Track three selections and collect all valid orders
Update counts through a three-stage history · Include each order for an exact colour count · Compare replacement with an exhausted-colour branch
Exam scope and assessment notes
Extended content; not a Core requirement. 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 trees
- Complementary probability
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
- Update counts through a three-stage history
- Include each order for an exact colour count
- Compare replacement with an exhausted-colour branch
Every stage follows its own history
From 4 red and 2 blue, draw three uniformly without replacement. Totals become 6,5,4. For exactly two red, use RRB,RBR,BRR. RRB has probability (4/6)(3/5)(2/4)=1/5. Each other order also gives 1/5, so the total is 3/5. Equal path values must be checked; they are not a universal tree property.
Zero branches and complements
Three blue is impossible because the bag begins with only two blue. At least one blue is the complement of RRR: 1−(4/6)(3/5)(2/4)=4/5. With replacement and fresh uniform mixing, all stages use red 2/3 and blue 1/3, including the now possible BBB path. Do not use a binomial-style product for an unreplaced sequence with changing counts.
Sources for this lesson
Pause and explain
Is BBB possible in three draws without replacement from 4 red and 2 blue?
Worked example
Find P(exactly two red) for three unreplaced draws from 4 red and 2 blue.
Show the worked solution
- RRB,RBR and BRR each have probability 1/5.
- Add the three disjoint paths.
Answer 3/5.
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.
From 4 red and 2 blue, draw three without replacement. Find P(RBR).
Give me a hint
Update the bag after each colour.
Compare my reasoning
- RBR=(4/6)(2/5)(3/4).
- The product simplifies to 1/5.
1/5.
Look for these in your work
- Use the correct stage totals
- Update the red count after the first draw
For the same three unreplaced draws, find P(exactly two blue).
Give me a hint
Use BBR,BRB,RBB.
Compare my reasoning
- Each order has probability 1/15, using its own remaining counts.
- Add three such paths.
1/5.
Look for these in your work
- Include all three orders
- Check each path separately
For three draws from 4 red and 2 blue, compare P(at least one blue) without replacement and with replacement plus fresh mixing.
Give me a hint
Complement RRR under each rule.
Compare my reasoning
- Without replacement: 1−(4/6)(3/5)(2/4)=4/5.
- With replacement: 1−(2/3)³=19/27.
Without replacement 4/5; with replacement 19/27.
Look for these in your work
- Use the specified sampling process
- Complement the complete no-blue path
Common mistakes
- Exactly two red has three orders.
- Track colour counts as well as the total.
What can you explain now?
Is BBB possible in three draws without replacement from 4 red and 2 blue?
Compare with the explanation
No; the third blue branch has probability zero.
Only two blue counters exist initially.
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Make it stick
Is BBB possible in three draws without replacement from 4 red and 2 blue?
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.