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Can these numbers be a probability distribution? Explain.

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 · 7.2 · PDF 61 / printed page 56

Revisit first: Conditional and total probability

TOPIC 01

Discrete random variables and distributions

Construct a distribution and compute event probabilities from its support.

What you will be able to explain

  • Construct a distribution and compute event probabilities from its support.
  • Justify the method and check the conditions in a new situation.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

Can these numbers be a probability distribution? Explain.

X=0,11,12;P=0.6,0.5,−0.1X=0,11,12;\quad P=0.6,0.5,-0.1
  • Every probability lies in [0,1]; values and events must not be double-counted.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
List the possible values and probabilities before computing moments.
Hint 2
Check nonnegativity as well as the total.
Worked solution
  1. List the possible values and probabilities before computing moments.

  2. Apply the stated relation and retain its conditions.

    0.6+0.5−0.1=10.6+0.5-0.1=1
  3. Apply the stated relation and retain its conditions.

    P(X=t+1)=−0.1<0P(X=t+1)=-0.1<0
  4. A sum of one alone is insufficient.

The requested relation or conclusion is shown below.

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Checks and common pitfalls: A sum of one alone is insufficient.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

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

Focus on one question

Teacher preparation and assessment

Question sequence

  • Construct a distribution and compute event probabilities from its support.
  • Which condition is essential in discrete random variables and distributions?
  • Can two different outcomes lead to the same random-variable value?

Board plan

  • Defining relation: Construct a distribution and compute event probabilities from its support.
    ∑xP(X=x)=1\sum_xP(X=x)=1
  • Conditions: Every probability lies in [0,1]; values and events must not be double-counted.

Anticipated thinking

  • A random variable value and its probability are not interchangeable.

Assessment checklist

  • 1 mark: choose the correct representation and conditions.
  • 1 mark: establish the intermediate relation.
  • 1 mark: complete a connected calculation or proof.
  • 1 mark: interpret and check the conclusion.

No sign-in. Work stays in this browser. Export before clearing browser data. Written reasoning is assessed with a checklist.

Curriculum and source notes ↗