Learning Objectives
By the end of this chapter, you will be able to:
- Apply the addition rule for mutually exclusive events
- Apply the general addition rule for non-mutually exclusive events
- Solve problems involving “or” probabilities
- Use Venn diagrams to visualize probability calculations
The Addition Rule
The addition rule is used when we want to find the probability of event A OR event B occurring (or both).
This answers questions like:
- What’s the probability of getting a King OR a Heart?
- What’s the probability that a student passed Math OR English?
- What’s the probability an employee is from HR OR Finance?
Case 1: Mutually Exclusive Events
Definition
Events are mutually exclusive if they cannot occur at the same time:
\[P(A \cap B) = 0\]flowchart LR
subgraph "Mutually Exclusive Events"
A["Event A"]
B["Event B"]
end
C["No overlap - cannot occur together"]
Addition Rule for Mutually Exclusive Events
\[P(A \cup B) = P(A) + P(B)\]Example 1: Rolling a Die
Problem: What is the probability of getting a 2 OR a 5?
Solution:
Events: A = {getting 2}, B = {getting 5}
These are mutually exclusive (can’t get both on one roll).
\[P(2 \text{ or } 5) = P(2) + P(5) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}\]Example 2: Employee Selection
Problem: In an office, 30% are in HR, 25% in Finance, and 20% in IT. What is the probability a random employee is in HR OR Finance?
Solution:
Since an employee can only be in one department (mutually exclusive):
\(P(\text{HR or Finance}) = P(\text{HR}) + P(\text{Finance})\) \(= 0.30 + 0.25 = 0.55 = 55\%\)
Example 3: Three Mutually Exclusive Events
Problem: A die is rolled. Find the probability of getting 1, 2, or 3.
Solution:
\(P(1 \text{ or } 2 \text{ or } 3) = P(1) + P(2) + P(3)\) \(= \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\)
Case 2: Non-Mutually Exclusive Events
Definition
Events are non-mutually exclusive if they CAN occur at the same time:
\[P(A \cap B) \neq 0\]flowchart LR
subgraph "Non-Mutually Exclusive Events"
A["Event A"]
B["A ∩ B"]
C["Event B"]
end
D["Overlap exists - can occur together"]
General Addition Rule
\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]We subtract $P(A \cap B)$ to avoid counting the overlap twice.
Why Subtract the Intersection?
flowchart TD
A["P(A) includes the overlap"]
B["P(B) also includes the overlap"]
C["If we just add P(A) + P(B)"]
D["Overlap is counted TWICE!"]
E["So we subtract P(A ∩ B) once"]
A --> C
B --> C
C --> D
D --> E
Step-by-Step Example 4: Cards
Problem: A card is drawn from a standard deck. Find the probability of drawing a King OR a Heart.
Solution:
Step 1: Identify the events
- A = Drawing a King (4 kings in deck)
- B = Drawing a Heart (13 hearts in deck)
- Are they mutually exclusive? NO! (King of Hearts exists)
Step 2: Calculate individual probabilities \(P(\text{King}) = \frac{4}{52}\) \(P(\text{Heart}) = \frac{13}{52}\)
Step 3: Find the intersection King AND Heart = King of Hearts = 1 card \(P(\text{King and Heart}) = \frac{1}{52}\)
Step 4: Apply general addition rule \(P(\text{King or Heart}) = P(\text{King}) + P(\text{Heart}) - P(\text{King and Heart})\) \(= \frac{4}{52} + \frac{13}{52} - \frac{1}{52}\) \(= \frac{16}{52} = \frac{4}{13} \approx 0.308\)
Step-by-Step Example 5: Survey Data
Problem: A survey of 100 government employees shows:
- 60 have a Master’s degree
- 45 have more than 5 years experience
- 30 have both
What is the probability that a randomly selected employee has a Master’s degree OR more than 5 years experience?
Solution:
Step 1: Define events
- A = Master’s degree
- B = More than 5 years experience
Step 2: Extract probabilities \(P(A) = \frac{60}{100} = 0.60\) \(P(B) = \frac{45}{100} = 0.45\) \(P(A \cap B) = \frac{30}{100} = 0.30\)
Step 3: Apply addition rule \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) \(= 0.60 + 0.45 - 0.30 = 0.75\)
Answer: 75% probability
Step-by-Step Example 6: Exam Passage
Problem: In a class of 50 students:
- 35 passed Statistics
- 28 passed Economics
- 20 passed both subjects
Find the probability that a randomly selected student: (a) Passed at least one subject (b) Failed both subjects
Solution:
Part (a): Passed at least one subject
\(P(\text{Stats}) = \frac{35}{50} = 0.70\) \(P(\text{Econ}) = \frac{28}{50} = 0.56\) \(P(\text{Both}) = \frac{20}{50} = 0.40\)
\(P(\text{At least one}) = P(\text{Stats or Econ})\) \(= 0.70 + 0.56 - 0.40 = 0.86\)
Part (b): Failed both subjects
Students who failed both = did NOT pass at least one \(P(\text{Failed both}) = 1 - P(\text{At least one})\) \(= 1 - 0.86 = 0.14\)
Answer: (a) 86% passed at least one; (b) 14% failed both
Venn Diagram Approach
Venn diagrams help visualize probability calculations:
Example 7: Using Venn Diagram
Problem: In a sample of 200 adults:
- 120 read newspapers
- 90 watch TV news
- 50 do both
Find probabilities using a Venn diagram.
Step 1: Fill in the Venn diagram
| Region | Calculation | Count |
|---|---|---|
| Newspaper only | 120 - 50 = 70 | 70 |
| Both | Given | 50 |
| TV only | 90 - 50 = 40 | 40 |
| Neither | 200 - 70 - 50 - 40 = 40 | 40 |
Step 2: Calculate probabilities
\(P(\text{Newspaper only}) = \frac{70}{200} = 0.35\) \(P(\text{TV only}) = \frac{40}{200} = 0.20\) \(P(\text{Both}) = \frac{50}{200} = 0.25\) \(P(\text{Neither}) = \frac{40}{200} = 0.20\)
Step 3: Verify \(P(\text{Newspaper or TV}) = 0.35 + 0.25 + 0.20 = 0.80\)
Or using formula: \(P(N \cup T) = 0.60 + 0.45 - 0.25 = 0.80\) ✓
Addition Rule for Three Events
\(P(A \cup B \cup C) = P(A) + P(B) + P(C)\) \(- P(A \cap B) - P(B \cap C) - P(A \cap C)\) \(+ P(A \cap B \cap C)\)
flowchart TD
A["Add individual probabilities"]
B["Subtract pairwise intersections"]
C["Add back triple intersection"]
A --> B --> C
Example 8: Three Events
Problem: For three events A, B, C:
- P(A) = 0.4, P(B) = 0.3, P(C) = 0.35
- P(A∩B) = 0.15, P(B∩C) = 0.12, P(A∩C) = 0.10
- P(A∩B∩C) = 0.05
Find P(A ∪ B ∪ C).
Solution:
\(P(A \cup B \cup C) = 0.4 + 0.3 + 0.35 - 0.15 - 0.12 - 0.10 + 0.05\) \(= 1.05 - 0.37 + 0.05 = 0.73\)
Common Mistakes to Avoid
Mistake 1: Adding Without Checking
❌ Wrong: $P(A \text{ or } B) = P(A) + P(B)$ always
✅ Right: Check if events are mutually exclusive first!
Mistake 2: Forgetting to Subtract Overlap
❌ Wrong: P(King or Heart) = $\frac{4}{52} + \frac{13}{52} = \frac{17}{52}$
✅ Right: P(King or Heart) = $\frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52}$
Mistake 3: Using Wrong Rule
| If events are… | Use… |
|---|---|
| Mutually exclusive | $P(A \cup B) = P(A) + P(B)$ |
| Not mutually exclusive | $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ |
Decision Flow for Addition Rule
flowchart TD
A["Need P(A or B)?"]
B["Can A and B occur together?"]
A --> B
B -->|No| C["Mutually Exclusive<br/>P(A∪B) = P(A) + P(B)"]
B -->|Yes| D["Not Mutually Exclusive<br/>P(A∪B) = P(A) + P(B) - P(A∩B)"]
Practice Problems
Problem 1
A card is drawn from a deck. Find the probability of getting: (a) An Ace or a King (b) A Queen or a Diamond (c) A face card (J, Q, K) or a red card
Problem 2
In a survey of 80 households:
- 55 have internet connection
- 40 have cable TV
- 25 have both
Find the probability that a randomly selected household has: (a) Internet or cable TV (b) Only internet (c) Neither internet nor cable TV
Problem 3
A die is rolled. Find the probability of getting: (a) An even number or a number less than 4 (b) A prime number or a number greater than 4
Problem 4
If P(A) = 0.5, P(B) = 0.4, and P(A ∪ B) = 0.7, find: (a) P(A ∩ B) (b) Are A and B mutually exclusive?
Problem 5
Three candidates X, Y, Z are running for office. P(X wins) = 0.3, P(Y wins) = 0.4, P(Z wins) = 0.3. Are these events mutually exclusive? Find P(X or Y wins).
Summary
| Rule | Formula | Condition |
|---|---|---|
| Simple Addition | $P(A \cup B) = P(A) + P(B)$ | Mutually exclusive events |
| General Addition | $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ | Any events |
| Complement | $P(\text{Neither}) = 1 - P(A \cup B)$ | Finding “neither” |
| Venn Diagram | Draw and fill regions | Complex problems |
Next Topic
In the next chapter, we will study the Multiplication Rule and Conditional Probability - how to calculate probabilities when events occur together (AND), and how prior information affects probabilities.
