← Senior Mathematics Studio

LEARN · EXPLAIN · REVISE

With priors 0.4,0.6 and defect rates 0.1,0.2, find P(A|defect).

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

Return to the lesson / paper ↗

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.

Examples 1–3: study a full demonstration; finish a solution from its first step; then choose a method independently before opening the solution. The remaining exercises may continue after class.

01 / Standard#Your turn

With priors 0.4,0.6 and defect rates 0.1,0.2, find P(A|defect).

P(A)=0.4; P(D∣A)=0.1; P(D∣B)=0.2P(A)=0.4;\ P(D|A)=0.1;\ P(D|B)=0.2
  • Cases must form a disjoint exhaustive partition; a posterior ratio requires positive evidence probability.
Skills and prerequisite lessons

Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.

Working and explanation

BUILD THE REASONING

Hint 1
Divide A’s defective contribution by all defective contributions.
Hint 2
Start from this intermediate relation.
P(A∩D)=0.04; P(D)=0.16P(A\cap D)=0.04;\ P(D)=0.16
Worked solution
  1. Divide A’s defective contribution by all defective contributions.

  2. Establish and simplify this mathematical relation.

    P(A∩D)=0.04; P(D)=0.16P(A\cap D)=0.04;\ P(D)=0.16
  3. Establish and simplify this mathematical relation.

    P(A∣D)=0.04/0.16=0.25P(A|D)=0.04/0.16=0.25
  4. The prior 0.4 changes after observing a defect.

The requested value is 0.25.

Checks and common pitfalls: The prior 0.4 changes after observing a defect.

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.

No sign-in. Work stays in this browser. Export before clearing browser data. Written reasoning is assessed with a checklist.

Curriculum and source notes ↗