AQA AS Level Mathematics Paper 2, June 2025: Question 20
4 marks · Medium difficulty · Multi-step Problem
Calculate the probabilities of stopping at all four traffic lights and at exactly one of four independent traffic lights given their individual probabilities of being red.
Practise this questionQuestion
Question text
20 An office worker has to pass through four sets of traffic lights when driving to work.
If a set of lights is red, the driver has to stop.
The probability that each set of lights is red when the office worker arrives at them is
given by
P(set 1 is red) = 0.54
P(set 2 is red) = 0.6
P(set 3 is red) = 0.7
P(set 4 is red) = 0.8
Assume that the event that a light is red is independent of whether the other lights
are red.
20 (a) Find the probability that the office worker is stopped at all four sets of lights.
[1 mark]
20 (b) Find the probability that the office worker is stopped at exactly one of the four sets
of lights.
[3 marks]
Mark scheme
Show the mark scheme
Q Marking instructions AO Marks Typical solution
20(a) Obtains AWRT 0.18 1.1b B1 P(all 4 are red)
= 0.54 × 0.6 × 0.7 × 0.8
= 0.181
Subtotal 1
20(b) Obtains at least one correct 3.1a M1
term P(exactly one set is red)
PI by correct answer
= 0.54 × 0.4 × 0.3 × 0.2
Obtains all four correct terms 1.1b A1 + 0.46 × 0.6 × 0.3 × 0.2
PI by correct answer
+ 0.46 × 0.4 × 0.7 × 0.2
Obtains AWRT 0.099 1.1b A1 + 0.46 × 0.4 × 0.3 × 0.8
= 0.01296 + 0.01656 + 0.02576 +
0.04416
= 0.0994
Subtotal 3
Question 20 Total 4
How to answer it
Independent Events: Traffic Lights Problem
What this question tests
This question assesses your ability to apply basic probability rules to compound events:
- Multiplication rule for independent events: P(A ∩ B) = P(A) × P(B) .
- Complementary events: P(not red) = 1 - P(red) .
- Combinations of mutually exclusive cases: Identifying all distinct scenarios where an outcome occurs exactly once when individual probabilities differ.
- Appropriate rounding: Giving answers to a sensible degree of accuracy (AWRT – Answers Which Round To).
💡 Key Data Given in the Scenario
Let Rᵢ be the event that light i is red, and Gᵢ be the event that light i is not red (green):
| Traffic Light | P(Red) = P(Rᵢ) | P(Not Red) = P(Gᵢ) = 1 - P(Rᵢ) |
|---|---|---|
| Set 1 | 0.54 | 1 - 0.54 = 0.46 |
| Set 2 | 0.60 | 1 - 0.60 = 0.40 |
| Set 3 | 0.70 | 1 - 0.70 = 0.30 |
| Set 4 | 0.80 | 1 - 0.80 = 0.20 |
Probability of being stopped at all four lights
Calculate P(All 4 sets are red)
📐 Step-by-Step Calculation
Because the events are stated to be independent, multiply the probabilities of each light being red directly:
- Identify probabilities: 0.54, 0.6, 0.7, 0.8
- Multiply:
P(all 4 red) = 0.54 × 0.6 × 0.7 × 0.8 - Compute the exact product:
= 0.18144 - Round to 3 significant figures:
0.181 (or 0.18)
✅ Correct Answer
0.181 or 0.18
• B1: Obtains AWRT 0.18 (e.g. 0.181, 0.18144).
❌ Common Errors
- Adding probabilities: Adding 0.54 + 0.6 + 0.7 + 0.8 instead of multiplying. The question asks for the intersection of independent events, so multiplication is required.
- Premature rounding: Rounding individual values before multiplying.
🧠 Exam Technique
Whenever you see "Assume that the event... is independent", this is your explicit cue that you can directly multiply the individual probabilities: P(A ∩ B ∩ C ∩ D) = P(A)P(B)P(C)P(D) .
Probability of being stopped at exactly one set of lights
Calculate P(Exactly 1 light is red out of 4)
📐 Step-by-Step Calculation
Because the probabilities for each set of lights are different, this is not a simple binomial distribution. You must evaluate all 4 mutually exclusive cases where exactly one light is red and the other three are green:
- Light 1 is red, others green:
R₁ ∩ G₂ ∩ G₃ ∩ G₄ = 0.54 × 0.4 × 0.3 × 0.2 = 0.01296 - Light 2 is red, others green:
G₁ ∩ R₂ ∩ G₃ ∩ G₄ = 0.46 × 0.6 × 0.3 × 0.2 = 0.01656 - Light 3 is red, others green:
G₁ ∩ G₂ ∩ R₃ ∩ G₄ = 0.46 × 0.4 × 0.7 × 0.2 = 0.02576 - Light 4 is red, others green:
G₁ ∩ G₂ ∩ G₃ ∩ R₄ = 0.46 × 0.4 × 0.3 × 0.8 = 0.04416 - Sum all four possibilities:
Total = 0.01296 + 0.01656 + 0.02576 + 0.04416
Total = 0.09944
✅ Correct Answer
0.0994 (or 0.099)
• M1 (AO 3.1a): Obtains at least one correct 4-factor product (e.g. 0.54 × 0.4 × 0.3 × 0.2).
• A1 (AO 1.1b): Obtains all four correct product terms added together.
• A1 (AO 1.1b): Final answer AWRT 0.099 (e.g. 0.0994 or 0.09944).
❌ Common Errors
- Using Binomial Formula: Trying to use ⁴C₁ × p¹ × (1 - p)³ . The binomial distribution requires the probability p to be constant, but here each light has a different probability.
- Forgetting the "Not Red" lights: Writing 0.54 + 0.6 + 0.7 + 0.8 . You must account for the other three lights simultaneously turning green!
- Omitting scenarios: Calculating only one case (e.g. only light 1 being red) and forgetting there are four separate lights that could be the one that is red.
🧠 Examiner Insights & Top Tips
- Systematic labelling: Use letters like R for Red and G for Green to list your cases: RGGG , GRGG , GGRG , GGGR . This ensures you never miss a term.
- "PI" in Mark Scheme: The mark scheme notes "PI by correct answer" (Process Implied). However, never rely on this! Write out all 4 terms clearly to secure the method marks even if you make a slight arithmetic typo in your final sum.
Topics
Statistics · M: Probability
Question and mark scheme from the AQA AS Level Mathematics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.