivity/Specificity** > > </strong> > > A test has 90% sensitivity and 85% specificity. Disease prevalence is 2%. Find: > > a) $P(\text{positive} | \text{disease})$ b) $P(\text{negative} | \text{no disease})$ c) $P(\text{disease} | \text{positive})$ > > <details> <strong>Solution</strong> > > a) Sensitivity = $P(\text{positive} | \text{disease}) = 0.90$ > > b) Specificity = $P(\text{negative} | \text{no disease}) = 0.85$ > > c) $P(\text{positive}) = P(\text{pos} | \text{disease})P(\text{disease}) + P(\text{pos} | \text{no disease})P(\text{no disease})$ $= 0.90(0.02) + 0.15(0.98) = 0.018 + 0.147 = 0.165$ > > $P(\text{disease} | \text{positive}) = \frac{0.018}{0.165} \approx 0.109 = 10.9\%$ > > $\boxed{0.90,\ 0.85,\ 0.109}$ </details> > **Q10: Sequential Probability** > > </strong> > > A box has 3 red and 7 blue chips. Draw chips with replacement until you get a red one. Find the probability that you need exactly 3 draws. > > <details> <strong>Solution</strong> > > For "exactly 3 draws," the first 2 must be blue and the 3rd must be red. > > $P(\text{blue}) = 7/10$, $P(\text{red}) = 3/10$ > > Since draws are with replacement (independent): > > $P(\text{first 2 blue, 3rd red}) = \frac{7}{10} \times \frac{7}{10} \times \frac{3}{10} = \frac{147}{1000} = 0.147$ > > This is a geometric distribution problem (Week 11). > > $\boxed{0.147}$ </details> * * * ## 🔗 Cross-References - **Next topic:** [Bayes' Theorem](/notes/01-foundation-bsma1002-stats-1-week08-14-bayes-theorem) — the reverse conditional probability - **Previous:** [Probability Rules](/notes/01-foundation-bsma1002-stats-1-week07-12-probability-rules) — multiplication rule foundation - **Week 11 (Geometric Distribution):** "Waiting time" problems use conditional probability - **BSMA1004 (Stats 2):** Bayesian inference - **BSMA1001-maths-1:** Partition of a set (law of total probability) [Join Discord](https://discord.gg/gE2m4Qrdqv) [Previous**Probability Rules**](/notes/01-foundation-bsma1002-stats-1-week07-12-probability-rules)[Next**Bayes' Theorem**](/notes/01-foundation-bsma1002-stats-1-week08-14-bayes-theorem)