9.2 Compound Events, Tree Diagrams & Independence

9.2 Compound Events, Tree Diagrams & Independence

Quick Summary: A compound event consists of two or more simple events occurring in sequence or simultaneously. Determining compound probabilities requires analyzing whether events are independent (the outcome of one does not affect another) or dependent (the outcome of one alters subsequent probabilities). The Multiplication Rule computes $P(A \text{ and } B)$, taking into account conditional probability $P(B|A)$ when sampling without replacement. The Addition Rule computes $P(A \text{ or } B)$, subtracting the intersection $P(A \text{ and } B)$ for overlapping events. Tree diagrams serve as essential visual tools for tracking multi-stage probabilistic decisions.

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Single vs. Compound Events: Categorization Framework

To solve probability problems accurately on the Ontario MPT, you must first classify the nature of the events involved.

• Single Event: An outcome evaluated from one single action or trial (e.g., drawing one card from a deck).
• Compound Event: An outcome evaluated from multiple actions or sequential stages (e.g., drawing two cards in sequence, or flipping a coin and rolling a die).

graph TD
    CE["Compound Events<br/>P(Multiple Outcomes)"]
    CE --> MULT["AND Logic: Sequential / Simultaneous<br/>Multiplication Rule: P(A and B)"]
    CE --> ADD["OR Logic: Combined Outcomes<br/>Addition Rule: P(A or B)"]
    MULT --> IND["Independent Events<br/>P(A and B) = P(A) × P(B)"]
    MULT --> DEP["Dependent Events (Conditional)<br/>P(A and B) = P(A) × P(B|A)"]
    ADD --> ME["Mutually Exclusive (Disjoint)<br/>P(A or B) = P(A) + P(B)"]
    ADD --> NME["Non-Mutually Exclusive (Overlapping)<br/>P(A or B) = P(A) + P(B) - P(A and B)"]

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The Multiplication Rule: $P(A \text{ and } B)$

The Multiplication Rule determines the probability that both event $A$ and event $B$ occur.

1. Independent Events

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

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

Examples of Independent Events:

• Flipping a coin twice in a row.
• Rolling two separate fair dice.
• Drawing a marble from a bag, replacing it, and drawing a second marble.

2. Dependent Events & Conditional Probability

Two events $A$ and $B$ are dependent if the occurrence of event $A$ changes the probability of event $B$. The probability of $B$ occurring given that $A$ has already occurred is written as $P(B|A)$ ("probability of $B$ given $A$").

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

Examples of Dependent Events:

• Drawing two cards sequentially from a deck without replacement.
• Choosing two committee members from a class without replacement.

Replacement vs. Non-Replacement Comparison

Consider a bag containing 5 Red and 5 Blue marbles ($10$ total). We want to find the probability of drawing 2 Red marbles in two consecutive draws.

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The Addition Rule: $P(A \text{ or } B)$

The Addition Rule determines the probability that event $A$ or event $B$ (or both) occurs.

1. Mutually Exclusive (Disjoint) Events

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

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

Example: Drawing a card from a standard deck that is either a Ace or a King. A card cannot be both an Ace and a King simultaneously. $P(\text{Ace or King}) = P(\text{Ace}) + P(\text{King}) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}$

2. Non-Mutually Exclusive (Overlapping) Events

Events are non-mutually exclusive if they can occur simultaneously ($P(A \text{ and } B) > 0$). Simply adding $P(A) + P(B)$ double-counts the overlapping outcomes. To correct for this, subtract the intersection $P(A \text{ and } B)$:

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

Example: Drawing a card that is either a King or a Heart.

• $P(\text{King}) = \frac{4}{52}$
• $P(\text{Heart}) = \frac{13}{52}$
• $P(\text{King and Heart}) = \frac{1}{52}$ (the King of Hearts) $P(\text{King or Heart}) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13}$

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Tree Diagrams for Multi-Stage Experiments

A tree diagram is a structured graphical representation of a multi-stage random experiment. Each stage corresponds to a decision or trial, with branches labeled by their respective probabilities.

Fundamental Tree Diagram Rules

1. Sum of Stage Branches = 1: The probabilities on all branches emanating from a single node must sum to exactly $1$.
2. Multiply Along Path (AND Rule): To calculate the probability of a complete multi-stage path from root to leaf, multiply the branch probabilities along that path.
3. Add Across Terminal Leaves (OR Rule): To calculate the probability of an event satisfied by multiple distinct paths, add the probabilities of those individual terminal paths.

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Worked Step-by-Step Problem Walkthroughs

Problem 1: Dependent Selection Without Replacement Using Tree Diagrams

Problem Statement: A drawer contains 6 Red socks and 4 Blue socks ($10$ total). A student randomly selects two socks, one after another, without replacement.

1. Construct the complete probability tree diagram.
2. Find the probability of drawing two socks of the same color (both Red OR both Blue).
3. Find the probability of drawing at least one Blue sock.

Step-by-Step Solution:

1. Analyze Stage 1 (First Draw):

   • $P(R_1) = \frac{6}{10} = \frac{3}{5}$
   • $P(B_1) = \frac{4}{10} = \frac{2}{5}$

2. Analyze Stage 2 (Second Draw Conditional Probabilities):

   • If First Sock was Red ($R_1$): Remaining = 5 Red, 4 Blue (9 total).
     • $P(R_2 | R_1) = \frac{5}{9}$
     • $P(B_2 | R_1) = \frac{4}{9}$
   • If First Sock was Blue ($B_1$): Remaining = 6 Red, 3 Blue (9 total).
     • $P(R_2 | B_1) = \frac{6}{9} = \frac{2}{3}$
     • $P(B_2 | B_1) = \frac{3}{9} = \frac{1}{3}$

3. Calculate All 4 Path Probabilities:

   • Path 1 ($R_1 \text{ and } R_2$): $\frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$
   • Path 2 ($R_1 \text{ and } B_2$): $\frac{6}{10} \times \frac{4}{9} = \frac{24}{90} = \frac{4}{15}$
   • Path 3 ($B_1 \text{ and } R_2$): $\frac{4}{10} \times \frac{6}{9} = \frac{24}{90} = \frac{4}{15}$
   • Path 4 ($B_1 \text{ and } B_2$): $\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$

   Check Sum: $\frac{30}{90} + \frac{24}{90} + \frac{24}{90} + \frac{12}{90} = \frac{90}{90} = 1.0$ (Valid tree!)

4. Part 2: $P(\text{Same Color}) = P(RR) + P(BB)$: $P(\text{Same Color}) = \frac{30}{90} + \frac{12}{90} = \frac{42}{90} = \frac{7}{15} \approx 0.467 = 46.7\%$

5. Part 3: $P(\text{At least one Blue sock})$ using Complement Rule:

   • The complement of "at least one Blue" is "zero Blue socks" (which means Both Red, $RR$): $P(\text{At least one Blue}) = 1 - P(RR) = 1 - \frac{30}{90} = \frac{60}{90} = \frac{2}{3} \approx 0.667 = 66.7\%$

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Problem 2: Overlapping Sets & Survey Addition Rule

Problem Statement: In a school cohort of 100 Grade 9 students, a survey reveals that 60 students participate in school sports, 45 participate in music programs, and 20 participate in both sports and music. A student is selected at random.

1. Calculate the probability that the student participates in sports OR music.
2. Calculate the probability that the student participates in NEITHER activity.

Step-by-Step Solution:

1. Identify Given Counts & Total Sample Size:

   • $n(S) = 100$
   • $n(\text{Sports}) = 60 \implies P(\text{Sports}) = \frac{60}{100} = 0.60$
   • $n(\text{Music}) = 45 \implies P(\text{Music}) = \frac{45}{100} = 0.45$
   • $n(\text{Both}) = 20 \implies P(\text{Sports and Music}) = \frac{20}{100} = 0.20$

2. Apply Non-Mutually Exclusive Addition Rule: $P(\text{Sports or Music}) = P(\text{Sports}) + P(\text{Music}) - P(\text{Sports and Music})$ $P(\text{Sports or Music}) = 0.60 + 0.45 - 0.20 = 0.85 = 85.0\%$

3. Calculate Probability of Neither Activity: $P(\text{Neither}) = 1 - P(\text{Sports or Music}) = 1 - 0.85 = 0.15 = 15.0\%$

   Conclusion: The probability of selecting a student in sports or music is 85%, while the probability of selecting a student in neither is 15%.