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

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 probabilityDistribution 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 / Foundation#Your turn

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

X∼Bin⁡(5,0.2)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)=npE(X)=np
Worked solution
  1. Add the expectations of five success indicators.

  2. Establish and simplify this mathematical relation.

    E(X)=npE(X)=np
  3. Establish and simplify this mathematical relation.

    E(X)=5(0.2)=1E(X)=5(0.2)=1
  4. 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.

Focus on one question

Teacher preparation and assessment

40-minute lesson

MinutesFocusTeaching and student action
5PredictPredict whether replacement increases the two-red probability from a mostly-red finite bag.
8Build the ideaSample response: without replacement the success fraction drops after a red draw.
10Guided comparisonSample response: 9/25 with replacement, 3/10 without; enumerate pairs to check.
12Independent practice and discussionSort scenarios by assumptions and use integer checks for reliability.
5Exit and next stepExit: 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)=(nk)pk(1−p)n−kP(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.

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Curriculum and source notes ↗