Read a two-selection tree with replacement
Label branches from the current state · Multiply along a complete path · Add separate paths for a requested event
Section Paper 1 · Historical Core · non-calculator
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
- Fractions
- Simple 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
- Label branches from the current state
- Multiply along a complete path
- Add separate paths for a requested event
Replacement restores the starting contents
A bag has 3 red and 2 blue counters. Draw uniformly, record the colour, replace the counter and mix before the second draw. Each selection has red probability 3/5 and blue 2/5. The branch probabilities are unchanged because the contents are restored; independence also requires the stated fresh uniform mixing.
At every split the outgoing probabilities sum to 1. Multiplying the first branch by the second branch probability for that state gives a complete path probability. This path rule uses probabilities along the history; it is not a licence to multiply unrelated marginal probabilities.
Different paths are separate outcomes
The four paths are RR, RB, BR, BB. Their probabilities are 9/25,6/25,6/25,4/25 and sum to 1. “One of each colour” includes RB and BR, so add those two disjoint paths. RR is just one path; do not multiply it by two.
Sources for this lesson
Pause and explain
For these replaced draws, does red on the first draw reduce the second red probability?
Worked example
For the stated replaced draws, find P(one of each colour).
Show the worked solution
- RB has probability (3/5)(2/5)=6/25.
- BR has probability (2/5)(3/5)=6/25.
Answer 12/25.
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 uniform draws with replacement from 3 red and 2 blue counters, find P(two red).
Give me a hint
Follow the RR path.
Compare my reasoning
- The first red branch is 3/5.
- The second red branch is also 3/5, giving 9/25.
9/25.
Look for these in your work
- Restore the bag contents
- Multiply along the path
Under the same replacement model, find P(exactly one blue).
Give me a hint
Use RB and BR.
Compare my reasoning
- Each mixed-colour path has probability 6/25.
- Add the two disjoint paths.
12/25.
Look for these in your work
- Include both orders
- Add path probabilities
A bag has 1 red and 3 blue counters. Draw twice uniformly with replacement and fresh mixing. Find P(same colour).
Give me a hint
Use RR and BB.
Compare my reasoning
- RR=(1/4)²=1/16.
- BB=(3/4)²=9/16, so their sum is 10/16.
5/8.
Look for these in your work
- Use the new bag probabilities
- Include both same-colour paths
Common mistakes
- Replacement includes returning and mixing before the next draw.
- One of each colour has two possible orders.
What can you explain now?
For these replaced draws, does red on the first draw reduce the second red probability?
Compare with the explanation
No; the replacement restores it to 3/5.
The contents and uniform mixing model are restored before the second draw.
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Make it stick
For these replaced draws, does red on the first draw reduce the second red probability?
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.