Probability Rules — Addition, Multiplication, and Trees

🎯 Learning Objectives

• Apply the addition rule for both disjoint and non-disjoint events
• Apply the multiplication rule for independent and dependent events
• Construct and use probability trees to solve multi-stage probability problems
• Determine whether events are independent
• Solve probability problems using systematic approaches

📋 Prerequisites

• Probability Intro (11-probability-intro) — sample spaces, events, basic probability
• Basic set theory: union, intersection, complement
• Conditional probability will be formalized in the next topic, but we'll use the intuition here

---

📖 Core Content

13.1 Intuition: Rules for Combining Probabilities

• Addition rule: What's the probability that A OR B happens?
• Multiplication rule: What's the probability that A AND B both happen?
• Probability trees: How do we compute probabilities over multiple sequential stages?

Everyday analogy: When planning a trip, you might need to know: probability of rain OR snow (addition), probability of rain AND your flight being delayed (multiplication), or the overall probability of a successful trip considering weather, traffic, and flight availability (tree). 🔑 Key Insight: The key distinction is always: are events disjoint (mutually exclusive) or independent? These terms sound similar but are very different.

13.2 The Addition Rule

13.2.1 For Disjoint (Mutually Exclusive) Events

$$P(A \cup B) = P(A) + P(B)$$

13.2.2 For Non-Disjoint Events (General Addition Rule)

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

Example

$$P(A \cup B) = 0.4 + 0.3 - 0.1 = 0.6$$

13.3 The Multiplication Rule

13.3.1 For Independent Events

$$P(A \cap B) = P(A) \times P(B)$$

• Flipping a coin twice: first flip doesn't affect second
• Rolling two dice: die A doesn't affect die B
• Drawing cards with replacement

13.3.2 For Dependent Events (General Multiplication Rule)

$$P(A \cap B) = P(A) \times P(B|A)$$

• Drawing cards without replacement: the first draw changes the deck composition

13.4 Checking Independence

1. $P(A \cap B) = P(A) \times P(B)$
2. $P(B|A) = P(B)$
3. $P(A|B) = P(A)$

Example

13.5 Probability Trees

How to Build a Tree

1. Start at the root
2. For each stage, draw branches for each possible outcome
3. Label each branch with its conditional probability
4. Multiply along paths to get joint probabilities
5. Sum relevant paths for the probability of interest

Example: Two Cards Without Replacement

13.6 "At Least One" Problems

$$P(\text{at least one 6}) = 1 - P(\text{no 6s}) = 1 - \left(\frac{5}{6}\right)^4 = 1 - \frac{625}{1296} = \frac{671}{1296} \approx 0.518$$

13.7 Worked Examples

Example 1: Addition Rule (Easy)

Example 2: Independence Check (Medium)

Example 3: Probability Tree (Harder)

$$P(D) = 0.03 + 0.032 = 0.062$$ $$P(M_1|D) = \frac{P(M_1 \cap D)}{P(D)} = \frac{0.03}{0.062} \approx 0.484$$

13.8 Edge Cases & Gotchas

Disjoint vs. Independent — Don't Confuse!

Key fact: Disjoint events (with positive probability) are NEVER independent! If A and B are disjoint and both have positive probability, then $P(A \cap B) = 0$ but $P(A)P(B) > 0$. Why? If A and B can't happen together, knowing that A happened tells you B definitely didn't happen — that's dependence.

"Or" in Everyday Language

13.9 Why This Matters

• Bayes' theorem (next topic): Based on the multiplication rule
• Diagnostic testing: Sensitivity, specificity, predictive values
• Risk analysis: Combining probabilities in chains of events
• Decision trees: Business decisions under uncertainty

---

📐 Key Formulas / Concepts

---

⚠️ Common Pitfalls

Pitfall 1: Adding Probabilities of Non-Disjoint Events Without Subtracting Overlap

Pitfall 2: Multiplying Probabilities of Dependent Events

Pitfall 3: Confusing Disjoint and Independent

---

📝 Practice Questions

Q1: Addition Rule

$P(A) = 0.35$, $P(B) = 0.25$, $P(A \cap B) = 0.10$. Find $P(A \cup B)$.

<details> <strong>Solution</strong>

$P(A \cup B) = 0.35 + 0.25 - 0.10 = 0.50$

$\boxed{0.50}$

</details> > **Q2: Multiplication Rule (Independent)** > > </strong> > > A fair coin is flipped and a fair die is rolled. Probability of heads AND a 6? > > <details> <strong>Solution</strong> > > These are independent events. > > $P(H \cap 6) = P(H) \times P(6) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$ > > $\boxed{1/12}$ </details> > **Q3: Without Replacement** > > </strong> > > Draw 2 cards from a deck without replacement. Probability both are aces? > > <details> <strong>Solution</strong> > > **Method 1:** $P(\text{first ace}) = 4/52$ $P(\text{second ace | first ace}) = 3/51$ $P(\text{both aces}) = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$ > > **Method 2 (combinations):** $\frac{C(4,2)}{C(52,2)} = \frac{6}{1326} = \frac{1}{221}$ > > $\boxed{1/221}$ </details> > **Q4: At Least One** > > A biased coin has P(H) = 0.6. Flip it 3 times. Probability of at least one head? > > <details> <strong>Solution</strong> > > $P(\text{at least one H}) = 1 - P(\text{no heads}) = 1 - P(\text{all tails})$ > > $P(\text{tails}) = 1 - 0.6 = 0.4$ > > $P(\text{all tails}) = 0.4^3 = 0.064$ > > $P(\text{at least one H}) = 1 - 0.064 = 0.936$ > > $\boxed{0.936}$ </details> > **Q5: Independence Check** > > </strong> > > $P(A) = 0.6$, $P(B) = 0.5$, $P(A \cap B) = 0.25$. Are A and B independent? > > <details> <strong>Solution</strong> > > $P(A) \times P(B) = 0.6 \times 0.5 = 0.30$ > > $P(A \cap B) = 0.25$ > > Since $0.25 \neq 0.30$, they are **not independent**. </details> > **Q6: Probability Tree** > > </strong> > > Bag A has 3 red, 2 blue. Bag B has 1 red, 4 blue. Flip a fair coin: if heads, draw from A; if tails, draw from B. Probability of drawing a red ball? > > <details> <strong>Solution</strong> > > **Tree:** > > - Heads (0.5) → Red from A: 3/5 → $0.5 \times 3/5 = 0.30$ > - Heads (0.5) → Blue from A: 2/5 → $0.5 \times 2/5 = 0.20$ > - Tails (0.5) → Red from B: 1/5 → $0.5 \times 1/5 = 0.10$ > - Tails (0.5) → Blue from B: 4/5 → $0.5 \times 4/5 = 0.40$ > > $P(\text{red}) = 0.30 + 0.10 = 0.40$ > > $\boxed{0.40}$ </details> > **Q7: Application — Diagnostic Test** > > </strong> > > A test for a disease has: > > - Sensitivity: 90% (correctly identifies those with the disease) > - Specificity: 95% (correctly identifies those without) > - Prevalence: 3% (the disease rate in the population) > > Find the probability that a randomly selected person tests positive. > > <details> <strong>Solution</strong> > > **Tree approach:** > > - Has disease (0.03) → positive (0.90): $0.03 \times 0.90 = 0.027$ > - Has disease (0.03) → negative (0.10): $0.03 \times 0.10 = 0.003$ > - No disease (0.97) → positive (0.05): $0.97 \times 0.05 = 0.0485$ > - No disease (0.97) → negative (0.95): $0.97 \times 0.95 = 0.9215$ > > $P(\text{positive}) = 0.027 + 0.0485 = 0.0755 \approx 7.55\%$ > > So about 7.55% of the population tests positive, but only a fraction (0.027/0.0755 ≈ 35.8%) actually have the disease. > > $\boxed{0.0755}$ </details> > **Q8: Extended Addition** > > </strong> > > $P(A) = 0.5$, $P(B) = 0.3$, $P(C) = 0.2$, $P(A \cap B) = 0.1$, $P(A \cap C) = 0.08$, $P(B \cap C) = 0.04$, $P(A \cap B \cap C) = 0.02$. Find $P(A \cup B \cup C)$. > > <details> <strong>Solution</strong> > > $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A\cap B) - P(A\cap C) - P(B\cap C) + P(A\cap B\cap C)$ > > $= 0.5 + 0.3 + 0.2 - 0.1 - 0.08 - 0.04 + 0.02 = 0.80$ > > $\boxed{0.80}$ </details> > **Q9: Three Events Independence** > > </strong> > > If $P(A) = 0.5$, $P(B) = 0.4$, $P(C) = 0.3$, and events are mutually independent, find $P(A \cap B \cap C)$. > > <details> <strong>Solution</strong> > > For mutually independent events: > > $P(A \cap B \cap C) = P(A) \times P(B) \times P(C) = 0.5 \times 0.4 \times 0.3 = 0.06$ > > $\boxed{0.06}$ </details> > **Q10: Application — System Reliability** > > </strong> > > A system works if at least one of two components works. Component A works with probability 0.9. Component B works with probability 0.8. They fail independently. What's the probability the system works? > > <details> <strong>Solution</strong> > > **Method 1 (Direct):** $P(\text{system works}) = P(A \cup B) = P(A) + P(B) - P(A \cap B)$ $= 0.9 + 0.8 - (0.9 \times 0.8) = 1.7 - 0.72 = 0.98$ > > **Method 2 (Complement):** $P(\text{system fails}) = P(\bar{A} \cap \bar{B}) = P(\bar{A}) \times P(\bar{B}) = 0.1 \times 0.2 = 0.02$ $P(\text{system works}) = 1 - 0.02 = 0.98$ > > $\boxed{0.98}$ </details> * * * ## 🔗 Cross-References - **Next topic:** [Conditional Probability](/notes/01-foundation-bsma1002-stats-1-week08-13-conditional-probability) — formalizing $P(A|B)$ - **After that:** [Bayes' Theorem](/notes/01-foundation-bsma1002-stats-1-week08-14-bayes-theorem) — the capstone of probability rules - **Week 9 (Random Variables):** Probability distributions build on these rules - **BSMA1004 (Stats 2):** Likelihood functions use the multiplication rule [Join Discord](https://discord.gg/gE2m4Qrdqv) [Previous**Probability Basics**](/notes/01-foundation-bsma1002-stats-1-week07-11-probability-intro)[Next**Conditional Probability**](/notes/01-foundation-bsma1002-stats-1-week08-13-conditional-probability)