Kenneth ChengMathematical modeling
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Conditional and total probability

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

高三選擇性必修 第三册(A版).pdf · 7.1 · PDF 49 / printed page 44

Revisit first: Permutations and combinations

TOPIC 01

Conditional and total probability

Use conditional denominators, disjoint partitions and Bayes inversion.

What you will be able to explain

  • Use conditional denominators, disjoint partitions and Bayes inversion.
  • Justify the method and check the conditions in a new situation.

Defining relation

Use conditional denominators, disjoint partitions and Bayes inversion.

P(A∣B)=P(A∩B)/P(B)P(A|B)=P(A\cap B)/P(B)

Conditions

The conditioning event must have positive probability; a total-probability partition is exhaustive and disjoint.

PREDICT → EXPLORE → EXPLAIN → TRANSFER

Make a prediction, then explore the relationship.

Lesson question: Can two events be independent even when both may occur together?

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: Construct a probability table showing independence and compare its two conditional directions.

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 P(A|B) from the given probabilities.

P(A∩B)=1/6,P(B)=2/6P(A\cap B)=1/6,\quad P(B)=2/6
  • The conditioning event must have positive probability; a total-probability partition is exhaustive and disjoint.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the event with a conditional probability or a disjoint partition.
Hint 2
Use this intermediate relation.
P(A∣B)=P(A∩B)/P(B)P(A|B)=P(A\cap B)/P(B)
Worked solution

  1. Express the event with a conditional probability or a disjoint partition.

  2. Apply the stated relation and retain its conditions.

    P(A∣B)=(1/u)/(2/u)=1/2P(A|B)=(1/u)/(2/u)=1/2

  3. Restrict the sample space to B.

The requested value is 0.5.

Checks and common pitfalls: Restrict the sample space to B.

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

02 / Standard#Worked example

Choose machine 1 with probability 1/(t+2), otherwise machine 2; their failure rates are 1/10 and 1/5. Given failure, find P(machine 1).

t=3t=3
  • The conditioning event must have positive probability; a total-probability partition is exhaustive and disjoint.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the event with a conditional probability or a disjoint partition.
Hint 2
Use this intermediate relation.
P(M1∣F)=P(M1∩F)/P(F)P(M_1|F)=P(M_1\cap F)/P(F)
Worked solution

  1. Express the event with a conditional probability or a disjoint partition.

  2. Apply the stated relation and retain its conditions.

    P(F)=9/[10(5)]P(F)=9/[10(5)]

  3. Apply the stated relation and retain its conditions.

    P(M1∣F)=1/[10(5)]9/[10(5)]=1/9P(M_1|F)=\frac{1/[10(5)]}{9/[10(5)]}=1/9

  4. The posterior denominator includes failures from both machines.

The requested value is 0.111111111111.

Checks and common pitfalls: The posterior denominator includes failures from both machines.

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

03 / Transfer#Worked example

A fair coin selects a bag: bag 1 has t red and 1 blue; bag 2 has 1 red and t blue. Given a red draw, find the probability bag 1 was chosen.

t=4t=4
  • The conditioning event must have positive probability; a total-probability partition is exhaustive and disjoint.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the event with a conditional probability or a disjoint partition.
Hint 2
Use Bayes with P(red)=1/2.
Worked solution

  1. Express the event with a conditional probability or a disjoint partition.

  2. Apply the stated relation and retain its conditions.

    P(R)=12[t/(5)+1/(5)]=1/2P(R)=\tfrac12[t/(5)+1/(5)]=1/2

  3. Apply the stated relation and retain its conditions.

    P(B1∣R)=t/5P(B_1|R)=t/5

  4. Equal bag-selection priors do not remain equal after observing the color.

The requested value is 0.8.

Checks and common pitfalls: Equal bag-selection priors do not remain equal after observing the color.

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

04 / Foundation#Your turn

Find P(A|B) from the given probabilities.

P(A∩B)=1/9,P(B)=2/9P(A\cap B)=1/9,\quad P(B)=2/9
  • The conditioning event must have positive probability; a total-probability partition is exhaustive and disjoint.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the event with a conditional probability or a disjoint partition.
Hint 2
Use this intermediate relation.
P(A∣B)=P(A∩B)/P(B)P(A|B)=P(A\cap B)/P(B)
Worked solution

  1. Express the event with a conditional probability or a disjoint partition.

  2. Apply the stated relation and retain its conditions.

    P(A∣B)=(1/u)/(2/u)=1/2P(A|B)=(1/u)/(2/u)=1/2

  3. Restrict the sample space to B.

The requested value is 0.5.

Checks and common pitfalls: Restrict the sample space to B.

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

05 / Foundation#Your turn

From a bag of t red and 2 blue balls, two are drawn without replacement. Given the first is red, find the probability the second is blue.

t=6t=6
  • The conditioning event must have positive probability; a total-probability partition is exhaustive and disjoint.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the event with a conditional probability or a disjoint partition.
Hint 2
After a red draw, total=t+1 and blue=2.
Worked solution

  1. Express the event with a conditional probability or a disjoint partition.

  2. Apply the stated relation and retain its conditions.

    P=2/7P=2/7

  3. The color count and denominator must reflect the information already given.

The requested value is 0.285714285714.

Checks and common pitfalls: The color count and denominator must reflect the information already given.

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

06 / Foundation#Your turn

Events A,B are independent with P(A)=1/3 and P(B)=t/(t+1). Find P(A∩B).

t=7t=7
  • The conditioning event must have positive probability; a total-probability partition is exhaustive and disjoint.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the event with a conditional probability or a disjoint partition.
Hint 2
Use this intermediate relation.
P(A∩B)=P(A)P(B)P(A\cap B)=P(A)P(B)
Worked solution

  1. Express the event with a conditional probability or a disjoint partition.

  2. Apply the stated relation and retain its conditions.

    P=7/[3(8)]P=7/[3(8)]

  3. Multiplication here is justified by the stated independence.

The requested value is 0.291666666667.

Checks and common pitfalls: Multiplication here is justified by the stated independence.

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

07 / Foundation#Your turn

Choose machine 1 with probability 1/(t+2) and machine 2 otherwise. Failure rates are 1/10 and 1/5. Find total failure probability.

t=8t=8
  • The conditioning event must have positive probability; a total-probability partition is exhaustive and disjoint.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the event with a conditional probability or a disjoint partition.
Hint 2
Use this intermediate relation.
P(F)=P(M1)P(F∣M1)+P(M2)P(F∣M2)P(F)=P(M_1)P(F|M_1)+P(M_2)P(F|M_2)
Worked solution

  1. Express the event with a conditional probability or a disjoint partition.

  2. Apply the stated relation and retain its conditions.

    P(F)=110110+91015=19100P(F)=\frac1{10}\frac1{10}+\frac{9}{10}\frac15=\frac{19}{100}

  3. The machine cases form a disjoint exhaustive partition.

The requested value is 0.19.

Checks and common pitfalls: The machine cases form a disjoint exhaustive partition.

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

08 / Standard#Your turn

Choose machine 1 with probability 1/(t+2) and machine 2 otherwise. Failure rates are 1/10 and 1/5. Find total failure probability.

t=9t=9
  • The conditioning event must have positive probability; a total-probability partition is exhaustive and disjoint.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the event with a conditional probability or a disjoint partition.
Hint 2
Use this intermediate relation.
P(F)=P(M1)P(F∣M1)+P(M2)P(F∣M2)P(F)=P(M_1)P(F|M_1)+P(M_2)P(F|M_2)
Worked solution

  1. Express the event with a conditional probability or a disjoint partition.

  2. Apply the stated relation and retain its conditions.

    P(F)=111110+101115=21110P(F)=\frac1{11}\frac1{10}+\frac{10}{11}\frac15=\frac{21}{110}

  3. The machine cases form a disjoint exhaustive partition.

The requested value is 0.190909090909.

Checks and common pitfalls: The machine cases form a disjoint exhaustive partition.

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

09 / Standard#Your turn

Choose machine 1 with probability 1/(t+2), otherwise machine 2; their failure rates are 1/10 and 1/5. Given failure, find P(machine 1).

t=10t=10
  • The conditioning event must have positive probability; a total-probability partition is exhaustive and disjoint.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the event with a conditional probability or a disjoint partition.
Hint 2
Use this intermediate relation.
P(M1∣F)=P(M1∩F)/P(F)P(M_1|F)=P(M_1\cap F)/P(F)
Worked solution

  1. Express the event with a conditional probability or a disjoint partition.

  2. Apply the stated relation and retain its conditions.

    P(F)=23/[10(12)]P(F)=23/[10(12)]

  3. Apply the stated relation and retain its conditions.

    P(M1∣F)=1/[10(12)]23/[10(12)]=1/23P(M_1|F)=\frac{1/[10(12)]}{23/[10(12)]}=1/23

  4. The posterior denominator includes failures from both machines.

The requested value is 0.0434782608696.

Checks and common pitfalls: The posterior denominator includes failures from both machines.

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

10 / Standard#Your turn

A and B are disjoint and both have positive probability. Can they be independent?

P(A)=1/15,P(B)=1/15P(A)=1/15,\quad P(B)=1/15
  • The conditioning event must have positive probability; a total-probability partition is exhaustive and disjoint.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the event with a conditional probability or a disjoint partition.
Hint 2
Compare the intersection with the product.
Worked solution

  1. Express the event with a conditional probability or a disjoint partition.

  2. Apply the stated relation and retain its conditions.

    P(A)P(B)=1/225>0P(A)P(B)=1/225>0

  3. Disjoint positive-probability events are dependent.

The requested relation or conclusion is shown below.

P(A∩B)=0≠P(A)P(B)P(A\cap B)=0\ne P(A)P(B)

Checks and common pitfalls: Disjoint positive-probability events are dependent.

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 coin selects a bag: bag 1 has t red and 1 blue; bag 2 has 1 red and t blue. Given a red draw, find the probability bag 1 was chosen.

t=12t=12
  • The conditioning event must have positive probability; a total-probability partition is exhaustive and disjoint.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the event with a conditional probability or a disjoint partition.
Hint 2
Use Bayes with P(red)=1/2.
Worked solution

  1. Express the event with a conditional probability or a disjoint partition.

  2. Apply the stated relation and retain its conditions.

    P(R)=12[t/(13)+1/(13)]=1/2P(R)=\tfrac12[t/(13)+1/(13)]=1/2

  3. Apply the stated relation and retain its conditions.

    P(B1∣R)=t/13P(B_1|R)=t/13

  4. Equal bag-selection priors do not remain equal after observing the color.

The requested value is 0.923076923077.

Checks and common pitfalls: Equal bag-selection priors do not remain equal after observing the color.

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

12 / Transfer#Your turn

A and B are disjoint and both have positive probability. Can they be independent?

P(A)=1/17,P(B)=1/17P(A)=1/17,\quad P(B)=1/17
  • The conditioning event must have positive probability; a total-probability partition is exhaustive and disjoint.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the event with a conditional probability or a disjoint partition.
Hint 2
Compare the intersection with the product.
Worked solution

  1. Express the event with a conditional probability or a disjoint partition.

  2. Apply the stated relation and retain its conditions.

    P(A)P(B)=1/289>0P(A)P(B)=1/289>0

  3. Disjoint positive-probability events are dependent.

The requested relation or conclusion is shown below.

P(A∩B)=0≠P(A)P(B)P(A\cap B)=0\ne P(A)P(B)

Checks and common pitfalls: Disjoint positive-probability events are dependent.

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 coin selects a bag: bag 1 has t red and 1 blue; bag 2 has 1 red and t blue. Given a red draw, find the probability bag 1 was chosen.

t=14t=14
  • The conditioning event must have positive probability; a total-probability partition is exhaustive and disjoint.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Express the event with a conditional probability or a disjoint partition.
Hint 2
Use Bayes with P(red)=1/2.
Worked solution

  1. Express the event with a conditional probability or a disjoint partition.

  2. Apply the stated relation and retain its conditions.

    P(R)=12[t/(15)+1/(15)]=1/2P(R)=\tfrac12[t/(15)+1/(15)]=1/2

  3. Apply the stated relation and retain its conditions.

    P(B1∣R)=t/15P(B_1|R)=t/15

  4. Equal bag-selection priors do not remain equal after observing the color.

The requested value is 0.933333333333.

Checks and common pitfalls: Equal bag-selection priors do not remain equal after observing the color.

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

Focus on one question

End-of-lesson check and correction

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    Teacher preparation and assessment

    Question sequence

    • Use conditional denominators, disjoint partitions and Bayes inversion.
    • Which condition is essential in conditional and total probability?
    • Can two events be independent even when both may occur together?

    Board plan

    • Defining relation: Use conditional denominators, disjoint partitions and Bayes inversion.
      P(A∣B)=P(A∩B)/P(B)P(A|B)=P(A\cap B)/P(B)
    • Conditions: The conditioning event must have positive probability; a total-probability partition is exhaustive and disjoint.

    Anticipated thinking

    • P(A|B) need not equal P(B|A), and independence differs from mutual exclusivity.

    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.

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