Test independence using a two-way frequency table
Read joint and marginal counts from the same population · Compare the joint probability with the product · Limit conclusions drawn from sample frequencies
Section Paper 2 · Historical Extended · non-calculator
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-event probabilities
- Frequency counts
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
- Read joint and marginal counts from the same population
- Compare the joint probability with the product
- Limit conclusions drawn from sample frequencies
Use one common denominator
The illustrated complete population has 100 equally selectable members: 18 both A and B, 12 A only, 42 B only and 28 neither. P(A)=30/100, P(B)=60/100 and P(both)=18/100. Their product is 0.3×0.6=0.18, matching the joint, so A and B are independent in this specified uniform-population model.
A sample is evidence, not the full process
Among the 30 A members, 18 have B: 18/30=0.6, matching the overall B probability. Restricting the denominator to A explains why A gives no new information about B here. If these counts were only a sample of a larger process, exact equality would not prove independence in that process, and a small mismatch could arise from sampling variation.
Sources for this lesson
Pause and explain
In the stated complete table, do A and B satisfy the independence condition?
Worked example
Test independence for the illustrated complete 100-member population.
Show the worked solution
- P(both)=18/100=0.18.
- P(A)P(B)=(30/100)(60/100)=0.18.
Answer Independent under uniform selection from this specified population.
Try a topic test
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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 the complete 100-member table with both 18, A only 12, B only 42 and neither 28, find P(B among A) and compare it with P(B).
Give me a hint
Use the A total for the restricted probability.
Compare my reasoning
- A total is 30, so the restricted probability is 18/30=0.6.
- Overall B probability is 60/100=0.6.
Both equal 0.6.
Look for these in your work
- Use the restricted denominator
- Compare like event probabilities
A different complete table has both 24, A only 6, B only 36 and neither 34. Test independence for uniform selection.
Give me a hint
The A and B totals are still 30 and 60.
Compare my reasoning
- P(A)P(B)=0.3×0.6=0.18.
- P(both)=24/100=0.24, so the condition fails.
Not independent in this model: 0.24≠0.18.
Look for these in your work
- Find marginals by adding cells
- Compare with the stated joint
A small random sample gives a joint relative frequency close to the product of its two marginal frequencies. Does this prove the unknown population events are independent?
Give me a hint
Distinguish the sample from the underlying population.
Compare my reasoning
- Sampling variability can produce agreement or disagreement.
- The result is evidence to assess with design and further data, not proof of the population property.
No; it is sample evidence, not a proof of population independence.
Look for these in your work
- Limit the inference to the evidence
- Acknowledge sampling variability
Common mistakes
- Use total population for joint and marginal probabilities.
- Do not promote sample evidence to a proof about an unknown process.
What can you explain now?
In the stated complete table, do A and B satisfy the independence condition?
Compare with the explanation
Yes; 0.18=0.3×0.6.
Their joint probability equals the product of their marginal probabilities.
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Make it stick
In the stated complete table, do A and B satisfy the independence condition?
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.