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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.

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 · 7.1 · 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.

Try it. Leave your reasoning visible.

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

01 / 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.

Focus on one question

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.

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

Curriculum and source notes ↗