P7: Possibility Spaces
Construct and use possibility spaces for single and combined experiments
📋 Key Concepts
Definition: A possibility space (or sample space) is a list or table showing all possible outcomes of an experiment.
Key Terms
| Term | Definition | Example |
|---|---|---|
| Sample Space | All possible outcomes | {1, 2, 3, 4, 5, 6} for a dice |
| Possibility Space | Same as sample space | Shows all outcomes |
| Combined Experiment | Two or more events together | Rolling two dice |
Total outcomes = Product of individual outcomes
For two dice: 6 × 6 = 36 outcomes
For two dice: 6 × 6 = 36 outcomes
📝 Single Event Sample Spaces
Method: List all outcomes in curly brackets { }. This is the sample space for a single event.
Example 1
Write the sample space for:
a) A fair 6-sided dice
b) A fair coin
c) A spinner with A, B, C
Solutions:
a) S = {1, 2, 3, 4, 5, 6}
b) S = {H, T} or {Heads, Tails}
c) S = {A, B, C}
📝 Combined Events - Grid Method
Method: Use a two-way table (grid) to show all outcomes when combining two events.
Example 2
Two fair dice are rolled. Create a possibility space showing the sum of the dice.
Solution:
| + | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Total outcomes: 36
📝 Finding Probabilities from Grids
Method: Count favourable outcomes from the grid, divide by total outcomes.
Example 3
Using the dice sum grid above, find:
a) P(sum = 7)
b) P(sum ≥ 10)
c) P(sum is even)
Solutions:
a) Sum = 7 appears: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 outcomes
P(sum = 7) = 6/36 = 1/6
b) Sum ≥ 10: (4,6), (5,5), (5,6), (6,4), (6,5), (6,6) = 6 outcomes
P(sum ≥ 10) = 6/36 = 1/6
c) Even sums: 2, 4, 6, 8, 10, 12
Count: 1 + 3 + 5 + 5 + 3 + 1 = 18 outcomes
P(even) = 18/36 = 1/2
📝 Two Coins Example
Example 4
Two fair coins are flipped. Create a possibility space and find P(at least one head).
Sample space:
| H | T | |
|---|---|---|
| H | HH | HT |
| T | TH | TT |
S = {HH, HT, TH, TT}
P(at least one head) = P(HH or HT or TH) = 3/4
📝 Product of Outcomes
Rule: Total outcomes = First event outcomes × Second event outcomes
Example 5
A spinner (4 sections) and a dice (6 sides) are used together. How many outcomes?
Solution:
Spinner outcomes × Dice outcomes = 4 × 6 = 24 outcomes
This saves listing all 24 combinations!
📝 Three or More Events
Method: For three events, extend the grid or list systematically. Total = m × n × p
Example 6
Three coins are flipped. How many outcomes? List them.
Solution:
Total outcomes = 2 × 2 × 2 = 8
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
🧮 Worked Examples — Exam Style
Possibility spaces: list or tabulate ALL outcomes, then count favourable ones. The sample space IS the denominator.
Two fair coins are flipped. List the sample space and find P(exactly one head).
Sample space: HH, HT, TH, TT (4 equally likely)
P(exactly one head) = 2/4 = 1/2
M1 for the four outcomes, A1 for 1/2. HT and TH are DIFFERENT outcomes — merging them is the classic error.
Two dice: find P(product is even).
P(odd product) requires both odd: 3 × 3 = 9 of 36 → P(odd) = 1/4
P(even) = 1 − 1/4 = 3/4
M1 for the complement, A1 for 3/4. 'At least one even' questions often fall to 1 − P(both odd).
Exam wording to expect: "list all the possible outcomes", "find the probability that …", or a grid to complete.
❓ Practice Questions
Q1: A spinner has 3 colours (R, G, B) and a coin is flipped. List all possible outcomes.
Q2: Two dice are rolled. Using the grid, find P(sum = 5).
Q3: A coin is flipped 3 times. Find P(exactly 2 heads).
Q4: A 4-sided spinner and 6-sided dice are used. How many possible outcomes?
Q5: Two dice are rolled. Find P(both show the same number).
Q6: Two fair coins and a die. How many outcomes, and P(all heads and a 6)?
Q7: Two dice: P(sum is a multiple of 4)?
✅ Answers
- 6 outcomes: RH, RT, GH, GT, BH, BT
- Sum = 5: (1,4), (2,3), (3,2), (4,1) = 4 outcomes. P = 4/36 = 1/9
- Exactly 2 heads: HHT, HTH, THH = 3 outcomes. P = 3/8
- 4 × 6 = 24 outcomes
- Same number: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6 outcomes. P = 6/36 = 1/6
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. 2 × 3 = 6 outcomes: RH, RT, GH, GT, BH, BT
Q2. (1,4), (2,3), (3,2), (4,1) → 4/36 = 1/9
Q3. HHT, HTH, THH → 3/8
Q4. 4 × 6 = 24 outcomes
Q5. 6 of 36 → 1/6
Q6. 2 × 2 × 6 = 24 outcomes; P = 1/24
Q7. Sums 4, 8, 12: (1,3),(2,2),(3,1); (2,6),(3,5),(4,4),(5,3),(6,2); (6,6) → 9 outcomes → 1/4
📌 Key Facts to Memorise
- Sample space = every possible outcome; count it for the denominator
- Grids: two dice = 6 × 6 = 36 equally likely cells
- Stages multiply: coins × dice → 2 × 6 = 12
- Order matters in lists: HT ≠ TH
- Complements: P(at least one…) = 1 − P(none)
🗺️ In the Exam
Possibility spaces: 3–5 marks on Paper 1. Drawing the grid earns marks even before probabilities are read off — always complete it systematically.
Links to: listing (N5), tree diagrams (P8), probability scale (P3).
🧭 Board Focus — OCR
Spec J560. Three papers of 100 marks each: Paper 1 non-calculator, Papers 2 and 3 calculator, at Foundation/Higher tiers. OCR marks problem-solving explicitly, and its papers tend to front-load multi-step worded contexts — practise extracting the mathematics from prose before calculating.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Draw the grid for two dice or two spinners - it makes counting easy
- Remember: total outcomes = multiply the outcomes for each event
- Always check your grid has the correct number of cells
- For "at least" questions, it's often easier to find P(none) and subtract from 1
- List systematically to avoid missing outcomes
🧠 Problem-Solving Strategies
Problem-Solving
For possibility space problems: (1) Draw the grid carefully with all row/column labels, (2) For "at least" questions, find P(none) and use 1 − P(none), (3) Count favourable outcomes in the grid carefully, (4) Total outcomes = rows × columns, so always verify your grid has the right number of cells, (5) When listing, be systematic — fix one variable first.
Multi-Step Problem
Two fair dice are rolled. (a) Draw a possibility space grid for the sum. (b) Find P(sum is prime). (c) Find P(sum > 9 or sum is even).
Solution: (b) Prime sums: 2(1 way), 3(2), 5(4), 7(6), 11(2) = 15 ways. P = 15/36 = 5/12. (c) Sum > 9: {10(3), 11(2), 12(1)} = 6 ways. Even sum: 18 ways. Even AND >9: 10(3) + 12(1) = 4 ways. P(sum>9 or even) = (6 + 18 − 4)/36 = 20/36 = 5/9.
⚠️ Common Errors
Watch Out!
1. Wrong: Counting (2,3) and (3,2) as the same outcome in a dice grid Correct: These are different outcomes — (2,3) means first die shows 2 and second shows 3. They must be counted separately
2. Wrong: Forgetting that "at least one" means one or more — calculating P(exactly one) instead Correct: P(at least one head) = 1 − P(no heads) — use the complement shortcut
3. Wrong: Miscounting favourable outcomes by not reading the grid carefully Correct: Systematically scan each row/column and tick off favourable outcomes to avoid missing any
✍️ 6-Mark Exam Question
Extended Answer
6 marks: Two fair dice are rolled. (a) Use a possibility space to find P(sum = 7). (b) Find P(sum ≤ 4). (c) A game costs £2 to play. You win £5 if the sum is 7, and £3 if the sum is ≤ 4. Calculate the expected profit/loss per game and state whether the game is fair.
(a) Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes. P(sum=7) = 6/36 = 1/6.
(b) Sum ≤ 4: sum=2(1), sum=3(2), sum=4(3) → 6 outcomes. P(sum≤4) = 6/36 = 1/6.
(c) Expected winnings = (1/6)×£5 + (1/6)×£3 = £5/6 + £3/6 = £8/6 ≈ £1.33. Cost = £2. Expected loss = £2 − £1.33 = £0.67 per game. Not fair — you lose on average.
Mark scheme: M1 for listing sum=7 outcomes, A1 for 1/6, M1 for listing sum≤4 outcomes, A1 for 1/6, M1 for expected value calculation, A1 for conclusion with £0.67 loss
📊 AO3: Reason & Interpret
Reasoning and Interpretation
Two spinners are spun. Spinner A has {1, 2, 3} and Spinner B has {1, 2, 3, 4}. The score is the product of the two numbers.
(a) How many outcomes are in the possibility space?
(b) Find P(score is greater than 6).
(c) Would adding the number 5 to Spinner B make P(score > 6) higher or lower? Justify your reasoning.
Answers: (a) 3 × 4 = 12 outcomes. (b) Products > 6: 2×4=8, 3×3=9, 3×4=12 → 3 outcomes. P = 3/12 = 1/4. (c) New space = 3 × 5 = 15. Products > 6: 2×4=8, 2×5=10, 3×3=9, 3×4=12, 3×5=15 → 5 outcomes. P = 5/15 = 1/3. Higher — adding 5 creates more large products.
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