P2: Fairness & Randomness
Understand that the probabilities of all possible outcomes sum to 1; randomness, fairness, equally likely events
📋 Key Concepts
Key Rule: The probabilities of all possible outcomes of an experiment always add up to 1. This is because one of the outcomes must happen.
Key Terms
| Term | Definition |
|---|---|
| Random | An outcome that cannot be predicted with certainty before it happens |
| Fair | All outcomes are equally likely; no bias |
| Equally Likely | Each outcome has the same probability |
| Bias | A systematic favouring of certain outcomes |
Probability Sum Rule:
P(Outcome₁) + P(Outcome₂) + ... + P(Outcomeₙ) = 1
P(Outcome₁) + P(Outcome₂) + ... + P(Outcomeₙ) = 1
📝 Fair Experiments
Fair dice/coin/spinner: Each outcome has equal probability. A fair dice has P(each number) = 1/6. A fair coin has P(Heads) = P(Tails) = 1/2.
Example 1
A fair 6-sided dice is rolled. What is the probability of each outcome?
Solution:
There are 6 equally likely outcomes: 1, 2, 3, 4, 5, 6
P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6
Check: 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6/6 = 1 ✓
Example 2
A fair coin is flipped. Prove the probabilities sum to 1.
Solution:
P(Heads) = 1/2 and P(Tails) = 1/2
P(Heads) + P(Tails) = 1/2 + 1/2 = 1 ✓
📝 Biased Experiments
Bias: When outcomes are NOT equally likely. Some results are more likely than others. The probabilities still sum to 1.
Example 3
A biased dice has P(6) = 0.3. All other outcomes are equally likely. Find the probability of rolling a 1.
Solution:
P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1
Let P(1) = P(2) = P(3) = P(4) = P(5) = x
5x + 0.3 = 1
5x = 0.7
x = 0.14
P(1) = 0.14
📝 Using Probability Sum to 1
Method: If you know some probabilities, subtract from 1 to find the remaining probability.
Example 4
A bag contains red, blue and green counters. P(red) = 0.4 and P(blue) = 0.35. Find P(green).
Solution:
P(red) + P(blue) + P(green) = 1
0.4 + 0.35 + P(green) = 1
0.75 + P(green) = 1
P(green) = 1 - 0.75 = 0.25
Example 5
A spinner has 5 sections. P(A) = 2/5, P(B) = 1/10, P(C) = 3/10. Find P(D).
Solution:
P(D) = 1 - (2/5 + 1/10 + 3/10)
= 1 - (4/10 + 1/10 + 3/10)
= 1 - 8/10
= 2/10 = 1/5
📝 Randomness
Random: Each trial is unpredictable, but over many trials patterns emerge that match theoretical probability.
Example 6
Is rolling a dice truly random? Explain your answer.
Answer: A fair dice roll is considered random because:
- We cannot predict the exact outcome before rolling
- Each number is equally likely (assuming a fair dice)
- Previous rolls don't affect future rolls (independent events)
🧮 Worked Examples — Exam Style
Fair = every outcome equally likely. Biased = some outcomes more likely. Probabilities of ALL outcomes must sum to 1 — the missing-probability question follows.
A bag has counters. P(red) = 0.4, P(blue) = 0.4, P(green) = 0.1. Find P(yellow).
1 − (0.4 + 0.4 + 0.1) = 1 − 0.9 = 0.1
B1 for 0.1. Sum-to-1 works for ANY number of exhaustive outcomes.
Explain why rolling a six-sided die 12 times and getting no six does not prove the die is biased.
Randomness produces runs: P(no six in 12 rolls) = (5/6)¹² ≈ 0.112 — happens about 11% of the time by chance. Short trials are unreliable evidence of bias.
M1 for acknowledging chance variation, A1 for the reasoning. Sample SIZE matters — expect the AO3 'is this evidence sufficient?' style.
Exam wording to expect: "Is the game fair?", "Find the missing probability", "explain whether the dice is biased".
❓ Practice Questions
Q1: A fair 8-sided dice is rolled. What is P(5)?
Q2: A bag has red, blue and yellow balls. P(red) = 0.5 and P(blue) = 0.3. Find P(yellow).
Q3: A biased coin has P(Heads) = 0.7. What is P(Tails)?
Q4: A spinner has 4 sections: A, B, C, D. P(A) = 0.25, P(B) = 0.25, P(C) = 0.25. Is this a fair spinner?
Q5: A biased dice has P(6) = 0.4. P(1) = P(2) = P(3) = P(4) = P(5). Find P(1).
Q6: A spinner has P(red) = 0.35, P(blue) = 0.25, rest green. Find P(green), and is it fair for 4 equal colours?
Q7: A biased coin shows heads 45 times in 100 flips. Estimate P(tails) from this data.
✅ Answers
- 1/8 (each of 8 outcomes equally likely)
- 1 - (0.5 + 0.3) = 1 - 0.8 = 0.2
- 1 - 0.7 = 0.3
- Yes - P(D) = 1 - 0.75 = 0.25, all probabilities are equal
- 5x + 0.4 = 1, so 5x = 0.6, x = 0.12
📋 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. 1/8 — each face equally likely
Q2. 1 − 0.8 = 0.2
Q3. 1 − 0.7 = 0.3
Q4. P(D) = 0.25 — all four equal → yes, fair
Q5. 5x + 0.4 = 1 → x = 0.12
Q6. P(green) = 1 − 0.6 = 0.4. For 4 equal colours each would need 0.25 — red and blue differ, so not fair
Q7. Relative frequency of tails = 55/100 = 0.55
📌 Key Facts to Memorise
- Fair: all outcomes equally likely (die: 1/6 each)
- Biased: unequal probabilities — found experimentally
- Exhaustive outcomes sum to 1 → missing probability = 1 − the rest
- Experimental estimates need LARGE samples to be trustworthy
- A single unusual run is not proof of bias — compare with expected variation
🗺️ In the Exam
Fairness/sum-to-1: 2–3 marks per paper. The AO3 evaluation ("is this evidence enough?") separates grade 4 from 5 — always mention sample size.
Links to: experimental probability (P5), probability scale (P3), trees (P8).
🧭 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
- Always check probabilities sum to 1 after calculating
- "Equally likely" means all probabilities are the same
- Fair = equally likely; biased = not equally likely
- If asked if something is fair, compare all probabilities
- Use algebra (like x) for unknown equal probabilities
🧠 Problem-Solving Strategies
Problem-Solving
When probabilities must sum to 1: (1) List all possible outcomes first, (2) Assign known probabilities, (3) Let x = unknown equal probabilities, (4) Set up equation: sum of all probabilities = 1, (5) Solve for x and check your answer is between 0 and 1.
Multi-Step Problem
A biased 5-section spinner has P(A) = 0.35 and P(B) = 0.35. P(C), P(D) and P(E) are all equal. Find P(C) and state whether the spinner is fair.
Solution: Let P(C) = P(D) = P(E) = x. Then 0.35 + 0.35 + 3x = 1, so 3x = 0.3, giving x = 0.1. Since P(C) = 0.1 ≠ P(A) = 0.35, the outcomes are not equally likely, so the spinner is not fair.
⚠️ Common Errors
Watch Out!
1. Wrong: Saying P(A') = P(A) when P(A) is not 0.5 Correct: P(A') = 1 − P(A) — the complement is found by subtracting from 1
2. Wrong: Assuming a spinner is fair just because it has equal-sized sections Correct: Fair means equally likely outcomes — you must verify probabilities, not just appearance
3. Wrong: Forgetting to check that all probabilities sum to 1 after calculating Correct: Always verify P(all outcomes) = 1 as a check — if not, recalculate
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A biased 4-sided spinner labelled A, B, C, D is spun. P(A) = 2 × P(B). P(C) = P(D) = 0.15. P(A) is twice the probability of B. Find P(A) and P(B), and explain whether the spinner is biased.
Let P(B) = x, then P(A) = 2x.
P(A) + P(B) + P(C) + P(D) = 1
2x + x + 0.15 + 0.15 = 1
3x + 0.3 = 1
3x = 0.7
x = 0.7/3 ≈ 0.233
So P(B) ≈ 0.233 and P(A) ≈ 0.467.
The spinner is biased because the probabilities are not equal (P(A) ≠ P(B) ≠ P(C) ≠ P(D)).
Mark scheme: M1 for letting P(B)=x and P(A)=2x, M1 for correct equation, M1 for solving 3x=0.7, A1 for P(B)≈0.233 and P(A)≈0.467, M1 for comparing probabilities, A1 for stating biased with reason
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A game uses a spinner with three outcomes: Win, Lose, and Draw. P(Win) = 0.2 and P(Lose) = 0.5.
(a) Find P(Draw).
(b) The game costs £1 to play. You win £3 if you win, £0 if you lose, and get your £1 back if you draw. Is the game fair? Show your working.
(c) The spinner is tested 500 times and "Draw" occurs 120 times. Does this support the claimed P(Draw)? Explain.
Answers: (a) P(Draw) = 1 − 0.2 − 0.5 = 0.3 (b) Expected value = 0.2×(3−1) + 0.5×(−1) + 0.3×0 = 0.4 − 0.5 + 0 = −0.1. Not fair — you lose 10p per game on average. (c) Experimental P(Draw) = 120/500 = 0.24. This is close to 0.3 but 500 trials is a reasonable sample — the difference may suggest the spinner is not exactly as claimed, or it could be natural variation.
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