LEARN · EXPLAIN · REVISE

Choosing binomial, hypergeometric and normal models

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

高三選擇性必修第三册(A版).pdf · 7.4 · PDF 77 / printed page 72

Revisit first: Conditions, partitions and posterior probability Distribution means, variances and dependence

TOPIC 01

Choosing binomial, hypergeometric and normal models

Choose a model from its assumptions and support before computing its probability or moments.

What you will be able to explain

Choose a model from its assumptions and support before computing its probability or moments.
Handle census and deterministic endpoints; standardize continuous normal variables.

Mechanism determines model

A model is a claim about how data are produced, not just a familiar formula.

Fixed independent common-probability trials differ from finite sampling without replacement.

P(X=k)=\binom nkp^k(1-p)^{n-k}

Fixed independent common-probability trials differ from finite sampling without replacement.
$P_{HG}(X=k)=\binom Kk\binom{N-K}{n-k}/\binom Nn$
Fixed independent common-probability trials differ from finite sampling without replacement.
$E_{HG}(X)=nK/N;\ \operatorname{Var}_{HG}(X)=n(K/N)(1-K/N)(N-n)/(N-1)$

When it applies

Finite-sampling sizes are integers with 0≤K,n≤N; normal σ>0 and its second parameter is variance.

Counterexample: Drawing every object gives a fixed count and zero variance; a binomial substitute would miss this boundary.

Mechanism diagram, exact mass table, support bounds and standard-score representation.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: For the same finite bag, predict how success-count distributions differ with and without replacement.

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

Without replacement the finite population reduces count variance; support can exclude some binomial counts.

Numerical data for the diagram; printable
Red count	Without replacement	With replacement
0	0.2	0.25
1	0.6	0.5
2	0.2	0.25

Compare a finite simulation with the exact probabilities above; simulated frequencies can differ.

Explain: Compare exact probabilities for n draws and note the finite-population variance reduction.

Transfer: Set n=N and explain why only K successes are possible without replacement.

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

Three independent trials share success probability 1/2. Find the probability of exactly two successes.

n=3;\ p=1/2

Trials are independent, have fixed count and share the stated success 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

Count positions of the two successes.

Hint 2

Start from this intermediate relation.

P(X=2)=\binom32(1/2)^2(1/2)

Worked solution

Count positions of the two successes.
Establish and simplify this mathematical relation.
$P(X=2)=\binom32(1/2)^2(1/2)$
Establish and simplify this mathematical relation.
$P(X=2)=3/8$
The combination coefficient counts distinct arrangements.

The requested value is 0.375.

Checks and common pitfalls: The combination coefficient counts distinct arrangements.

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

02 / Standard#Worked example

A bag has 3 red and 2 blue. Draw two without replacement. Find the probability of two red.

N=5;\ K=3;\ n=2

First step provided — complete the reasoning:Count favorable pairs over all pairs.

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

Count favorable pairs over all pairs.

Hint 2

Start from this intermediate relation.

P(X=2)=\binom32/\binom52

Worked solution

Count favorable pairs over all pairs.
Establish and simplify this mathematical relation.
$P(X=2)=\binom32/\binom52$
Establish and simplify this mathematical relation.
$P(X=2)=3/10$
Without replacement changes the second probability.

The requested value is 0.3.

Checks and common pitfalls: Without replacement changes the second probability.

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

03 / Transfer#Worked example

From 3 red and 2 blue, compare two-red probability with replacement and without replacement. Explain the gap.

N=5;\ K=3;\ n=2

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

Inspect whether the first success reduces the success fraction.

Hint 2

Start from this intermediate relation.

P_{with}=(3/5)^2=9/25

Worked solution

Inspect whether the first success reduces the success fraction.
Establish and simplify this mathematical relation.
$P_{with}=(3/5)^2=9/25$
Establish and simplify this mathematical relation.
$P_{without}=(3/5)(2/4)=3/10$
The marginal first-draw rate is identical but dependence changes the joint rate.

With replacement 9/25; without replacement 3/10.

Checks and common pitfalls: The marginal first-draw rate is identical but dependence changes the joint rate.

Reasoning checklist · self / teacher assessment

Inspect whether the first success reduces the success fraction.
Give the intermediate relations and explain the conclusion with its restrictions.
The marginal first-draw rate is identical but dependence changes the joint rate.

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

04 / Foundation#Your turn

X~Bin(5,0.2). Find E(X).

X\sim\operatorname{Bin}(5,0.2)

Trials are independent, have fixed count and share the stated success 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

Add the expectations of five success indicators.

Hint 2

Start from this intermediate relation.

E(X)=np

Worked solution

Add the expectations of five success indicators.
Establish and simplify this mathematical relation.
$E(X)=np$
Establish and simplify this mathematical relation.
$E(X)=5(0.2)=1$
The expected count need not describe every realization.

The requested value is 1.

Checks and common pitfalls: The expected count need not describe every realization.

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

05 / Foundation#Your turn

Four independent fair trials are made. Find the probability of no success.

n=4;\ p=1/2

Trials are independent, have fixed count and share the stated success 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

Every trial must fail.

Hint 2

Start from this intermediate relation.

P(X=0)=(1-p)^n

Worked solution

Every trial must fail.
Establish and simplify this mathematical relation.
$P(X=0)=(1-p)^n$
Establish and simplify this mathematical relation.
$P(X=0)=1/16$
There is one all-failure sequence, not four.

The requested value is 0.0625.

Checks and common pitfalls: There is one all-failure sequence, not four.

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

06 / Foundation#Your turn

Ten draws are without replacement from a small finite bag. Must their success count be binomial?

\text{finite population; no replacement}

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

Check independence and common probability, not only fixed trial count.

Hint 2

Start from this intermediate relation.

P(R_2|R_1)=(K-1)/(N-1)

Worked solution

Check independence and common probability, not only fixed trial count.
Establish and simplify this mathematical relation.
$P(R_2|R_1)=(K-1)/(N-1)$
Establish and simplify this mathematical relation.
$P(R_1)=K/N$
A large-population approximation needs separate justification.

No; draws are generally dependent and their conditional success probabilities change.

Checks and common pitfalls: A large-population approximation needs separate justification.

Reasoning checklist · self / teacher assessment

Check independence and common probability, not only fixed trial count.
Give the intermediate relations and explain the conclusion with its restrictions.
A large-population approximation needs separate justification.

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

07 / Foundation#Your turn

X~Bin(4,1/4). Find Var(X).

X\sim\operatorname{Bin}(4,1/4)

Trials are independent, have fixed count and share the stated success 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

Use the Bernoulli variance for each independent trial.

Hint 2

Start from this intermediate relation.

\operatorname{Var}(X)=np(1-p)

Worked solution

Use the Bernoulli variance for each independent trial.
Establish and simplify this mathematical relation.
$\operatorname{Var}(X)=np(1-p)$
Establish and simplify this mathematical relation.
$\operatorname{Var}(X)=4(1/4)(3/4)=3/4$
The factor 1−p must be retained.

The requested value is 0.75.

Checks and common pitfalls: The factor 1−p must be retained.

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

08 / Standard#Your turn

From 4 red and 1 blue, choose 3 without replacement. What values can the red count take?

N=5;\ K=4;\ n=3

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

Respect available failures and the draw size.

Hint 2

Start from this intermediate relation.

\max(0,n-(N-K))=2

Worked solution

Respect available failures and the draw size.
Establish and simplify this mathematical relation.
$\max(0,n-(N-K))=2$
Establish and simplify this mathematical relation.
$\min(n,K)=3$
The formal range 0,…,n includes impossible counts in this bag.

Only 2 or 3.

Checks and common pitfalls: The formal range 0,…,n includes impossible counts in this bag.

Reasoning checklist · self / teacher assessment

Respect available failures and the draw size.
Give the intermediate relations and explain the conclusion with its restrictions.
The formal range 0,…,n includes impossible counts in this bag.

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

09 / Standard#Your turn

X~N(10,4). Find z for x=14.

X\sim N(10,4);\ x=14

The normal model has positive standard deviation; its second parameter is variance.

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

Take the square root of the variance.

Hint 2

Start from this intermediate relation.

\sigma=2

Worked solution

Take the square root of the variance.
Establish and simplify this mathematical relation.
$\sigma=2$
Establish and simplify this mathematical relation.
$z=(14-10)/2=2$
The second normal parameter is variance, not standard deviation.

The requested value is 2.

Checks and common pitfalls: The second normal parameter is variance, not standard deviation.

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

10 / Standard#Your turn

For a continuous normal distribution, find P(X=μ).

X\sim N(\mu,\sigma^2);\ \sigma>0

The normal model has positive standard deviation; its second parameter is variance.

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

Distinguish density height from point probability.

Hint 2

Start from this intermediate relation.

P(X=\mu)=\int_{\mu}^{\mu}f(x)\,dx

Worked solution

Distinguish density height from point probability.
Establish and simplify this mathematical relation.
$P(X=\mu)=\int_{\mu}^{\mu}f(x)\,dx$
Establish and simplify this mathematical relation.
$P(X=\mu)=0$
The peak density does not give positive mass at one point.

The requested value is 0.

Checks and common pitfalls: The peak density does not give positive mass at one point.

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

11 / Standard#Your turn

X~N(10,4), Φ(1)=0.8413. Find P(X≤12).

X\sim N(10,4);\ \Phi(1)=0.8413

The normal model has positive standard deviation; its second parameter is variance.

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

Convert the threshold to a standard score.

Hint 2

Start from this intermediate relation.

(12-10)/2=1

Worked solution

Convert the threshold to a standard score.
Establish and simplify this mathematical relation.
$(12-10)/2=1$
Establish and simplify this mathematical relation.
$P(X\le12)=\Phi(1)=0.8413$
Use the provided table value rather than inventing extra precision.

The requested value is 0.8413.

Checks and common pitfalls: Use the provided table value rather than inventing extra precision.

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

12 / Transfer#Your turn

Independent attempts succeed with probability 1/2. Find the least n making at least one success probability ≥0.95.

p=1/2;\ P(X\ge1)\ge0.95

Trials are independent, have fixed count and share the stated success 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

Use the complement and check neighboring integer values.

Hint 2

Start from this intermediate relation.

1-2^{-n}\ge0.95

Worked solution

Use the complement and check neighboring integer values.
Establish and simplify this mathematical relation.
$1-2^{-n}\ge0.95$
Establish and simplify this mathematical relation.
$n=4:15/16<0.95;\ n=5:31/32\ge0.95$
A real-valued bound must be rounded up and checked.

The requested value is 5.

Checks and common pitfalls: A real-valued bound must be rounded up and checked.

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

13 / Transfer#Your turn

All 6 objects contain exactly 2 red; draw all 6 without replacement. What are the red-count mean and variance?

N=6;\ K=2;\ n=6

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

At a census boundary, no uncertainty about the count remains.

Hint 2

Start from this intermediate relation.

P(X=2)=1

Worked solution

At a census boundary, no uncertainty about the count remains.
Establish and simplify this mathematical relation.
$P(X=2)=1$
Establish and simplify this mathematical relation.
$E(X)=2;\ \operatorname{Var}(X)=0$
The binomial variance would be positive and is the wrong model.

Mean 2, variance 0: the count is fixed.

Checks and common pitfalls: The binomial variance would be positive and is the wrong model.

Reasoning checklist · self / teacher assessment

At a census boundary, no uncertainty about the count remains.
Give the intermediate relations and explain the conclusion with its restrictions.
The binomial variance would be positive and is the wrong model.

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 replacement increases the two-red probability from a mostly-red finite bag.
8	Build the idea	Sample response: without replacement the success fraction drops after a red draw.
10	Guided comparison	Sample response: 9/25 with replacement, 3/10 without; enumerate pairs to check.
12	Independent practice and discussion	Sort scenarios by assumptions and use integer checks for reliability.
5	Exit and next step	Exit: when drawing all six objects with two reds, X=2 surely and variance is zero.

Question sequence

Predict whether replacement increases the two-red probability from a mostly-red finite bag.
Sample response: 9/25 with replacement, 3/10 without; enumerate pairs to check.

Board development

Mechanism determines model
$P(X=k)=\binom nkp^k(1-p)^{n-k}$

Expected correct and incorrect responses

Correct sample: Hypergeometric
Anticipated wrong response: Binomial without checking conditions — Uses fixed trial count alone to justify a binomial model.
Anticipated wrong response: 9 — Reads the normal variance parameter as standard deviation.
Choosing binomial from fixed n alone.
Using normal variance as its standard deviation.

Follow-up questions

Follow-up: Track the finite population without replacement. Then explain why Binomial without checking conditions fails.
Follow-up: Read the second parameter as variance. Then explain why 9 fails.
Set n=N and explain why only K successes are possible without replacement.

Remediation

Sample response: without replacement the success fraction drops after a red draw.
Return to the first example, identify the missing condition, then solve one fresh task without viewing its solution.

Exit check

Exit: when drawing all six objects with two reds, X=2 surely and variance is zero.

Assessment checklist and sample responses

Representation and conditions
No replacement changes the next success fraction, so finite count is hypergeometric.
Connected mathematical reasoning
Two red probability from three red of five is (3/5)(2/4)=3/10.
Interpretation and boundary check
Drawing the full bag with two reds gives X=2 surely and variance zero.