Recover missing probabilities and retain every valid model
Use branch totals and joint probabilities to recover missing values · Solve a probability equation with its range constraints · Recognise when an event probability does not identify a unique model
Section E9.4 · Extended · Addition, multiplication and independence
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
- Independent paths
- Simple quadratic equations
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
- Use branch totals and joint probabilities to recover missing values
- Solve a probability equation with its range constraints
- Recognise when an event probability does not identify a unique model
Use the tree constraints
Every branch split sums to 1, and a complete path is the product along its history. If P(first red)=0.3 and P(first red and second blue)=0.12, then P(second blue given first red)=0.12/0.3=0.4. The complementary second-red branch is 0.6. This tells you nothing about the other first-stage branch unless more information is supplied.
Enforce 0≤p≤1 without discarding valid alternatives
For fresh independent replaced draws with red probability p, two red has probability p². If p²=0.36, take p=0.6; the negative root is not a probability. But exactly one red has probability 2p(1−p). If it equals 0.48, the equation p²−p+0.24=0 gives p=0.4 or 0.6. Both are valid: swapping red and blue preserves the mixed-colour chance. Additional information is needed to distinguish them.
Sources for this lesson
Pause and explain
If 2p(1−p)=0.48, is p=0.6 the only valid answer?
Worked example
Independent replaced draws have P(two red)=0.36. Find P(red on one draw).
Show the worked solution
- p²=0.36 gives roots ±0.6.
- Only the non-negative root lies in the probability range.
Answer p=0.6.
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.
A tree gives first red probability 0.3 and red-then-blue path probability 0.12. Find the second blue and red branch probabilities after first red.
Give me a hint
Divide the path by the first branch, then complement.
Compare my reasoning
- Blue-after-red=0.12/0.3=0.4.
- Red-after-red=1−0.4=0.6.
Blue 0.4; red 0.6 after first red.
Look for these in your work
- Use a positive first-branch denominator
- Check the split sums to 1
Independent replaced draws have P(two blue)=0.49. Find P(blue) and P(one of each colour).
Give me a hint
Use the valid root, then both mixed orders.
Compare my reasoning
- Blue probability is 0.7, so red is 0.3.
- Mixed probability=2×0.7×0.3=0.42.
P(blue)=0.7; mixed colours 0.42.
Look for these in your work
- Reject the negative root
- Include both orders
A five-counter bag is sampled twice uniformly with replacement and fresh mixing. P(exactly one red)=0.48. How many red counters could the bag contain?
Give me a hint
Solve for p, then use p=red count/5.
Compare my reasoning
- 2p(1−p)=0.48 gives p=0.4 or 0.6.
- Multiplying by five gives two or three red counters; both satisfy the stated event probability.
2 or 3 red counters; the information does not distinguish them.
Look for these in your work
- Retain both valid models
- Check integer counts and the original probability
Common mistakes
- A recovered branch depends on the specified history.
- Range checks can reject invalid roots without rejecting valid ambiguity.
What can you explain now?
If 2p(1−p)=0.48, is p=0.6 the only valid answer?
Compare with the explanation
No; p=0.4 is also valid.
The probability of mixed colours is unchanged by exchanging p and 1−p.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
If 2p(1−p)=0.48, is p=0.6 the only valid answer?
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.