5.3 Simple, Compound, Independent, and Dependent Probability

Simple, Compound, Independent, and Dependent Probability

OpenExamPrep provides this probabilistic and statistical reasoning review to help students master probability principles tested on the Texas Success Initiative Assessment 2.0 (TSIA2) Mathematics section. Probability quantifies the likelihood of random events occurring and serves as the mathematical foundation for decision-making under uncertainty, risk analysis, and statistical inference.

---

1. Fundamental Principles of Probability

An experiment is any structured, repeatable process that yields an observable outcome.

• Outcome: A single specific result of an experiment.
• Sample Space (S): The set of all possible distinct outcomes of an experiment.
• Event (E): Any specified subset of outcomes from the sample space.

The Axioms of Probability

For any event A defined on a sample space S:

1. Bounded Probability Range: The probability of any event must lie between 0 and 1 inclusive:

   0 ≤ P(A) ≤ 1
   

   • P(A) = 0 indicates an impossible event (it can never occur under the defined conditions).
   • P(A) = 1 indicates a certain event (it is guaranteed to occur on every trial).
2. Total Sample Space Probability: The sum of probabilities for all elementary outcomes in sample space S equals exactly 1:

   P(S) = ∑ P(eᵢ) = 1
   

3. Classical Theoretical Probability (Equally Likely Outcomes): If all outcomes in a finite sample space S have an equal likelihood of occurring, the probability of event A is the ratio of favorable outcomes to total outcomes:

   P(A) = n(A) ÷ n(S) = (Number of outcomes favorable to A) ÷ (Total number of possible outcomes in S)
   

---

2. The Complement Rule & "At Least One" Strategies

The complement of event A (denoted Aᶜ, A', or "not A") consists of all outcomes in the sample space that are not contained within A.

Complement Rule: P(not A) = 1 - P(A)   ⟺   P(A) + P(not A) = 1

The "At Least One" Shortcut in Multi-Stage Trials

In multi-trial probability problems, computing the probability of obtaining "at least one" success directly can require summing numerous mutually exclusive cases. Because the complement of "at least one success" is "zero successes (complete failure)", you can apply the complement rule:

P(at least one occurrence) = 1 - P(zero occurrences)

Worked Scenario: The "At Least One" Shortcut
A quality assurance system tests computer memory chips where the probability 
that any single chip is defective is 0.10. If 4 chips are sampled independently, 
what is the probability that AT LEAST ONE chip is defective?

Step 1: Determine the probability that a single chip is NOT defective:
  P(Non-defective) = 1 - 0.10 = 0.90

Step 2: Calculate the probability that ALL 4 chips are non-defective (zero defective):
  Because selections are independent:
  P(All 4 Non-defective) = 0.90 × 0.90 × 0.90 × 0.90 = (0.90)⁴ = 0.6561

Step 3: Apply the complement rule:
  P(At least one defective) = 1 - P(Zero defective)
  = 1 - 0.6561 = 0.3439 (34.39%)

Note: Calculating directly would require computing P(exactly 1) + P(exactly 2) 
+ P(exactly 3) + P(all 4), requiring substantially more arithmetic.

---

3. Compound Events: The Addition Rule ("Or")

A compound event combines two or more simple events. The word "or" indicates the union of events (A ∪ B), signifying that event A occurs, event B occurs, or both occur.

Mutually Exclusive (Disjoint) Events

Two events A and B are mutually exclusive (disjoint) if they cannot occur simultaneously. They share no common outcomes: P(A and B) = 0.

Addition Rule for Mutually Exclusive Events:
P(A or B) = P(A) + P(B)

• Example: Rolling a single standard six-sided die. Let event A = rolling a 2, and event B = rolling an odd number {1, 3, 5}. Because 2 is not odd, A and B are disjoint: P(A or B) = 1/6 + 3/6 = 4/6 = 2/3.

Non-Mutually Exclusive (Overlapping) Events: The General Addition Rule

When events A and B can occur simultaneously, simply adding their probabilities counts the shared outcomes twice. You must subtract the intersection (overlap):

General Addition Rule:
P(A or B) = P(A) + P(B) - P(A and B)

Worked Scenario: Overlapping Card Probability
From a standard, thoroughly shuffled 52-card deck, what is the probability of 
drawing a King OR a Heart?

Step 1: Identify individual event probabilities:
  Total cards in deck = 52
  P(King) = 4 ÷ 52  (there are 4 Kings)
  P(Heart) = 13 ÷ 52  (there are 13 Hearts)

Step 2: Identify the overlapping intersection:
  The King of Hearts belongs to both categories:
  P(King and Heart) = 1 ÷ 52

Step 3: Apply the General Addition Rule:
  P(King or Heart) = P(King) + P(Heart) - P(King and Heart)
  = (4 ÷ 52) + (13 ÷ 52) - (1 ÷ 52) = 16 ÷ 52 = 4 ÷ 13 ≈ 0.3077 (30.8%)

---

4. Compound Events: The Multiplication Rule ("And")

The word "and" indicates the intersection of events (A ∩ B), requiring both event A and event B to occur in sequence or simultaneously.

Independent Events

Two events A and B are independent if the occurrence of event A has absolutely no influence on the probability of event B occurring: P(B|A) = P(B).

Multiplication Rule for Independent Events:
P(A and B) = P(A) × P(B)

• Typical independent contexts include tossing coins, rolling dice, spinning wheels, or sampling with replacement.

Dependent Events & Conditional Probability

Two events are dependent if the occurrence of event A alters the probability of event B.

General Multiplication Rule:
P(A and B) = P(A) × P(B|A)

Where P(B|A) represents the conditional probability of event B occurring given that event A has already occurred.

Conditional Probability Formula:
P(B|A) = P(A and B) ÷ P(A)   (provided P(A) > 0)

Sampling With Replacement vs. Without Replacement

The mechanic of replacement is a foundational TSIA2 testing theme:

Worked Scenario: Sequential Marble Draws
A bag contains 7 Green marbles, 5 Blue marbles, and 8 Yellow marbles (Total = 20 marbles).
Two marbles are drawn sequentially at random. Find the probability that both 
selected marbles are Green under both sampling conditions.

Condition 1: With Replacement (Independent Events)
  - Draw 1: P(Green₁) = 7 ÷ 20
  - Marble is replaced; bag still has 7 Green and 20 total.
  - Draw 2: P(Green₂) = 7 ÷ 20
  - P(Both Green) = (7/20) × (7/20) = 49 / 400 = 0.1225 (12.25%)

Condition 2: Without Replacement (Dependent Events)
  - Draw 1: P(Green₁) = 7 ÷ 20
  - Marble is kept out; bag now has 6 Green and 19 total.
  - Draw 2: P(Green₂ | Green₁) = 6 ÷ 19
  - P(Both Green) = (7/20) × (6/19) = 42 / 380 = 21 / 190 ≈ 0.1105 (11.05%)

---

5. Conditional Probability with Two-Way Tables

Contingency tables make conditional probability calculations intuitive by physically bounding the restricted sample space to a single row or column.

Worked Scenario: Medical Diagnostic Screening Table
A clinic records diagnostic screening results for 1,000 patients tested for Condition X:

| Screening Result | Has Condition (C) | Condition Absent (Cᶜ) | Marginal Row Total |
| :--- | :--- | :--- | :--- |
| **Positive Test (+)** | 92 | 28 | 120 |
| **Negative Test (-)** | 8 | 872 | 880 |
| **Marginal Column Total** | 100 | 900 | 1,000 (Grand Total) |

Problem 1: Find P(Positive | Has Condition)
  - Given Condition: "Has Condition" restricts our universe to the C column (Total = 100).
  - Favorable in that column: 92 positive results.
  - P(+ | C) = 92 ÷ 100 = 0.92 (Sensitivity = 92.0%)

Problem 2: Find P(Has Condition | Positive)
  - Given Condition: "Positive Test" restricts our universe to the Positive row (Total = 120).
  - Favorable in that row: 92 individuals with the condition.
  - P(C | +) = 92 ÷ 120 = 23 ÷ 30 ≈ 0.7667 (Positive Predictive Value = 76.67%)

⚠️ Key Distinction: P(+ | C) = 92.0%, whereas P(C | +) = 76.67%! Reversing the 
conditioning condition completely changes the denominator and the statistical meaning.

---

6. TSIA2 Exam Traps & Strategic Checkpoints

• Trap 1: The Gambler's Fallacy: Believing that independent events "balance out" over short runs. If a fair coin lands on Tails 8 times in a row, the probability of Tails on the 9th toss is still exactly 1/2.
• Trap 2: Forgetting to Subtract Overlap: Whenever computing an "or" probability, always ask: Can both events occur simultaneously? If they can, you must subtract the intersection P(A and B).
• Trap 3: Denominator Neglect in Consecutive Non-Replacement: When items are drawn without replacement, remember to decrease the denominator (and the numerator if drawing from the same group) for every subsequent draw.
• Trap 4: Inverting Conditionals: P(A|B) is not equal to P(B|A). Always look for the word "given", "among", or "of those who" to identify the condition that forms the denominator.