11.2 Compound Events, Independent & Dependent Probability

Compound Events, Independent & Dependent Probability

Quick Summary: A compound event combines two or more simple events. On the HiSET exam, you will calculate total possibilities using the Fundamental Counting Principle, evaluate joint probabilities using the Multiplication Rule (distinguishing between independent events and dependent events sampled without replacement), and determine compound probabilities using the Addition Rule (distinguishing between mutually exclusive and overlapping events).

Understanding whether events occur in sequence ("AND" $\implies$ multiply) or as alternative options ("OR" $\implies$ add) is the central diagnostic skill for solving high-level probability problems.

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The Fundamental Counting Principle

The Fundamental Counting Principle states that if an experiment consists of a sequence of choices where stage 1 can occur in $n_1$ ways, stage 2 in $n_2$ ways, stage 3 in $n_3$ ways, and so on through stage $k$, then the total number of distinct possible outcomes is the product of the number of choices at each stage:

$\text{Total Possible Outcomes} = n_1 \times n_2 \times n_3 \times \dots \times n_k$

   Tree Diagram for a 3-Stage Decision (Appetizer, Entree, Dessert)
   
   Appetizer (2)         Entree (3)           Dessert (2)       Total Combos
                                            ┌─ Cake (D1)  ───►  A1-E1-D1
                       ┌─ Chicken (E1) ─────┤
                       │                    └─ Fruit (D2) ───►  A1-E1-D2
                       │                    ┌─ Cake (D1)  ───►  A1-E2-D1
   ┌─ Soup (A1) ───────┼─ Salmon (E2) ──────┤
   │                   │                    └─ Fruit (D2) ───►  A1-E2-D2
   │                   │                    ┌─ Cake (D1)  ───►  A1-E3-D1
   │                   └─ Pasta (E3) ───────┤
   │                                        └─ Fruit (D2) ───►  A1-E3-D2
───┤                                                             (6 combos)
   │                                        ┌─ Cake (D1)  ───►  A2-E1-D1
   │                   ┌─ Chicken (E1) ─────┤
   │                   │                    └─ Fruit (D2) ───►  A2-E1-D2
   │                   │                    ┌─ Cake (D1)  ───►  A2-E2-D1
   └─ Salad (A2) ──────┼─ Salmon (E2) ──────┤
                       │                    └─ Fruit (D2) ───►  A2-E2-D2
                       │                    ┌─ Cake (D1)  ───►  A2-E3-D1
                       └─ Pasta (E3) ───────┤
                                            └─ Fruit (D2) ───►  A2-E3-D2
                                                                 (6 combos)
   Total Possible Meal Combinations = 2 × 3 × 2 = 12

Counting with Constraints (Restrictions & Repetition)

HiSET counting problems frequently impose specific conditions on choices:

• With Repetition Allowed: Each position retains the full pool of options.
• Without Repetition (No Duplicates): Each successive choice reduces the available pool by 1 ($n, n-1, n-2, \dots$).
• Positional Restrictions: Handle restricted positions first before multiplying remaining slots.

Worked Example 1: Creating Security Passcodes

A security passcode consists of 4 digits chosen from $0$ through $9$ (10 possible digits). a) How many codes can be formed if digits can be repeated? b) How many codes can be formed if digits CANNOT be repeated? c) How many 4-digit codes can be formed if the code cannot start with $0$ and no repetition is allowed?

1. With repetition: $10 \times 10 \times 10 \times 10 = 10^4 = 10,000\text{ possible codes}$.
2. Without repetition: $10 \times 9 \times 8 \times 7 = 5,040\text{ possible codes}$.
3. Restricted first digit without repetition:
   • Position 1 (digits 1–9 only): $9$ choices
   • Position 2 (remaining 9 unused digits including 0): $9$ choices
   • Position 3 (remaining 8 unused digits): $8$ choices
   • Position 4 (remaining 7 unused digits): $7$ choices
   • Total $= 9 \times 9 \times 8 \times 7 = 4,536\text{ codes}$.

Compound Events: The "AND" Rule (Multiplication)

When calculating the probability that two events $A$ AND $B$ both occur, use the Multiplication Rule.

1. Independent Events (Sampling With Replacement)

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

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

Common Independent Scenarios:

• Flipping multiple coins or flipping the same coin multiple times.
• Rolling multiple independent dice.
• Selecting an item from a container, recording its value, and replacing it before drawing the next item.

Worked Example 2: Independent Compound Probability

A fair 6-sided die is rolled twice. What is the probability of rolling a 3 on the first roll AND an even number on the second roll?

1. First event: $P(\text{roll a } 3) = \frac{1}{6}$.
2. Second event: Even numbers on a die are ${2, 4, 6}$, so $P(\text{even}) = \frac{3}{6} = \frac{1}{2}$.
3. Apply multiplication rule: $P(\text{3 and even}) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$

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2. Dependent Events (Sampling Without Replacement & Conditional Probability)

Two events $A$ and $B$ are dependent if the occurrence of event $A$ alters the probability that event $B$ will occur. The updated probability of $B$ given that $A$ has already happened is called the conditional probability, written $P(B|A)$.

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

Common Dependent Scenarios:

• Drawing playing cards from a deck without putting them back.
• Selecting raffle tickets, marble tokens, or jury members sequentially without replacement.

   Sampling With vs. Without Replacement Comparison
   Container has 5 Blue (B) and 3 Red (R) Marbles (Total = 8)
   
   WITH REPLACEMENT (Independent)          WITHOUT REPLACEMENT (Dependent)
   Draw 1: P(Blue) = 5/8                  Draw 1: P(Blue) = 5/8
     │ [Marble is returned to bag]          │ [Marble stays out; Bag now has 4B, 3R]
     ▼                                      ▼
   Draw 2: P(Blue) = 5/8                  Draw 2: P(Blue|First Blue) = 4/7
   P(Both Blue) = 5/8 × 5/8 = 25/64       P(Both Blue) = 5/8 × 4/7 = 20/56 = 5/14
   (≈ 39.1%)                              (≈ 35.7%)

Worked Example 3: Drawing Without Replacement

A box contains 7 green markers and 3 yellow markers (total 10 markers). A student selects 2 markers at random one after another without replacement. What is the probability that both markers are yellow?

1. First draw: $P(\text{first yellow}) = \frac{3}{10}$.
2. Second draw: After removing 1 yellow marker, 2 yellow markers remain out of 9 total markers in the box. Thus, $P(\text{second yellow} | \text{first yellow}) = \frac{2}{9}$.
3. Multiply probabilities: $P(\text{both yellow}) = \frac{3}{10} \times \frac{2}{9} = \frac{6}{90} = \frac{1}{15}$

Compound Events: The "OR" Rule (Addition)

When calculating the probability that event $A$ OR event $B$ occurs (meaning either $A$ occurs, $B$ occurs, or both occur), use the Addition Rule.

1. Mutually Exclusive (Disjoint) Events

Events $A$ and $B$ are mutually exclusive if they cannot occur at the same time ($P(A \text{ and } B) = 0$).

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

Examples: Rolling an odd number OR rolling a 6 on a single die; drawing a Heart OR drawing a Spade from a deck.

2. Non-Mutually Exclusive (Overlapping) Events

Events $A$ and $B$ are overlapping if they can both occur simultaneously ($P(A \text{ and } B) > 0$). To prevent double-counting outcomes that belong to both events, subtract their intersection:

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

   Venn Diagram: Non-Mutually Exclusive Events
   ┌───────────────────────────────────────────────┐
   │                                               │
   │       Event A                Event B          │
   │     ╭──────────────┬──────────────╮           │
   │     │              │              │           │
   │     │   Only A     │   A and B    │   Only B  │
   │     │              │ (Intersection│           │
   │     │              │   Overlap)   │           │
   │     ╰──────────────┴──────────────╯           │
   │                                               │
   │  P(A or B) = P(A) + P(B) - P(A and B)         │
   └───────────────────────────────────────────────┘

Worked Example 4: The Addition Rule with Cards

A card is drawn at random from a standard 52-card deck. What is the probability that the card is a Queen OR a Diamond?

1. Find $P(\text{Queen})$: There are 4 Queens in a 52-card deck: $P(Q) = \frac{4}{52}$.
2. Find $P(\text{Diamond})$: There are 13 Diamonds in a 52-card deck: $P(D) = \frac{13}{52}$.
3. Find the overlap $P(Q \text{ and } D)$: The Queen of Diamonds is counted in both groups: $P(Q \text{ and } D) = \frac{1}{52}$.
4. Apply non-mutually exclusive addition rule: $P(Q \text{ or } D) = P(Q) + P(D) - P(Q \text{ and } D) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$

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Two-Way Frequency Tables (Contingency Tables)

Two-way tables display bivariate data across rows and columns. They are ideal for reading joint probabilities, marginal probabilities, and conditional probabilities on the HiSET exam.

Sample Survey: 100 Students Transport to School

Reading Probabilities from the Table

• Simple Marginal Probability: Probability a randomly selected student takes the Bus: $P(\text{Bus}) = \frac{\text{Bus Column Total}}{\text{Grand Total}} = \frac{40}{100} = 0.40$
• Joint Probability (AND): Probability a student is in 9th–10th grade AND takes the bus: $P(\text{9-10 and Bus}) = \frac{30}{100} = 0.30$
• Union Probability (OR): Probability a student is in 11th–12th grade OR takes public transit: $P(\text{11-12 or Transit}) = P(\text{11-12}) + P(\text{Transit}) - P(\text{11-12 and Transit}) = \frac{50}{100} + \frac{20}{100} - \frac{15}{100} = \frac{55}{100} = 0.55$
• Conditional Probability (GIVEN THAT): Given that a student takes the bus, what is the probability they are in 9th–10th grade? $P(\text{9-10} | \text{Bus}) = \frac{\text{9-10 Bus Count}}{\text{Total Bus Count}} = \frac{30}{40} = \frac{3}{4} = 0.75$

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HiSET Problem-Solving Traps to Avoid

1. Forgetting to Subtract the Overlap: When adding probabilities for "OR" questions, always check if the two events can happen together. If you calculate $P(\text{King or Spade}) = \frac{4}{52} + \frac{13}{52} = \frac{17}{52}$, you double-counted the King of Spades; the correct answer is $\frac{16}{52} = \frac{4}{13}$.
2. Failing to Reduce the Denominator in Sampling Without Replacement: If 2 cards are drawn sequentially from 52 cards, the second probability must be over 51, not 52.
3. Confusing 'AND' with 'OR': 'AND' implies joint simultaneous occurrence (multiply probabilities); 'OR' implies union of possibilities (add probabilities, subtract overlap).