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Conditions, partitions and posterior probability

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

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.

Change the denominator

A flag can come from true positives or false flags, so its meaning depends on prevalence.

Given B, the eligible outcomes are those in B; reverse conditioning uses a different group.

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

Given B, the eligible outcomes are those in B; reverse conditioning uses a different group.
$P(F)=p\alpha+(1-p)\beta$
Given B, the eligible outcomes are those in B; reverse conditioning uses a different group.
$P(D|F)=p\alpha/[p\alpha+(1-p)\beta]$

When it applies

P(B)>0; mixture cases are disjoint and exhaustive.

Counterexample: Equal prior bags need not remain equally likely after a red draw; positive-probability disjoint events are dependent.

Two-way count table, probability tree and prior/posterior bar comparison.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Keep flagging likelihoods fixed and predict how changing the defective prior changes the posterior.

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

The flagged group includes both joint contributions; posterior is 0.5, rather than the likelihood 0.9.

Numerical data for the diagram; printable
State	Flag joint probability	No-flag joint probability
Defective / 次品	0.09	0.01
Good / 良品	0.09	0.81
Total / 合計	0.18	0.82

Explain: Read both joint contributions; the denominator is all flags.

Transfer: Find a prior giving a half posterior and explain the zero-flag-probability boundary.

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 / Foundation#Worked example

In a class of 30, 12 play music and 5 of those play chess. Given music, find P(chess).

N=30;\ n(M)=12;\ n(C\cap M)=5

Select uniformly among the named people; the conditioning group is nonempty.

Skills and prerequisite lessons

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

Working and explanation

BUILD THE REASONING

Hint 1

Conditioning restricts the denominator to the music group.

Hint 2

Start from this intermediate relation.

P(C|M)=n(C\cap M)/n(M)

Worked solution

Conditioning restricts the denominator to the music group.
Establish and simplify this mathematical relation.
$P(C|M)=n(C\cap M)/n(M)$
Establish and simplify this mathematical relation.
$P(C|M)=5/12$
The whole-class denominator would answer a joint-probability question.

The requested value is 0.4166666666666667.

Checks and common pitfalls: The whole-class denominator would answer a joint-probability question.

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

02 / Standard#Worked example

Select machine A with probability 0.4, otherwise B. Defect rates are 0.1 and 0.2. Find the defect probability.

P(A)=0.4;\ P(D|A)=0.1;\ P(D|B)=0.2

First step provided — complete the reasoning:Weight the two conditional rates by the selection probabilities.

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

Weight the two conditional rates by the selection probabilities.

Hint 2

Start from this intermediate relation.

P(D)=0.4(0.1)+0.6(0.2)

Worked solution

Weight the two conditional rates by the selection probabilities.
Establish and simplify this mathematical relation.
$P(D)=0.4(0.1)+0.6(0.2)$
Establish and simplify this mathematical relation.
$P(D)=0.16$
An unweighted average ignores the unequal priors.

The requested value is 0.16.

Checks and common pitfalls: An unweighted average ignores the unequal priors.

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

03 / Transfer#Worked example

A nonmedical screening process flags 90% of the 10% genuinely defective units and 10% of good units. Given a flag, find P(defective).

P(D)=0.1;\ P(F|D)=0.9;\ P(F|D^c)=0.1

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

Create a two-branch joint-probability tree before reversing the condition.

Hint 2

Start from this intermediate relation.

P(D\cap F)=0.09;\ P(F)=0.09+0.09=0.18

Worked solution

Create a two-branch joint-probability tree before reversing the condition.
Establish and simplify this mathematical relation.
$P(D\cap F)=0.09;\ P(F)=0.09+0.09=0.18$
Establish and simplify this mathematical relation.
$P(D|F)=0.09/0.18=0.5$
A 90% detection rate is not a 90% posterior probability.

The requested value is 0.5.

Checks and common pitfalls: A 90% detection rate is not a 90% posterior probability.

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

04 / Foundation#Your turn

P(A∩B)=0.12 and P(B)=0.3. Find P(A|B).

P(A\cap B)=0.12;\ P(B)=0.3

The elementary conditional ratio requires positive conditioning probability; explicitly report when this fails.

Skills and prerequisite lessons

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

Working and explanation

BUILD THE REASONING

Hint 1

Divide by the probability of the conditioning event.

Hint 2

Start from this intermediate relation.

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

Worked solution

Divide by the probability of the conditioning event.
Establish and simplify this mathematical relation.
$P(A|B)=P(A\cap B)/P(B)$
Establish and simplify this mathematical relation.
$P(A|B)=0.4$
Check the conditioning probability is positive.

The requested value is 0.4.

Checks and common pitfalls: Check the conditioning probability is positive.

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

05 / Foundation#Your turn

P(A∩B)=0.12, P(A)=0.2. Find P(B|A).

P(A\cap B)=0.12;\ P(A)=0.2

The elementary conditional ratio requires positive conditioning probability; explicitly report when this fails.

Skills and prerequisite lessons

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

Working and explanation

BUILD THE REASONING

Hint 1

The denominator changes when the condition is reversed.

Hint 2

Start from this intermediate relation.

P(B|A)=0.12/0.2

Worked solution

The denominator changes when the condition is reversed.
Establish and simplify this mathematical relation.
$P(B|A)=0.12/0.2$
Establish and simplify this mathematical relation.
$P(B|A)=0.6$
Conditional probabilities need not be symmetric.

The requested value is 0.6.

Checks and common pitfalls: Conditional probabilities need not be symmetric.

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

06 / Foundation#Your turn

A bag has 4 red and 3 blue. Given the first draw was red without replacement, find the second-blue probability.

R=4;\ B=3

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

Skills and prerequisite lessons

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

Working and explanation

BUILD THE REASONING

Hint 1

Remove the observed red ball before counting.

Hint 2

Start from this intermediate relation.

R_{remaining}=3;\ B_{remaining}=3

Worked solution

Remove the observed red ball before counting.
Establish and simplify this mathematical relation.
$R_{remaining}=3;\ B_{remaining}=3$
Establish and simplify this mathematical relation.
$P(B_2|R_1)=3/6=1/2$
The second denominator is 6 rather than 7.

The requested value is 0.5.

Checks and common pitfalls: The second denominator is 6 rather than 7.

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

07 / Foundation#Your turn

Can P(A|B) be computed by the elementary ratio when P(B)=0?

P(B)=0

The elementary conditional ratio requires positive conditioning probability; explicitly report when this fails.

Skills and prerequisite lessons

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

Working and explanation

BUILD THE REASONING

Hint 1

Inspect the denominator before division.

Hint 2

Start from this intermediate relation.

P(A\cap B)=0

Worked solution

Inspect the denominator before division.
Establish and simplify this mathematical relation.
$P(A\cap B)=0$
Establish and simplify this mathematical relation.
$P(A|B)=0/0\text{ is undefined by this formula}$
Do not assign zero to an undefined conditional ratio.

No; the elementary ratio has a zero denominator.

Checks and common pitfalls: Do not assign zero to an undefined conditional ratio.

Reasoning checklist · self / teacher assessment

Inspect the denominator before division.
Give the intermediate relations and explain the conclusion with its restrictions.
Do not assign zero to an undefined conditional ratio.

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

08 / Standard#Your turn

P(A)=0.4, P(B)=0.5, P(A∩B)=0.2. Are A and B independent?

P(A)=0.4;\ P(B)=0.5;\ P(A\cap B)=0.2

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.

Working and explanation

BUILD THE REASONING

Hint 1

Use the definition rather than the ability to occur together.

Hint 2

Start from this intermediate relation.

P(A)P(B)=0.4(0.5)=0.2

Worked solution

Use the definition rather than the ability to occur together.
Establish and simplify this mathematical relation.
$P(A)P(B)=0.4(0.5)=0.2$
Establish and simplify this mathematical relation.
$P(A\cap B)=P(A)P(B)$
Independent events can overlap.

Yes, because the intersection probability equals the product.

Checks and common pitfalls: Independent events can overlap.

Reasoning checklist · self / teacher assessment

Use the definition rather than the ability to occur together.
Give the intermediate relations and explain the conclusion with its restrictions.
Independent events can overlap.

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

09 / Standard#Your turn

Can overlapping A and B be used directly as a two-case total-probability partition?

P(A\cap B)>0

Cases must form a disjoint exhaustive partition. Evaluate conditional factors only for positive-probability cells; a zero-probability cell contributes zero through its joint 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

A partition must not count the same outcome twice.

Hint 2

Start from this intermediate relation.

\Omega=A\mathbin{\dot\cup}A^c

Worked solution

A partition must not count the same outcome twice.
Establish and simplify this mathematical relation.
$\Omega=A\mathbin{\dot\cup}A^c$
Establish and simplify this mathematical relation.
$P(F)=P(F\cap A)+P(F\cap A^c)$
Establish and simplify this mathematical relation.
$P(F\cap C)=P(F|C)P(C)\quad\text{if }P(C)>0$
Establish and simplify this mathematical relation.
$P(C)=0\implies P(F\cap C)=0$
For a cell of probability zero, use the zero joint contribution rather than an undefined conditional. Overlapping case products can double-count outcomes.

No; first make a disjoint exhaustive partition.

Checks and common pitfalls: For a cell of probability zero, use the zero joint contribution rather than an undefined conditional. Overlapping case products can double-count outcomes.

Reasoning checklist · self / teacher assessment

A partition must not count the same outcome twice.
Give the intermediate relations and explain the conclusion with its restrictions.
For a cell of probability zero, use the zero joint contribution rather than an undefined conditional. Overlapping case products can double-count outcomes.

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

10 / 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)=0.2

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\cap D)=0.04;\ P(D)=0.16

Worked solution

Divide A’s defective contribution by all defective contributions.
Establish and simplify this mathematical relation.
$P(A\cap D)=0.04;\ P(D)=0.16$
Establish and simplify this mathematical relation.
$P(A|D)=0.04/0.16=0.25$
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.

11 / Standard#Your turn

Among 40 day pupils and 20 boarders, 10 and 12 respectively use the library. Given library use, find P(boarder).

n(D)=40;\ n(B)=20;\ n(L\cap D)=10;\ n(L\cap B)=12

Select uniformly among the named people; the conditioning group is nonempty.

Skills and prerequisite lessons

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

Working and explanation

BUILD THE REASONING

Hint 1

Keep only the library column of the two-way table.

Hint 2

Start from this intermediate relation.

n(L)=10+12=22

Worked solution

Keep only the library column of the two-way table.
Establish and simplify this mathematical relation.
$n(L)=10+12=22$
Establish and simplify this mathematical relation.
$P(B|L)=12/22=6/11$
The conditioning group has 22 students, not all 60.

The requested value is 0.5454545454545454.

Checks and common pitfalls: The conditioning group has 22 students, not all 60.

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

12 / 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.

13 / Transfer#Your turn

A defective-unit flag has likelihoods 0.8 for defective and 0.2 for good. For which defective prior p is P(defective|flag)=1/2?

P(F|D)=0.8;\ P(F|D^c)=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

Equate the two joint contributions for a half posterior.

Hint 2

Start from this intermediate relation.

0.8p=0.2(1-p)

Worked solution

Equate the two joint contributions for a half posterior.
Establish and simplify this mathematical relation.
$0.8p=0.2(1-p)$
Establish and simplify this mathematical relation.
$p=0.2$
The prior need not be one half when the likelihoods differ.

The requested value is 0.2.

Checks and common pitfalls: The prior need not be one half when the likelihoods differ.

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

End-of-lesson check and correction

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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.