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A,B are disjoint and each has probability 1/4. Are they independent?

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

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高三選擇性必修 第三册(A版).pdf · 7.1 · PDF 49 / printed page 44

TOPIC 01

Conditions, partitions and posterior probability

Identify the conditioning group, verify partitions, and update priors with observed evidence.

What you will be able to explain

  • Identify the conditioning group, verify partitions, and update priors with observed evidence.
  • Distinguish reverse conditionals, disjointness and independence.

Try it. Leave your reasoning visible.

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

01 / Foundation#Your turn

A,B are disjoint and each has probability 1/4. Are they independent?

P(A)=P(B)=1/4; A∩B=∅P(A)=P(B)=1/4;\ A\cap B=\varnothing
  • Use P(A∩B)=P(A)P(B) as the independence criterion, including the stated overlap or disjointness.
Skills and prerequisite lessons

Scope: JM01 / JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.

  1. Yes: disjoint means independent.
  2. Yes: equal probabilities imply independence.
  3. No: 0 differs from 1/16.
  4. Impossible to decide.
Working and explanation
BUILD THE REASONING
Hint 1
Compare intersection and product.
Hint 2
Start from this intermediate relation.
P(A∩B)=0P(A\cap B)=0
Worked solution
  1. Compare intersection and product.
  2. Establish and simplify this mathematical relation.
    P(A∩B)=0P(A\cap B)=0
  3. Establish and simplify this mathematical relation.
    P(A)P(B)=1/16P(A)P(B)=1/16
  4. Positive-probability disjoint events are dependent.
No: 0 differs from 1/16.
Checks and common pitfalls: Positive-probability disjoint events are dependent.
Think first. Reveal a hint when the class is ready.
Focus on one question
Teacher preparation and assessment
40-minute lesson
MinutesFocusTeaching and student action
5PredictPredict whether a 90% flagging rate guarantees a 90% defective probability.
8Build the ideaSample response: among 100 units, 9 defective and 9 good units are flagged; posterior is one half.
10Guided comparisonSample response: music is given, so 5/12 rather than 5/30.
12Independent practice and discussionStudents annotate denominators and compare two-draw versus one-draw evidence.
5Exit and next stepExit: state the given event, its positive probability and the disjoint contributions.
Question sequence
  • Predict whether a 90% flagging rate guarantees a 90% defective probability.
  • Sample response: music is given, so 5/12 rather than 5/30.
Board development
  • Change the denominator
    P(A∣B)=P(A∩B)/P(B)P(A|B)=P(A\cap B)/P(B)
Expected correct and incorrect responses
  • Correct sample: 0.4
  • Anticipated wrong response: 0.16 — Retains the whole population after conditioning on a subgroup.
  • Anticipated wrong response: Yes: disjoint means independent. — Equates disjointness with independence.
  • Using the whole-population denominator after conditioning.
  • Equating likelihood with posterior probability.
Follow-up questions
  • Follow-up: Restrict to the cycling group. Then explain why 0.16 fails.
  • Follow-up: Compare intersection and product. Then explain why Yes: disjoint means independent. fails.
  • Find a prior giving a half posterior and explain the zero-flag-probability boundary.
Remediation
  • Sample response: among 100 units, 9 defective and 9 good units are flagged; posterior is one half.
  • Return to the first example, identify the missing condition, then solve one fresh task without viewing its solution.
Exit check
  • Exit: state the given event, its positive probability and the disjoint contributions.
Assessment checklist and sample responses
  • Representation and conditions
    Given music, the eligible group has 12, so probability is 5/12.
  • Connected mathematical reasoning
    The defective joint contribution is 0.09 and all flags total 0.18, so posterior is 1/2.
  • Interpretation and boundary check
    If all flag contributions vanish, P(flag)=0 and the elementary posterior is undefined.
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Curriculum and source notes ↗