Update a tree after selection without replacement
Reduce the total after removing a counter · Update the relevant colour count on each branch · Check complete path probabilities sum to 1
Exam scope and assessment notes
Core and Extended foundations. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Two-stage tree diagrams
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Reduce the total after removing a counter
- Update the relevant colour count on each branch
- Check complete path probabilities sum to 1
Track what remains
Draw two counters uniformly without replacement from 3 red and 2 blue. After red first, the remaining bag is 2 red and 2 blue, so second branches are 2/4 and 2/4. After blue first, it is 3 red and 1 blue, so the branches are 3/4 and 1/4. The same second-draw denominator is 4, while the numerators depend on the first outcome.
Path probabilities use their own state
RR=(3/5)(2/4)=3/10; RB=(3/5)(2/4)=3/10; BR=(2/5)(3/4)=3/10; BB=(2/5)(1/4)=1/10. These four probabilities sum to 1. A requested mixed-colour event has two paths. A colour absent from the remaining bag has branch probability zero; it is not an additional positive possibility.
Sources for this lesson
Pause and explain
After blue is removed from 3 red and 2 blue, is P(red next)=3/4?
Worked example
For the stated unreplaced draws, find P(two red).
Show the worked solution
- First red has probability 3/5.
- After red, only two red remain out of four.
Answer (3/5)(2/4)=3/10.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For two unreplaced draws from 3 red and 2 blue, find P(two blue).
Give me a hint
Follow blue then blue with the updated count.
Compare my reasoning
- First blue=2/5.
- Second blue after blue=1/4; product=2/20.
1/10.
Look for these in your work
- Update the blue count
- Update the total
Under the same model, find P(one of each colour).
Give me a hint
Add red-blue and blue-red paths.
Compare my reasoning
- RB=(3/5)(2/4)=3/10.
- BR=(2/5)(3/4)=3/10; total=6/10.
3/5.
Look for these in your work
- Use each branch history
- Include both orders
A bag has 1 red and 3 blue counters. Draw twice without replacement. Find P(same colour).
Give me a hint
After the only red is taken, RR is impossible.
Compare my reasoning
- RR=(1/4)(0/3)=0.
- BB=(3/4)(2/3)=1/2.
1/2.
Look for these in your work
- Recognise an exhausted colour
- Keep the zero-probability path distinct
Common mistakes
- Decrease both the total and the removed colour count.
- Do not use one unchanged second-stage probability everywhere.
What can you explain now?
After blue is removed from 3 red and 2 blue, is P(red next)=3/4?
Compare with the explanation
Yes; there are 3 red among 4 remaining counters.
Only the blue count has been reduced.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
After blue is removed from 3 red and 2 blue, is P(red next)=3/4?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.