Build a possibility diagram and count ordered outcomes
Place one experiment on each grid axis · Keep ordered outcomes distinct · Count favourable cells only under equal-likelihood assumptions
Section Paper 2 · Historical Extended · 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
- Single-trial sample spaces
- Inclusive or
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
- Place one experiment on each grid axis
- Keep ordered outcomes distinct
- Count favourable cells only under equal-likelihood assumptions
One row and one column per outcome
For two independent fair dice, make six rows for the first die and six columns for the second. Each of the 36 cells represents an ordered pair with probability 1/36. The sums 2 to 12 are not equally likely: only (1,1) gives 2, while six pairs give 7.
The outcomes (2,5) and (5,2) differ because the dice are labelled first and second. A grid for biased or dependent experiments may still list possibilities, but its cells do not automatically have equal probability.
Select complete events
For total 7, mark (1,6),(2,5),(3,4),(4,3),(5,2),(6,1). For a double, mark the six diagonal pairs. Total 7 and a double have no common cell, so either event has twelve favourable cells. Do not count a shared cell twice when other events overlap.
Sources for this lesson
Pause and explain
Is P(total 2)=1/11 because there are eleven possible totals?
Worked example
Find P(total 7) for two independent fair dice.
Show the worked solution
- Six of the 36 ordered pairs sum to 7.
- 6/36=1/6.
Answer 1/6.
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.
For two independent fair dice, list the pairs giving total 5 and find its probability.
Give me a hint
Read the four anti-diagonal cells.
Compare my reasoning
- The pairs are (1,4),(2,3),(3,2),(4,1).
- 4/36=1/9.
The four listed pairs; probability 1/9.
Look for these in your work
- Keep the order of the dice
- Count out of 36
For two independent fair dice, find the probability that the first number is greater than the second.
Give me a hint
Count 0,1,2,3,4,5 favourable cells in the successive rows.
Compare my reasoning
- There are 15 favourable pairs.
- 15/36=5/12.
5/12.
Look for these in your work
- Use a strict inequality
- Exclude the six diagonal ties
For two independent fair dice, find P(total 7 or a double). Explain whether any pair is counted in both groups.
Give me a hint
Compare the six sum-7 cells with the diagonal.
Compare my reasoning
- There are six sum-7 cells and six doubles.
- A double has an even total, so none totals 7; favourable count=12.
1/3; there is no overlap.
Look for these in your work
- Check common cells
- Count distinct favourable pairs
Common mistakes
- The possible sums are not equally likely.
- Labelled dice produce ordered pairs.
What can you explain now?
Is P(total 2)=1/11 because there are eleven possible totals?
Compare with the explanation
No; only one of 36 equally likely pairs gives total 2.
Different totals contain different numbers of ordered pairs.
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Make it stick
Is P(total 2)=1/11 because there are eleven possible totals?
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.