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

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 · 7.4 · 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.

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

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.

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.