2017 H2 Maths Paper 2 Q9 — Worked Solution

Part (i)

2 marks

For the number of faulty lights in a box to be well modelled by a binomial distribution, we need, in this context:

Whether any one light is faulty is independent of whether any other light in the box is faulty.
Each light has the same constant probability, $0.08$ , of being faulty.

Part (ii)

1 mark

Let $X$ be the number of faulty lights in a box of 12. Then $X \sim B(12,\ 0.08)$ .

P(X \geq 1)

\Rightarrow 1 - P(X = 0)

= 1 - (0.92)^{12}

= 0.632\ \left( 3\ \text{s.f.} \right)\ \ \text{(by GC)}

Part (iii)

2 marks

Let $W$ be the number of boxes, out of the 20 in a carton, that contain at least one faulty light. Since the boxes are independent, each box has probability $P(X \geq 1) = 0.63233$ (5 s.f.) of containing at least one faulty light, so we require every one of the 20 boxes to satisfy this.

P\left( \text{every box has} \geq 1\ \text{faulty light} \right)

\Rightarrow \left\lbrack P(X \geq 1) \right\rbrack^{20}

= (0.63233\ldots)^{20}

= 0.000104\ \left( 3\ \text{s.f.} \right)\ \ \text{(by GC)}

Part (iv)

2 marks

Approach

A carton contains $20 \times 12 = 240$ lights in total, and each light is independently faulty with probability $0.08$ , so the total number of faulty lights in the carton is itself binomial.

Let $Y$ be the number of faulty lights in a randomly selected carton of $240$ lights. Then $Y \sim B(240,\ 0.08)$ .

P(Y \geq 20)

\Rightarrow 1 - P(Y \leq 19)

= 1 - 0.54167\ldots

= 0.458\ \left( 3\ \text{s.f.} \right)\ \ \text{(by GC)}

Part (v)

Explanation: The event in (iv) only requires the total count of faulty lights across the whole carton to reach $20$ ; it does not matter how those faulty lights are distributed among the 20 boxes — they could all be concentrated in a handful of boxes. The event in (iii) is far more restrictive: it requires faulty lights to be spread out so that every single one of the 20 boxes has at least one. Since (iii)’s event is a much stronger requirement than (iv)’s, $P\left( \text{(iv)} \right) > P\left( \text{(iii)} \right)$ .

Part (vi)

3 marks

Let $F$ be the event that a light is faulty, and $T$ be the event that the quick test identifies a light as faulty.

P(F) = 0.08,\quad\quad P(T \mid F) = 0.95,\quad\quad P(T \mid F') = 0.06

Approach

We want $P(F'\mid T)$ , the reverse of the conditional probabilities given, so we use Bayes’ theorem, finding $P(T)$ first.

P(T)

= P(F)P(T \mid F) + P(F')P(T \mid F')

= (0.08)(0.95) + (0.92)(0.06)

= 0.1312

P(F'\mid T)

= \frac{P(F')P(T \mid F')}{P(T)}

= \frac{(0.92)(0.06)}{0.1312}

= 0.421\ \left( 3\ \text{s.f.} \right)

Part (vii)

1 mark

The test correctly identifies a light either by (correctly) flagging a faulty light as faulty, or by (correctly) not flagging a non-faulty light as faulty.

P\left( \text{correct identification} \right)

= P(F)P(T \mid F) + P(F')P(T'\mid F')

= (0.08)(0.95) + (0.92)(0.94)

= 0.941\ \left( 3\ \text{s.f.} \right)

Part (viii)

1 mark

The test correctly identifies lights $94.1\%$ of the time, which sounds good, but a large fraction — $42.1\%$ — of lights the test flags as faulty are in fact not faulty, so good lights would be needlessly discarded. Whether the test is worthwhile depends on whether the cost of discarding these good lights is outweighed by the benefit of catching most of the genuinely faulty ones.

2017 H2 Maths Paper 2, Question 9

Part (i)

Part (ii)

Part (iii)

Part (iv)

Part (v)

Part (vi)

Part (vii)

Part (viii)

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