Kenneth ChengMathematical modeling
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Independence of events

Read the idea, work independently, then explain what changed.

高一必修 第二册(A版).pdf · 10.2 · PDF 256 / printed page 249

Revisit first: Random events and probability

TOPIC 01

Independence of events

Build understanding of independence of events through definitions, contrasting cases and justified applications.

What you will be able to explain

  • Use independence of events to state a justified conclusion.
  • Connect representations and justify the steps, including boundary cases.
  • Explain a solution and apply the idea to a changed situation.

Independence

Independent events satisfy the product rule.

P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B)

Model check

Shared causes and sampling without replacement can create dependence.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Before calculating, predict how the conclusion changes when one defining condition in independence of events changes. Record a reason.

Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.

Binomial model: fixed n=5, independent trials, common p=0.5. E(X)=2.5, Var(X)=1.25. Bars show exact probabilities.

Binomial model: fixed n=5, independent trials, common p=0.5. E(X)=2.5, Var(X)=1.25. Bars show exact probabilities.

Explain: Compare two admissible cases and one boundary or invalid case. Explain the observed difference using the stated definition.

Transfer: Construct a new example and a tempting incorrect solution. Repair the solution by naming the missing condition.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Foundation#Worked example

Independent A,B have probabilities 1/2 and 1/(k+2). Find P(A∩B).

k=2k=2
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use independence only when justified and keep it distinct from disjointness.
Hint 2
Independent event probabilities multiply.
Worked solution

  1. Use independence only when justified and keep it distinct from disjointness.

  2. Calculate or simplify this relation.

    P(A∩B)=1/2⋅1/4=1/8P(A\cap B)=1/2\cdot1/4=1/8

  3. The multiplication is supported by the independence assumption.

The requested value is 0.125.

Checks and common pitfalls: The multiplication is supported by the independence assumption.

Think first. Reveal a hint when the class is ready.

02 / Standard#Worked example

Two independent components each work with probability k/(k+1). A series system needs both. Find success probability.

k=2k=2
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use independence only when justified and keep it distinct from disjointness.
Hint 2
Series success is an intersection.
Worked solution

  1. Use independence only when justified and keep it distinct from disjointness.

  2. Calculate or simplify this relation.

    P=(k/(k+1))2=4/9P=(k/(k+1))^2=4/9

  3. “At least one” would describe a different system.

The requested value is 0.44444444.

Checks and common pitfalls: “At least one” would describe a different system.

Think first. Reveal a hint when the class is ready.

03 / Transfer#Worked example

The probability of each component failing is 1/(k+1), but common power failures can affect both. Can the independent-series formula be used without further evidence?

k=2k=2
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use independence only when justified and keep it distinct from disjointness.
Hint 2
Inspect shared causes, not just marginal probabilities.
Worked solution

  1. Use independence only when justified and keep it distinct from disjointness.

  2. Calculate or simplify this relation.

    P(A∩B)≠P(A)P(B) in generalP(A\cap B)\ne P(A)P(B)\text{ in general}

  3. A numerical marginal rate does not establish independence.

No; a shared cause may create dependence.

Checks and common pitfalls: A numerical marginal rate does not establish independence.

Reasoning checklist · self / teacher assessment
  • State the relevant definition, condition or model.
  • Show a valid calculation, proof or counterexample.
  • Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

Independent A,B have probabilities 1/2 and 1/(k+2). Find P(A∩B).

k=3k=3
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use independence only when justified and keep it distinct from disjointness.
Hint 2
Independent event probabilities multiply.
Worked solution

  1. Use independence only when justified and keep it distinct from disjointness.

  2. Calculate or simplify this relation.

    P(A∩B)=1/2⋅1/5=1/10P(A\cap B)=1/2\cdot1/5=1/10

  3. The multiplication is supported by the independence assumption.

The requested value is 0.1.

Checks and common pitfalls: The multiplication is supported by the independence assumption.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

Two independent components each work with probability k/(k+1). A series system needs both. Find success probability.

k=3k=3
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use independence only when justified and keep it distinct from disjointness.
Hint 2
Series success is an intersection.
Worked solution

  1. Use independence only when justified and keep it distinct from disjointness.

  2. Calculate or simplify this relation.

    P=(k/(k+1))2=9/16P=(k/(k+1))^2=9/16

  3. “At least one” would describe a different system.

The requested value is 0.5625.

Checks and common pitfalls: “At least one” would describe a different system.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

Two independent components each fail with probability 1/(k+1). A parallel system works if at least one works. Find success probability.

k=3k=3
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use independence only when justified and keep it distinct from disjointness.
Hint 2
Use the complement of both failing.
Worked solution

  1. Use independence only when justified and keep it distinct from disjointness.

  2. Calculate or simplify this relation.

    P=1−P(both fail)=1−1/16P=1-P(\text{both fail})=1-1/16

  3. Independence must hold for the failure events too.

The requested value is 0.9375.

Checks and common pitfalls: Independence must hold for the failure events too.

Think first. Reveal a hint when the class is ready.

07 / Foundation#Your turn

A,B are independent and P(B)>0. Compare P(A|B) with P(A).

P(B)=1/4P(B)=1/4
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use independence only when justified and keep it distinct from disjointness.
Hint 2
Use the product definition of independence.
Worked solution

  1. Use independence only when justified and keep it distinct from disjointness.

  2. Calculate or simplify this relation.

    P(A∣B)=P(A∩B)/P(B)=P(A)P(A|B)=P(A\cap B)/P(B)=P(A)

  3. Conditioning on B does not change A’s probability in this model.

They are equal.

Checks and common pitfalls: Conditioning on B does not change A’s probability in this model.

Reasoning checklist · self / teacher assessment
  • State the relevant definition, condition or model.
  • Show a valid calculation, proof or counterexample.
  • Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

08 / Standard#Your turn

A and B are disjoint with P(A)=P(B)=1/(k+2). Are they independent?

k=3k=3
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use independence only when justified and keep it distinct from disjointness.
Hint 2
Compare the intersection probability with the product of probabilities.
Worked solution

  1. Use independence only when justified and keep it distinct from disjointness.

  2. Calculate or simplify this relation.

    0=P(A∩B)≠P(A)P(B)=1/(k+2)20=P(A\cap B)\ne P(A)P(B)=1/(k+2)^2

  3. Positive-probability disjoint events are dependent.

No, their intersection probability is zero but the product of their probabilities is positive.

Checks and common pitfalls: Positive-probability disjoint events are dependent.

Reasoning checklist · self / teacher assessment
  • State the relevant definition, condition or model.
  • Show a valid calculation, proof or counterexample.
  • Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

09 / Standard#Your turn

A bag has k red and 1 blue; draw twice without replacement. Are the two red-draw events independent?

k=3k=3
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use independence only when justified and keep it distinct from disjointness.
Hint 2
Update the bag after the first outcome.
Worked solution

  1. Use independence only when justified and keep it distinct from disjointness.

  2. Calculate or simplify this relation.

    (k−1)/k≠k/(k+1)(k-1)/k\ne k/(k+1)

  3. Without replacement, the first draw changes the second distribution.

No; P(second red|first red)=(k−1)/k differs from k/(k+1).

Checks and common pitfalls: Without replacement, the first draw changes the second distribution.

Reasoning checklist · self / teacher assessment
  • State the relevant definition, condition or model.
  • Show a valid calculation, proof or counterexample.
  • Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

10 / Standard#Your turn

The probability of each component failing is 1/(k+1), but common power failures can affect both. Can the independent-series formula be used without further evidence?

k=3k=3
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use independence only when justified and keep it distinct from disjointness.
Hint 2
Inspect shared causes, not just marginal probabilities.
Worked solution

  1. Use independence only when justified and keep it distinct from disjointness.

  2. Calculate or simplify this relation.

    P(A∩B)≠P(A)P(B) in generalP(A\cap B)\ne P(A)P(B)\text{ in general}

  3. A numerical marginal rate does not establish independence.

No; a shared cause may create dependence.

Checks and common pitfalls: A numerical marginal rate does not establish independence.

Reasoning checklist · self / teacher assessment
  • State the relevant definition, condition or model.
  • Show a valid calculation, proof or counterexample.
  • Interpret the conclusion with its restrictions.

Think first. Reveal a hint when the class is ready.

11 / Standard#Your turn

Independent A,B have probabilities 1/2 and 1/(k+2). Find P(A∩B).

k=4k=4
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use independence only when justified and keep it distinct from disjointness.
Hint 2
Independent event probabilities multiply.
Worked solution

  1. Use independence only when justified and keep it distinct from disjointness.

  2. Calculate or simplify this relation.

    P(A∩B)=1/2⋅1/6=1/12P(A\cap B)=1/2\cdot1/6=1/12

  3. The multiplication is supported by the independence assumption.

The requested value is 0.08333333.

Checks and common pitfalls: The multiplication is supported by the independence assumption.

Think first. Reveal a hint when the class is ready.

12 / Transfer#Your turn

Two independent components each work with probability k/(k+1). A series system needs both. Find success probability.

k=4k=4
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use independence only when justified and keep it distinct from disjointness.
Hint 2
Series success is an intersection.
Worked solution

  1. Use independence only when justified and keep it distinct from disjointness.

  2. Calculate or simplify this relation.

    P=(k/(k+1))2=16/25P=(k/(k+1))^2=16/25

  3. “At least one” would describe a different system.

The requested value is 0.64.

Checks and common pitfalls: “At least one” would describe a different system.

Think first. Reveal a hint when the class is ready.

13 / Transfer#Your turn

Two independent components each fail with probability 1/(k+1). A parallel system works if at least one works. Find success probability.

k=4k=4
  • Use the domain, units and sampling assumptions stated in the question.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Use independence only when justified and keep it distinct from disjointness.
Hint 2
Use the complement of both failing.
Worked solution

  1. Use independence only when justified and keep it distinct from disjointness.

  2. Calculate or simplify this relation.

    P=1−P(both fail)=1−1/25P=1-P(\text{both fail})=1-1/25

  3. Independence must hold for the failure events too.

The requested value is 0.96.

Checks and common pitfalls: Independence must hold for the failure events too.

Think first. Reveal a hint when the class is ready.

Focus on one question

End-of-lesson check and correction

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    Teacher preparation and assessment

    Question sequence

    • What must be true before using the main rule for independence of events?
    • Which representation makes this task easier, and why?
    • Change one assumption. Does the conclusion survive?

    Board plan

    • Use independence of events to state a justified conclusion.
    • Independence: Independent events satisfy the product rule.
      P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B)
    • Model check: Shared causes and sampling without replacement can create dependence.
    • Close with: conditions → representation → reasoning → check.

    Anticipated thinking

    • Expected reasoning: Independent events satisfy the product rule.
    • Expected reasoning: Shared causes and sampling without replacement can create dependence.
    • Expected correction: Known marginal probabilities do not establish independence.

    Assessment checklist

    • 1: identify the givens and required quantity.
    • 1: choose a valid definition, representation or method.
    • 1: present connected, correct reasoning.
    • 1: check conditions and explain the result.

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    Curriculum and source notes ↗