Learn on PengiSaxon Algebra 1Chapter 4: Linear Equations and Proportions

Lesson 33: Finding the Probability of Independent and Dependent Events

In this Grade 9 Saxon Algebra 1 lesson, students learn to distinguish between independent and dependent events and calculate the probability of each using multiplication rules such as P(A and B) = P(A) · P(B). The lesson covers how removing an item without replacement changes the sample space and affects the probability of subsequent draws. Students also practice building tree diagrams to map out all possible outcomes for compound events.

Section 1

📘 Independent and Dependent Events

New Concept

Events can be independent (unaffected by previous outcomes) or dependent (affected by previous outcomes), each with a distinct method for calculating combined probability.

Type of EventsDefinitionCalculating the Probability
Independent EventsThe outcome of the first event does not affect the second event.P(A and B)=P(A)P(B)P(A \text{ and } B) = P(A) \cdot P(B)
Dependent EventsThe outcome of the first event does affect the second event.P(A and B)=P(A)P(B)P(A \text{ and } B) = P(A) \cdot P(B), where P(B)P(B) is calculated under the new conditions.

What’s next

Next, you'll use this foundation to solve problems with tree diagrams, direct calculation, and the concept of odds.

Section 2

Independent events

Property

The outcome of the first event does not affect the second event. The probability is calculated as P(A and B)=P(A)P(B)P(A \text{ and } B) = P(A) \cdot P(B).

Examples

  • Rolling a 5 on a die and flipping heads: P(5 and heads)=1612=112P(5 \text{ and heads}) = \frac{1}{6} \cdot \frac{1}{2} = \frac{1}{12}.
  • Drawing a card, replacing it, then drawing another: P(King and then Queen)=452452=1169P(\text{King and then Queen}) = \frac{4}{52} \cdot \frac{4}{52} = \frac{1}{169}.
  • A spinner with 8 sectors is spun twice: P(red and then blue)=1818=164P(\text{red and then blue}) = \frac{1}{8} \cdot \frac{1}{8} = \frac{1}{64}.

Explanation

Think of this as totally separate actions, like flipping a coin and then rolling a die. What happens first has zero impact on what happens next! To find the combined chance of both happening, you simply multiply their individual probabilities together. It’s like a probability team-up where each member acts alone.

Section 3

Dependent events

Property

The outcome of the first event does affect the second event. The probability is P(A and B)=P(A)P(B)P(A \text{ and } B) = P(A) \cdot P(B), where P(B)P(B) is calculated under the new conditions.

Examples

  • From a bag with 5 red and 4 blue marbles, drawing two red without replacement: P(red, then red)=5948=2072=518P(\text{red, then red}) = \frac{5}{9} \cdot \frac{4}{8} = \frac{20}{72} = \frac{5}{18}.
  • Drawing two face cards from a deck without replacement: P(Face Card, then Face Card)=12521151=1322652=11221P(\text{Face Card, then Face Card}) = \frac{12}{52} \cdot \frac{11}{51} = \frac{132}{2652} = \frac{11}{221}.
  • Choosing two students from 10 girls and 8 boys: P(girl, then boy)=1018817=80306=40153P(\text{girl, then boy}) = \frac{10}{18} \cdot \frac{8}{17} = \frac{80}{306} = \frac{40}{153}.

Explanation

Imagine grabbing a doughnut from a box and eating it. Now there's one less! The next person's choice is different because of what you did. The events are linked, so the probability changes for the second event based on the first outcome. It's a chain reaction of chances!

Book overview

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Continue this chapter

Chapter 4: Linear Equations and Proportions

  1. Lesson 1

    Lesson 31: Using Rates, Ratios, and Proportions

  2. Lesson 2

    Lesson 32: Simplifying and Evaluating Expressions with Integer and Zero Exponents

  3. Lesson 3Current

    Lesson 33: Finding the Probability of Independent and Dependent Events

  4. Lesson 4

    Lesson 34: Recognizing and Extending Arithmetic Sequences

  5. Lesson 5

    Lesson 35: Locating and Using Intercepts

  6. Lesson 6

    Lesson 36: Writing and Solving Proportions

  7. Lesson 7

    Lesson 37: Using Scientific Notation

  8. Lesson 8

    Lesson 38: Simplifying Expressions Using the GCF

  9. Lesson 9

    Lesson 39: Using the Distributive Property to Simplify Rational Expressions

  10. Lesson 10

    Lesson 40: Simplifying and Evaluating Expressions Using the Power Property of Exponents

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Independent and Dependent Events

New Concept

Events can be independent (unaffected by previous outcomes) or dependent (affected by previous outcomes), each with a distinct method for calculating combined probability.

Type of EventsDefinitionCalculating the Probability
Independent EventsThe outcome of the first event does not affect the second event.P(A and B)=P(A)P(B)P(A \text{ and } B) = P(A) \cdot P(B)
Dependent EventsThe outcome of the first event does affect the second event.P(A and B)=P(A)P(B)P(A \text{ and } B) = P(A) \cdot P(B), where P(B)P(B) is calculated under the new conditions.

What’s next

Next, you'll use this foundation to solve problems with tree diagrams, direct calculation, and the concept of odds.

Section 2

Independent events

Property

The outcome of the first event does not affect the second event. The probability is calculated as P(A and B)=P(A)P(B)P(A \text{ and } B) = P(A) \cdot P(B).

Examples

  • Rolling a 5 on a die and flipping heads: P(5 and heads)=1612=112P(5 \text{ and heads}) = \frac{1}{6} \cdot \frac{1}{2} = \frac{1}{12}.
  • Drawing a card, replacing it, then drawing another: P(King and then Queen)=452452=1169P(\text{King and then Queen}) = \frac{4}{52} \cdot \frac{4}{52} = \frac{1}{169}.
  • A spinner with 8 sectors is spun twice: P(red and then blue)=1818=164P(\text{red and then blue}) = \frac{1}{8} \cdot \frac{1}{8} = \frac{1}{64}.

Explanation

Think of this as totally separate actions, like flipping a coin and then rolling a die. What happens first has zero impact on what happens next! To find the combined chance of both happening, you simply multiply their individual probabilities together. It’s like a probability team-up where each member acts alone.

Section 3

Dependent events

Property

The outcome of the first event does affect the second event. The probability is P(A and B)=P(A)P(B)P(A \text{ and } B) = P(A) \cdot P(B), where P(B)P(B) is calculated under the new conditions.

Examples

  • From a bag with 5 red and 4 blue marbles, drawing two red without replacement: P(red, then red)=5948=2072=518P(\text{red, then red}) = \frac{5}{9} \cdot \frac{4}{8} = \frac{20}{72} = \frac{5}{18}.
  • Drawing two face cards from a deck without replacement: P(Face Card, then Face Card)=12521151=1322652=11221P(\text{Face Card, then Face Card}) = \frac{12}{52} \cdot \frac{11}{51} = \frac{132}{2652} = \frac{11}{221}.
  • Choosing two students from 10 girls and 8 boys: P(girl, then boy)=1018817=80306=40153P(\text{girl, then boy}) = \frac{10}{18} \cdot \frac{8}{17} = \frac{80}{306} = \frac{40}{153}.

Explanation

Imagine grabbing a doughnut from a box and eating it. Now there's one less! The next person's choice is different because of what you did. The events are linked, so the probability changes for the second event based on the first outcome. It's a chain reaction of chances!

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 4: Linear Equations and Proportions

  1. Lesson 1

    Lesson 31: Using Rates, Ratios, and Proportions

  2. Lesson 2

    Lesson 32: Simplifying and Evaluating Expressions with Integer and Zero Exponents

  3. Lesson 3Current

    Lesson 33: Finding the Probability of Independent and Dependent Events

  4. Lesson 4

    Lesson 34: Recognizing and Extending Arithmetic Sequences

  5. Lesson 5

    Lesson 35: Locating and Using Intercepts

  6. Lesson 6

    Lesson 36: Writing and Solving Proportions

  7. Lesson 7

    Lesson 37: Using Scientific Notation

  8. Lesson 8

    Lesson 38: Simplifying Expressions Using the GCF

  9. Lesson 9

    Lesson 39: Using the Distributive Property to Simplify Rational Expressions

  10. Lesson 10

    Lesson 40: Simplifying and Evaluating Expressions Using the Power Property of Exponents