10.2 Compound Probability: Independent Events, Dependent Events, & Addition Rules

10.2 Compound Probability: Independent Events, Dependent Events, & Addition Rules

While single-event probability examines individual isolated outcomes, compound probability evaluates the likelihood of two or more events occurring in combination or succession. On the TABE Mathematics assessment, compound probability problems test your understanding of two core structural dimensions:

1. Sequential / Joint Occurrence ("AND"): Governed by the Multiplication Rules (evaluating whether events are independent or dependent).
2. Alternative / Disjunctive Occurrence ("OR"): Governed by the Addition Rules (evaluating whether events are mutually exclusive or overlapping).

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Compound Operations: "AND" vs. "OR"

Understanding the verbal conjunction in a word problem immediately identifies the required mathematical operation:

Compound Probability Framework:
├── Joint Events ("AND" / Intersection ∩) ──> MULTIPLICATION RULES
│   ├── Independent (With Replacement): P(A and B) = P(A) × P(B)
│   └── Dependent (Without Replacement): P(A and B) = P(A) × P(B|A)
└── Alternative Events ("OR" / Union ∪) ───> ADDITION RULES
    ├── Mutually Exclusive (Disjoint): P(A or B) = P(A) + P(B)
    └── Overlapping (Inclusive): P(A or B) = P(A) + P(B) - P(A and B)

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Independent Events & The Multiplication Rule

Two events $A$ and $B$ are independent if the occurrence or outcome of Event $A$ has absolutely no effect on the probability of Event $B$.

Hallmarks of Independent Events

• Tossing a coin multiple times (each toss is an isolated physical event).
• Rolling multiple dice or rolling the same die consecutively.
• Spinning a game spinner repeatedly.
• Selecting objects from a container WITH replacement (the chosen item is returned before the next draw, restoring the exact original sample space).

The Multiplication Rule for Independent Events

$\mathbf{P(A \text{ and } B) = P(A) \times P(B)}$ $\mathbf{P(A \text{ and } B \text{ and } C) = P(A) \times P(B) \times P(C)}$

Worked Example 1: Repeated Independent Coin Flips

Problem: A fair coin is tossed three times in succession. What is the probability of obtaining Heads on all three tosses?

$P(H_1 \text{ and } H_2 \text{ and } H_3) = P(H_1) \times P(H_2) \times P(H_3) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} = \mathbf{0.125 = 12.5\%}$

Worked Example 2: Card Drawing With Replacement & Die Roll

Problem: A card is drawn from a standard 52-card deck, recorded, and returned to the deck. Then, a standard 6-sided die is rolled. What is the probability of drawing a Diamond and rolling a number greater than $4$?

1. Probability of a Diamond: $P(\text{Diamond}) = \frac{13}{52} = \frac{1}{4}$.
2. Probability of rolling $> 4$ (rolling a $5$ or $6$): $P(> 4) = \frac{2}{6} = \frac{1}{3}$.
3. Multiply independent probabilities: $P(\text{Diamond and } > 4) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12} \approx \mathbf{0.0833 = 8.33\%}$

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Dependent Events, Conditional Probability, & Sampling Without Replacement

Two events $A$ and $B$ are dependent if the occurrence of Event $A$ alters the probability that Event $B$ will occur.

Conditional Probability: $P(B|A)$

The notation $P(B|A)$ is read as "the probability of event $B$ occurring given that event $A$ has already occurred."

The Multiplication Rule for Dependent Events

$\mathbf{P(A \text{ and } B) = P(A) \times P(B|A)}$

The Hallmark of Dependency: Sampling WITHOUT Replacement

When an item is selected from a group and not replaced, two critical changes occur for the second selection:

1. The total denominator (sample space) decreases by $1$.
2. The target numerator decreases by $1$ if the second selection targets the same category as the first.

Worked Example 3: Tool Selection Without Replacement

Problem: A maintenance technician has a pouch containing $10$ drill bits: $6$ masonry bits and $4$ wood bits. The technician randomly pulls out two bits consecutively without replacement. What is the probability that both selected items are masonry bits?

1. First Selection ($A$): Pulling a masonry bit from the initial pool of $10$: $P(M_1) = \frac{6}{10}$
2. Second Selection ($B|A$): Now $9$ bits remain in the pouch, and $5$ are masonry bits: $P(M_2 | M_1) = \frac{5}{9}$
3. Apply the Multiplication Rule for Dependent Events: $P(M_1 \text{ and } M_2) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3} \approx \mathbf{0.3333 = 33.33\%}$

Worked Example 4: Sequential Dependent Draws of Different Items

Problem: Using the same pouch ($6$ masonry, $4$ wood bits), what is the probability of selecting a masonry bit first and a wood bit second without replacement?

1. First pick (Masonry): $P(M_1) = \frac{6}{10}$.
2. Second pick (Wood given 1st was Masonry): $9$ bits remain total, of which all $4$ wood bits are still present $\implies P(W_2 | M_1) = \frac{4}{9}$.
3. Multiply: $P(M_1 \text{ and } W_2) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90} = \frac{4}{15} \approx \mathbf{0.2667 = 26.67\%}$

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Mutually Exclusive (Disjoint) Events & The Basic Addition Rule

Two events $A$ and $B$ are mutually exclusive (disjoint) if they cannot happen at the same time. The occurrence of one event completely precludes the occurrence of the other ($P(A \text{ and } B) = 0$).

The Addition Rule for Mutually Exclusive Events

$\mathbf{P(A \text{ or } B) = P(A) + P(B)}$

Mutually Exclusive Events (Disjoint Venn Diagram):

     ┌──────────────┐     ┌──────────────┐
     │   Event A    │     │   Event B    │
     │  (e.g., Odd) │     │ (e.g., Even) │
     │   {1, 3, 5}  │     │  {2, 4, 6}   │
     └──────────────┘     └──────────────┘
            No Common Elements: P(A ∩ B) = 0

Worked Example 5: Rolling a Single Die

Problem: A standard 6-sided die is rolled. What is the probability of rolling a number less than $3$ OR rolling a $6$?

• Event $A$ ($< 3$): outcomes ${1, 2} \implies P(A) = \frac{2}{6}$.
• Event $B$ ($6$): outcome ${6} \implies P(B) = \frac{1}{6}$.
• Since a single roll cannot be both $< 3$ and $6$, the events are mutually exclusive: $P(A \text{ or } B) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} = \mathbf{0.50 = 50\%}$

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Overlapping (Inclusive) Events & The General Addition Rule

Two events $A$ and $B$ are overlapping (inclusive) if they can occur simultaneously ($P(A \text{ and } B) > 0$).

The Double-Counting Trap

If you simply add $P(A) + P(B)$, any outcomes that belong to both events are counted twice! To correct for this double-counting, you must subtract the intersection ($P(A \text{ and } B)$).

The General Addition Rule

$\mathbf{P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)}$

Overlapping Events (Inclusive Venn Diagram):

        ┌─────────────────────────┐
        │ Event A       Event B   │
        │ (King)       (Hearts)   │
        │   ┌───────┬───────┐     │
        │   │ 3 K's │King of│12   │     │
        │   │(other)│Hearts │other│     │
        │   │       │(Both) │H's  │     │
        │   └───────┴───────┘     │
        └─────────────────────────┘
          Overlap: Subtract P(A ∩ B) to avoid double-counting

Worked Example 6: Card Drawing (King OR Heart)

Problem: A single card is drawn from a standard 52-card deck. What is the probability of drawing a King OR a Heart?

1. Probability of a King: $P(\text{King}) = \frac{4}{52}$.
2. Probability of a Heart: $P(\text{Heart}) = \frac{13}{52}$.
3. Overlap (King of Hearts): $P(\text{King and Heart}) = \frac{1}{52}$.
4. Apply the General Addition Rule: $P(\text{King or Heart}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \approx \mathbf{0.3077 = 30.77\%}$

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Two-Way Frequency Contingency Tables

A two-way frequency table (contingency table) organizes bivariate categorical data, allowing rapid calculation of joint, marginal, and conditional probabilities.

Workplace Training Survey Dataset ($100$ Employees)

Calculating Probabilities from the Table

1. Marginal Probability: Probability of selecting an employee who completed training: $P(C) = \frac{\text{Column Total for } C}{\text{Grand Total}} = \frac{80}{100} = \mathbf{0.80 = 80\%}$
2. Joint Probability ("AND"): Probability of selecting an employee who is Full-Time AND completed training: $P(FT \text{ and } C) = \frac{\text{Cell Intersection}}{N} = \frac{54}{100} = \mathbf{0.54 = 54\%}$
3. Conditional Probability ("GIVEN"): Probability that an employee completed training given that the employee is Part-Time:
   • Rule: Restrict the denominator to the condition's marginal row total ($40$ Part-Time employees): $P(C | PT) = \frac{\text{Cell }(PT \text{ and } C)}{\text{Row Total for } PT} = \frac{26}{40} = \frac{13}{20} = \mathbf{0.65 = 65\%}$