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Two bags are selected with equal probability. A has 3 red/1 blue and B has 1 red/3 blue. Given two red draws without replacement from the chosen bag, find P(A).

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.

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 / Transfer#Your turn

Two bags are selected with equal probability. A has 3 red/1 blue and B has 1 red/3 blue. Given two red draws without replacement from the chosen bag, find P(A).

P(A)=P(B)=1/2;\ A:(3R,1B);\ B:(1R,3B)

Draw uniformly from the remaining finite bag without replacement, except the explicitly compared replacement model.

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

Use the impossibility of two red draws from bag B.

Hint 2

Start from this intermediate relation.

P(RR|A)=(3/4)(2/3)=1/2;\ P(RR|B)=0

Worked solution

Use the impossibility of two red draws from bag B.
Establish and simplify this mathematical relation.
$P(RR|A)=(3/4)(2/3)=1/2;\ P(RR|B)=0$
Establish and simplify this mathematical relation.
$P(A|RR)=[(1/2)(1/2)]/[(1/2)(1/2)]=1$
Dependence of draws within a bag does not invalidate Bayes; use the correct likelihood.

The requested value is 1.

Checks and common pitfalls: Dependence of draws within a bag does not invalidate Bayes; use the correct likelihood.

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

Teacher preparation and assessment

40-minute lesson

Minutes	Focus	Teaching and student action
5	Predict	Predict whether a 90% flagging rate guarantees a 90% defective probability.
8	Build the idea	Sample response: among 100 units, 9 defective and 9 good units are flagged; posterior is one half.
10	Guided comparison	Sample response: music is given, so 5/12 rather than 5/30.
12	Independent practice and discussion	Students annotate denominators and compare two-draw versus one-draw evidence.
5	Exit and next step	Exit: 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
$P(A|B)=P(A\cap B)/P(B)$

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.