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}
❓ 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).
✅ 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
🎯 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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