9.3 Probability of Simple & Compound Events

9.3 Probability of Simple & Compound Events

Quick Answer: The probability $P(E)$ of an event is a numerical value on a scale from $0$ (impossible) to $1$ (certain), with $0.5$ indicating an event that is equally likely to occur or not occur. Theoretical probability is determined by mathematical reasoning ($P = \frac{\text{favorable}}{\text{total}}$), whereas experimental probability records empirical outcomes (relative frequency). The Law of Large Numbers guarantees that experimental results converge to theoretical probability over many trials. For simple events, the complement rule states $P(\text{not } A) = 1 - P(A)$. For compound events, independent trials multiply directly ($P(A \text{ and } B) = P(A) \times P(B)$), while dependent trials (sampling without replacement) require updating the sample space ($P(A \text{ and } B) = P(A) \times P(B \mid A)$).

---

The Foundations of Probability & The Probability Scale

Under Florida B.E.S.T. benchmark MA.7.DP.2.1, probability quantitatively measures the likelihood that a specific event will occur during a random process or experiment. Every probability value is strictly bounded on the closed interval $[0, 1]$:

$0 \le P(E) \le 1$

Probabilities can be expressed interchangeably as simplified fractions, decimals, or percentages:

$\begin{array}{ccccc} 0 & 0.25 & 0.50 & 0.75 & 1.0 \\[4pt] \mathbf{0} & \mathbf{\frac{1}{4}} & \mathbf{\frac{1}{2}} & \mathbf{\frac{3}{4}} & \mathbf{1} \\[4pt] 0\% & 25\% & 50\% & 75\% & 100\% \\[6pt] \text{Impossible} & \text{Unlikely} & \text{Equally Likely as Not} & \text{Likely} & \text{Certain} \end{array}$

The Fundamental Terminology

• Experiment (Trial): An activity or process involving chance that leads to observable results (e.g., rolling a fair six-sided die).
• Outcome: A single possible result of a trial (e.g., rolling a $4$).
• Sample Space ($\mathcal{S}$): The comprehensive set of all possible distinct outcomes of an experiment. For a standard die: $\mathcal{S} = {1, 2, 3, 4, 5, 6}$, so the total number of outcomes $n(\mathcal{S}) = 6$.
• Event ($E$): Any specific outcome or subset of outcomes from the sample space (e.g., rolling an even number: $E = {2, 4, 6}$).

---

Theoretical vs. Experimental Probability & The Law of Large Numbers

Under Florida benchmark MA.7.DP.2.2 and MA.7.DP.2.3, students must distinguish between mathematical anticipation and empirical observation.

Theoretical Probability

Theoretical probability assumes all elementary outcomes in a sample space are equally likely. It is computed deductively before any physical trial takes place:

$P(E) = \frac{n(E)}{n(\mathcal{S})} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes in } \mathcal{S}}$

Example: On a standard six-sided die, the theoretical probability of rolling a prime number ($\mathcal{S}_{\text{prime}} = {2, 3, 5}$) is: $P(\text{prime}) = \frac{3}{6} = \frac{1}{2} = 0.5 = 50\%$

Experimental (Empirical) Probability

Experimental probability (also known as relative frequency) is computed inductively based on the actual recorded results of a series of repeated empirical trials:

$P_{\text{exp}}(E) = \frac{\text{Number of times event } E \text{ occurred}}{\text{Total number of trials conducted}}$

Example: If a student tosses a coin $50$ times and records $29$ heads, the experimental probability is: $P_{\text{exp}}(\text{Heads}) = \frac{29}{50} = 0.58 = 58\%$

The Law of Large Numbers (LLN)

Why does the experimental probability of $58%$ differ from the theoretical probability of $50%$? In a small number of trials, random chance causes substantial short-run fluctuations. However, the Law of Large Numbers establishes a vital mathematical bridge:

[!NOTE] The Law of Large Numbers: As the number of repetitions of an experiment increases ($n \to \infty$), the experimental relative frequency of an outcome gets progressively closer to its theoretical probability.

If the student continues flipping that same coin for $10,000$ trials, the proportion of heads will converge extremely close to $0.500$ (e.g., $5,012$ heads $\implies 50.12%$). Short-term variability diminishes into long-term stability.

---

Simple Events and the Complement Rule

A simple event describes an event consisting of a single condition or outcome (e.g., drawing a Queen from a standard deck of $52$ cards).

The Complement of an Event

The complement of an event $A$, denoted as $\text{not } A$, $A'$, or $A^c$, consists of all outcomes in the sample space $\mathcal{S}$ that are not in event $A$.

Because an event must either occur or not occur, event $A$ and its complement are mutually exclusive and exhaustive. Their probabilities must sum to $1$:

$P(A) + P(\text{not } A) = 1$

From this axiom, we derive the Complement Rule:

$P(\text{not } A) = 1 - P(A)$

• Example: The probability of randomly drawing a heart from a standard $52\text{-card}$ deck is $P(\text{Heart}) = \frac{13}{52} = \frac{1}{4}$. The probability of drawing a non-heart is: $P(\text{not Heart}) = 1 - P(\text{Heart}) = 1 - \frac{1}{4} = \frac{3}{4} = 75\%$

Strategic Application: The "At Least One" Shortcut

On the FAST assessment, questions asking for the probability that an event occurs "at least once" across multiple trials can be solved effortlessly using the complement rule:

$P(\text{at least one success}) = 1 - P(\text{zero successes})$

---

Compound Events & Sample Space Representations

Under Florida B.E.S.T. benchmark MA.7.DP.2.4 and MA.8.DP.2.1, a compound event combines two or more simple events occurring simultaneously or in chronological sequence (e.g., flipping a coin AND rolling a number cube).

The Fundamental Counting Principle

If a compound experiment consists of $k$ consecutive stages, where Stage 1 has $n_1$ possible outcomes, Stage 2 has $n_2$ possible outcomes, and Stage $k$ has $n_k$ possible outcomes, the total number of distinct outcomes in the sample space is the product:

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

• Example: A school uniform requires choosing $1$ of $4$ shirts, $1$ of $3$ pants, and $1$ of $2$ shoes. Total possible uniform combinations $= 4 \times 3 \times 2 = 24$ outcomes.

Methods for Representing Sample Spaces

1. Organized Systematic Lists: Exhaustively listing all permutations using systematic codes. For flipping a coin twice: ${HH, HT, TH, TT}$.
2. Two-Way Grids / Tables: Highly effective when combining exactly two events. Rows represent Event 1 and columns represent Event 2, creating an $m \times n$ grid of intersection cells.
3. Tree Diagrams: A visual branching diagram that illustrates every sequential decision path. Each branch represents a possible outcome, and tracing a path from trunk to tip generates one complete compound outcome.

Tree Diagram: Tossing a Coin and Rolling a 1-4 Spinner

Coin         Spinner      Sample Space Outcome
               -- 1 ----------> (H, 1)
             / -- 2 ----------> (H, 2)
        -- H
       /     \ -- 3 ----------> (H, 3)
      /        -- 4 ----------> (H, 4)
Start
      \        -- 1 ----------> (T, 1)
       \     / -- 2 ----------> (T, 2)
        -- T
             \ -- 3 ----------> (T, 3)
               -- 4 ----------> (T, 4)
Total Outcomes = 2 × 4 = 8 equally likely outcomes.

---

Independent vs. Dependent Events (B.E.S.T. MA.8.DP.2)

When calculating the probability that two events $A$ and $B$ both occur ($P(A \text{ and } B)$), you must determine whether the occurrence of the first event alters the probability of the second.

1. Independent Events (With Replacement)

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

• Classic Examples: Tossing a coin multiple times, rolling dice, spinning spinners, or sampling items with replacement (where an item is drawn, recorded, and returned to the pool before the next draw).
• The Multiplication Rule for Independent Events: $P(A \text{ and } B) = P(A) \times P(B)$

2. Dependent Events (Without Replacement)

Two events $A$ and $B$ are dependent if the occurrence of Event $A$ changes the probability of Event $B$.

• Classic Example (Sampling Without Replacement): When an item is drawn from a container and not returned, both the count of favorable items and the total sample size decrease before the second selection.
• The Multiplication Rule for Dependent Events: $P(A \text{ and } B) = P(A) \times P(B \mid A)$ (where $P(B \mid A)$ represents the conditional probability that event $B$ occurs given that event $A$ has already occurred).

---

Step-by-Step Worked Example: Marble Sampling With vs. Without Replacement

Scenario: An urn contains $6$ red marbles, $4$ blue marbles, and $2$ green marbles ($6 + 4 + 2 = 12\text{ total marbles}$). Two marbles are drawn at random.

Case 1: Sampling With Replacement

Find the probability of drawing a red marble, replacing it, and then drawing a blue marble.

• Event $A$ (Red on 1st draw): $P(\text{Red}_1) = \frac{6}{12} = \frac{1}{2}$.
• Marble is replaced: total marbles remain $12$; blue marbles remain $4$.
• Event $B$ (Blue on 2nd draw): $P(\text{Blue}_2) = \frac{4}{12} = \frac{1}{3}$.
• Since events are independent: $P(\text{Red}_1 \text{ and } \text{Blue}_2) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \approx 0.167 \quad (16.7\%)$

Case 2: Sampling Without Replacement

Find the probability of drawing two red marbles in a row without replacement.

• Event $A$ (Red on 1st draw): $P(\text{Red}_1) = \frac{6}{12} = \frac{1}{2}$.
• The first red marble is kept: total marbles decrease to $12 - 1 = 11$; remaining red marbles decrease to $6 - 1 = 5$.
• Event $B$ (Red on 2nd draw given red first): $P(\text{Red}_2 \mid \text{Red}_1) = \frac{5}{11}$.
• Since events are dependent: $P(\text{Red}_1 \text{ and } \text{Red}_2) = \frac{6}{12} \times \frac{5}{11} = \frac{1}{2} \times \frac{5}{11} = \frac{5}{22} \approx 0.227 \quad (22.7\%)$

---

Common Exam Traps & Misconceptions

[!WARNING]

Exam Trap 1: The Gambler's Fallacy

A perennial trap on standardized exams asks: "A fair coin is flipped 5 times, landing on Heads every time. What is the probability that the next flip will land on Tails?" Many students incorrectly answer that Tails is "due" or has a higher probability (e.g., $75%$ or $90%$). Coins have no memory. Each flip of a fair coin is an independent event with a fixed probability of $P(\text{Tails}) = 0.50$ ($50%$), regardless of past outcomes.

[!WARNING]

Exam Trap 2: Forgetting to Decrement the Denominator in "Without Replacement" Problems

In multi-stage sampling without replacement, students frequently remember to subtract $1$ from the favorable numerator but forget to subtract $1$ from the total sample size denominator. If you draw from $15$ cards without replacement, the second draw must have a denominator of $14$, not $15$.

[!WARNING]

Exam Trap 3: Adding Probabilities Instead of Multiplying for Compound "AND" Events

When asked for the probability of Event A AND Event B occurring, students often add $P(A) + P(B)$. Remember: AND means multiply; OR means add. Rolling a $6$ on a die ($1/6$) and flipping heads on a coin ($1/2$) yields $\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$, not $\frac{1}{6} + \frac{1}{2} = \frac{4}{6}$.