Kenneth ChengMathematical modeling
← Senior Mathematics Studio

LEARN · EXPLAIN · REVISE

Discrete random variables and distributions

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

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.

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.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Can two different outcomes lead to the same random-variable value?

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

Binomial model: fixed n=5, independent trials, common p=0.5. E(X)=2.5, Var(X)=1.25. Bars show exact probabilities.

Binomial model: fixed n=5, independent trials, common p=0.5. E(X)=2.5, Var(X)=1.25. Bars show exact probabilities.

Explain: Calculate two valid cases and explain the change using the defining relation.

Transfer: Group outcomes by their value before adding probabilities.

Try it. Leave your reasoning visible.

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

01 / Foundation#Worked example

Find the missing probability p.

X∈{0,2,3};P=(1/4,p,1/2)X\in\{0,2,3\};\quad P=(1/4,p,1/2)
  • 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
Use this intermediate relation.
1/4+p+1/2=11/4+p+1/2=1
Worked solution

  1. List the possible values and probabilities before computing moments.

  2. Apply the stated relation and retain its conditions.

    p=1/4p=1/4

  3. Normalisation determines the remaining probability.

The requested value is 0.25.

Checks and common pitfalls: Normalisation determines the remaining probability.

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

02 / Standard#Worked example

Find p when P(X=0)=p, P(X=1)=2p and P(X=2)=tp.

t=3t=3
  • 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
Use this intermediate relation.
p+2p+tp=1p+2p+tp=1
Worked solution

  1. List the possible values and probabilities before computing moments.

  2. Apply the stated relation and retain its conditions.

    6p=1⇒p=1/66p=1\Rightarrow p=1/6

  3. Check all resulting probabilities are nonnegative.

The requested value is 0.166666666667.

Checks and common pitfalls: Check all resulting probabilities are nonnegative.

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

03 / Transfer#Worked example

A fair die is rolled; X=1 for a face at most d, and X=0 otherwise. Find P(X=0).

d=5d=5
  • 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
Count faces strictly above d.
Worked solution

  1. List the possible values and probabilities before computing moments.

  2. Apply the stated relation and retain its conditions.

    P(X=0)=(6−5)/6=1/6P(X=0)=(6-5)/6=1/6

  3. Different die faces are grouped into one binary value.

The requested value is 0.166666666667.

Checks and common pitfalls: Different die faces are grouped into one binary value.

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

04 / Foundation#Your turn

Find the missing probability p.

X∈{0,5,6};P=(1/4,p,1/2)X\in\{0,5,6\};\quad P=(1/4,p,1/2)
  • 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
Use this intermediate relation.
1/4+p+1/2=11/4+p+1/2=1
Worked solution

  1. List the possible values and probabilities before computing moments.

  2. Apply the stated relation and retain its conditions.

    p=1/4p=1/4

  3. Normalisation determines the remaining probability.

The requested value is 0.25.

Checks and common pitfalls: Normalisation determines the remaining probability.

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

05 / Foundation#Your turn

Find P(X≥t).

X∈{0,6,7};P=(1/4,1/4,1/2)X\in\{0,6,7\};\quad P=(1/4,1/4,1/2)
  • 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
Add probabilities at t and t+1.
Worked solution

  1. List the possible values and probabilities before computing moments.

  2. Apply the stated relation and retain its conditions.

    P(X≥t)=1/4+1/2=3/4P(X\ge t)=1/4+1/2=3/4

  3. The threshold selects values, then their probabilities are added.

The requested value is 0.75.

Checks and common pitfalls: The threshold selects values, then their probabilities are added.

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

06 / Foundation#Your turn

Toss r independent fair coins; X counts heads. Find P(X=1).

r=5r=5
  • 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
Group the equally likely outcomes with one head.
Worked solution

  1. List the possible values and probabilities before computing moments.

  2. Apply the stated relation and retain its conditions.

    P(X=1)=5/32P(X=1)=5/32

  3. There is one favorable outcome for each possible position of the sole head.

The requested value is 0.15625.

Checks and common pitfalls: There is one favorable outcome for each possible position of the sole head.

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

07 / Foundation#Your turn

Give the distribution of Y=2X+1.

X=0,8;P=1/3,2/3X=0,8;\quad P=1/3,2/3
  • 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
Map each support value through Y=2X+1.
Worked solution

  1. List the possible values and probabilities before computing moments.

  2. Apply the stated relation and retain its conditions.

    0↦1,8↦170\mapsto1,\quad 8\mapsto17

  3. An injective value transformation keeps the corresponding probabilities.

The requested relation or conclusion is shown below.

Y=1,17;P=1/3,2/3Y=1,17;\quad P=1/3,2/3

Checks and common pitfalls: An injective value transformation keeps the corresponding probabilities.

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.

08 / Standard#Your turn

Give the distribution of Y=2X+1.

X=0,9;P=1/3,2/3X=0,9;\quad P=1/3,2/3
  • 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
Map each support value through Y=2X+1.
Worked solution

  1. List the possible values and probabilities before computing moments.

  2. Apply the stated relation and retain its conditions.

    0↦1,9↦190\mapsto1,\quad 9\mapsto19

  3. An injective value transformation keeps the corresponding probabilities.

The requested relation or conclusion is shown below.

Y=1,19;P=1/3,2/3Y=1,19;\quad P=1/3,2/3

Checks and common pitfalls: An injective value transformation keeps the corresponding probabilities.

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.

09 / Standard#Your turn

Find p when P(X=0)=p, P(X=1)=2p and P(X=2)=tp.

t=10t=10
  • 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
Use this intermediate relation.
p+2p+tp=1p+2p+tp=1
Worked solution

  1. List the possible values and probabilities before computing moments.

  2. Apply the stated relation and retain its conditions.

    13p=1⇒p=1/1313p=1\Rightarrow p=1/13

  3. Check all resulting probabilities are nonnegative.

The requested value is 0.0769230769231.

Checks and common pitfalls: Check all resulting probabilities are nonnegative.

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

10 / 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.

invalid\text{invalid}

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.

11 / Standard#Your turn

A fair die is rolled; X=1 for a face at most d, and X=0 otherwise. Find P(X=0).

d=1d=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
Count faces strictly above d.
Worked solution

  1. List the possible values and probabilities before computing moments.

  2. Apply the stated relation and retain its conditions.

    P(X=0)=(6−1)/6=5/6P(X=0)=(6-1)/6=5/6

  3. Different die faces are grouped into one binary value.

The requested value is 0.833333333333.

Checks and common pitfalls: Different die faces are grouped into one binary value.

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

12 / Transfer#Your turn

Can these numbers be a probability distribution? Explain.

X=0,13,14;P=0.6,0.5,−0.1X=0,13,14;\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.

invalid\text{invalid}

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.

13 / Transfer#Your turn

A fair die is rolled; X=1 for a face at most d, and X=0 otherwise. Find P(X=0).

d=3d=3
  • 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
Count faces strictly above d.
Worked solution

  1. List the possible values and probabilities before computing moments.

  2. Apply the stated relation and retain its conditions.

    P(X=0)=(6−3)/6=3/6P(X=0)=(6-3)/6=3/6

  3. Different die faces are grouped into one binary value.

The requested value is 0.5.

Checks and common pitfalls: Different die faces are grouped into one binary value.

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

Focus on one question

End-of-lesson check and correction

Review your latest checked answers and explanations. A draft change requires a fresh check. Written work needs your self-assessment or a teacher’s review.

Enable JavaScript for a summary of your local work.

    Choose a foundation skill to revisit ↗

    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 ↗