AQA AS Level Mathematics Paper 2, June 2025: Question 18

✅ Acceptable Answers (Any ONE)

• Whether one battery is faulty is independent of any other battery being faulty.
• The probability of a battery being faulty is constant (at 3% or 0.03).
• There are only two possible outcomes for each battery: faulty or not faulty.

💡 Key Knowledge: The Four Binomial Conditions

Remember the acronym BINS:

• Binary: Two outcomes (faulty / not faulty).
• Independent: One trial does not affect another.
• Number of trials fixed: Already fixed by the question (packs of 12).
• Success probability constant: p = 0.03 remains identical throughout.

🧠 Exam Technique: Always State Context

The question strictly specifies "in context". Simply writing "trials are independent" or "constant probability" without mentioning batteries or faulty will lose the mark. Always name the object and the condition being tested!

❌ Common Errors

• Writing "Fixed number of trials": The mark scheme explicitly rejects this because the question has already set n = 12 ("Batteries are sold in packs of 12"). This is a fact of the scenario, not an assumption.
• Generic textbook quotes: Quoting "events are independent" with zero reference to batteries.

📐 Step-by-Step Calculation

Let X = number of faulty batteries in a pack of 12.

X ~ B(12, 0.03)

1. Identify parameters: n = 12, p = 0.03, r = 1.
2. Use the probability mass formula or calculator (Binomial PD):
   P(X = 1) = ¹²C₁ × (0.03)¹ × (0.97)¹¹
3. Calculate value:
   P(X = 1) = 12 × 0.03 × 0.7153... = 0.2575 (to 4 d.p.)

✅ Correct Answer

0.2575 (Accept any value in range [0.2575, 0.26])

Giving 0.258 (3 s.f.) or 0.26 (2 s.f.) also scores full marks, but it is best practice to keep 4 decimal places for subsequent calculations.

🧠 Calculator Shortcut

On the Casio ClassWiz: Go to Menu 7 (Distribution) → scroll down to Binomial PD → choose Variable. Enter x = 1 , N = 12 , p = 0.03 .

❌ Common Errors

• Using Binomial CD instead of Binomial PD, which calculates P(X ≤ 1) = 0.9515 rather than P(X = 1).
• Typing p = 0.3 instead of p = 0.03 (confusing 3% with 30%).

📐 Step-by-Step Calculation

Let Y = number of packs (out of 10) that have exactly one faulty battery.

1. Define the new distribution:
   Number of trials: n = 10 packs.
   Probability of success: p = P(pack has 1 faulty battery) = 0.2575...
   Y ~ B(10, 0.2575)
2. Express "at least two":
   P(Y ≥ 2) = 1 − P(Y ≤ 1)
   = 1 − [P(Y = 0) + P(Y = 1)]
3. Calculate cumulative value:
   Using Binomial CD with x = 1, N = 10, p = 0.2575:
   P(Y ≤ 1) = 0.22754...
4. Subtract from 1:
   P(Y ≥ 2) = 1 − 0.22754... = 0.772 (3 s.f.)

✅ Correct Answer

0.772 (Accept any value in range [0.77, 0.78])

Awarded for correctly applying the complement rule to the secondary binomial model.

❌ Common Calculation Traps

• Wrong Complement: Writing P(Y ≥ 2) = 1 − P(Y ≤ 2). Remember, because Y is discrete, the complement of {2, 3, 4, ..., 10} is {0, 1}, which is P(Y ≤ 1).
• Premature Rounding: Using p = 0.26 from part (b) instead of the unrounded value. While AWFW [0.77, 0.78] gives leeway here, premature rounding can lose accuracy marks in multi-step questions.
• Using original p = 0.03: Mistakenly setting Y ~ B(10, 0.03). The "trial" here is a pack of 12, not an individual battery!

🧠 Exam Technique: Recognising "Two-Stage" Problems

Whenever an exam question calculates a probability in part (b) and then introduces a group of items (e.g. 10 packs, 8 samples, 15 days) in part (c), alarm bells should ring! You are almost certainly setting up a second binomial distribution where the answer to (b) is your new parameter p .