2023 H2 Maths Paper 2 Q10 — Worked Solution

Part (a)(i)

2 marks

Assumption 1: Each ornament is faulty independently of whether any other ornament is faulty.

Assumption 2: Each ornament has the same, constant probability of being faulty.

To check your answer: a binomial model needs a fixed number of independent trials, each with the same probability of “success” (here, “faulty”) — the two assumptions above supply exactly independence and constant probability. ✓

Part (a)(ii)

1 mark

Let $X$ be the number of faulty ornaments in a day: $X \sim B(50,\ 0.04)$ .

E(X) = np = 50(0.04) = 2

\text{Var}(X) = np(1 - p) = 50(0.04)(0.96) = 1.92

E(X) - \text{Var}(X) = 2 - 1.92 = 0.08\ \ \text{(shown)}

To check your answer: in general

\text{mean} - \text{variance} = np - np(1 - p) = np^{2}

; here

np^{2} = 50(0.04)^{2} = 0.08

, matching directly. ✓

Part (a)(iii)

1 mark

P(X \leq 2) = 0.677\ \left( 3\text{ s.f.} \right)

Explanation: by GC, summing the $B(50,\ 0.04)$ probabilities from $X = 0$ to $X = 2$ gives $P(X \leq 2) = 0.676714\ldots \approx 0.677$ .

Part (a)(iv)

3 marks

Approach

“No more than 2 faulty ornaments” is itself a Bernoulli-type event with the probability found in (iii), repeated independently over the 5 days of the week. This calls for a second binomial distribution built on top of the first, not a fresh calculation from the ornament counts.

Let $Y$ be the number of days, out of 5, on which no more than 2 faulty ornaments are produced. Each day is independent with the same probability $P(X \leq 2) = 0.676714$ (unrounded value, from (iii)), so

Y \sim B(5,\ 0.676714)

We require $P(Y \geq 3)$ , converting to cumulative form first:

P(Y \geq 3) = 1 - P(Y \leq 2)

By GC, $P(Y \leq 2) = 0.195219$ , so

P(Y \geq 3) = 1 - 0.195219 = 0.805\ \left( 3\text{ s.f.} \right)

To check your answer: the per-day probability

0.677

is above

0.5

, and

3

out of

5

is close to the expected count

5(0.677) \approx 3.38

, so a fairly high probability is reasonable. ✓

Part (a)(v)

2 marks

Approach

the week’s total fault count is a sum of $5$ independent $B(50,\ 0.04)$ variables. Since they share the same $p$ , this sum is itself binomial with $n = 5 \times 50$ , rather than a normal approximation or a fresh distribution.

Let $T$ be the total number of faulty ornaments produced in the 5-day week. Since the $5 \times 50 = 250$ ornaments made in the week are each independently faulty with probability $0.04$ ,

T \sim B(250,\ 0.04)

P(T \leq 10) = 0.583\ \left( 3\text{ s.f.} \right)

Explanation: by GC, $P(T \leq 10) = 0.583057\ldots \approx 0.583$ .

To check your answer:

E(T) = 250(0.04) = 10

, so

P(T \leq 10)

sitting just above

0.5

is consistent with

10

being the mean of

T

. ✓

Part (b)

6 marks

Approach

we cannot compare $P\left( \text{Ming accepts} \right)$ and $P\left( \text{Lu accepts} \right)$ by evaluating them at particular values of $p$ — the question asks for all $0 < p < 1$ . The way in is to write both probabilities as functions of $p$ , subtract, and factorise the difference so that its sign is visible for every $p$ in the interval.

Let $p = P\left( \text{pen is not faulty} \right)$ , so a pen is faulty with probability $1 - p$ , where $0 < p < 1$ .

Mr Lu’s method: Let $X_{L}$ be the number of faulty pens in his sample of 6, so $X_{L} \sim B(6,\ 1 - p)$ . He accepts the box if $X_{L} = 0$ or $X_{L} = 1$ .

P\left( \text{Lu accepts} \right)

= P\left( X_{L} = 0 \right) + P\left( X_{L} = 1 \right)

= p^{6} + \binom{6}{1}(1 - p)p^{5}

= p^{6} + 6(1 - p)p^{5}

= p^{5}\left\lbrack p + 6(1 - p) \right\rbrack

= p^{5}(6 - 5p)

Mrs Ming’s method: Let $X_{1}$ be the number of faulty pens in her first sample of 3, so $X_{1} \sim B(3,\ 1 - p)$ .

If $X_{1} = 0$ (probability $p^{3}$ ), she accepts immediately. If $X_{1} = 1$ (probability $\binom{3}{1}(1 - p)p^{2} = 3(1 - p)p^{2}$ ), she takes a second sample of 3, accepting only if that sample has 0 faulty pens (probability $p^{3}$ ). If $X_{1} = 2$ or $3$ , she rejects.

P\left( \text{Ming accepts} \right)

= p^{3} + \left\lbrack 3(1 - p)p^{2} \right\rbrack \times p^{3}

= p^{3} + 3(1 - p)p^{5}

To check your answer: at

p = 0.9

,

P\left( \text{Lu accepts} \right) \approx 0.886

and

P\left( \text{Ming accepts} \right) \approx 0.906

, so Mrs Ming does accept a greater proportion at this value. ✓

To compare the two methods for every $p$ , we consider the difference $P\left( \text{Ming accepts} \right) - P\left( \text{Lu accepts} \right)$ .

P\left( \text{Ming accepts} \right) - P\left( \text{Lu accepts} \right)

= p^{3} + 3(1 - p)p^{5} - p^{5}(6 - 5p)

= p^{3}\left\lbrack 1 + 3(1 - p)p^{2} - p^{2}(6 - 5p) \right\rbrack

= p^{3}\left\lbrack 1 + 3p^{2} - 3p^{3} - 6p^{2} + 5p^{3} \right\rbrack

= p^{3}\left( 2p^{3} - 3p^{2} + 1 \right)

Factorising $2p^{3} - 3p^{2} + 1$ : substituting $p = 1$ gives $2 - 3 + 1 = 0$ , so $(p - 1)$ is a factor.

2p^{3} - 3p^{2} + 1 = (p - 1)\left( 2p^{2} - p - 1 \right)

The quadratic factor $2p^{2} - p - 1$ has discriminant $1 + 8 = 9$ , giving roots $\frac{1 \pm 3}{4} = 1$ or $- \frac{1}{2}$ , so

2p^{2} - p - 1 = (p - 1)(2p + 1)

To check your answer: expanding

(p - 1)(2p + 1) = 2p^{2} + p - 2p - 1 = 2p^{2} - p - 1

, matching. ✓

Combining the two factorisations,

2p^{3} - 3p^{2} + 1 = (p - 1)(p - 1)(2p + 1) = (p - 1)^{2}(2p + 1)

so

P\left( \text{Ming accepts} \right) - P\left( \text{Lu accepts} \right) = p^{3}(p - 1)^{2}(2p + 1)

For $0 < p < 1$ : $p^{3} > 0$ since $p > 0$ ; $(p - 1)^{2} > 0$ since $p \neq 1$ on this open interval; and $2p + 1 > 0$ since $p > 0$ . Hence the product $p^{3}(p - 1)^{2}(2p + 1) > 0$ for all $0 < p < 1$ .

To check your answer: at

p = 0.5

,

p^{3}(p - 1)^{2}(2p + 1) = (0.125)(0.25)(2) = 0.0625

, matching the direct computation

0.171875 - 0.109375 = 0.0625

. ✓

$\therefore$ $P\left( \text{Ming accepts} \right) - P\left( \text{Lu accepts} \right) > 0$ for all $0 < p < 1$ , so Mrs Ming always accepts a strictly greater proportion of boxes than Mr Lu does, regardless of the value of $p$ .

2023 H2 Maths Paper 2, Question 10

Part (a)(i)

Part (a)(ii)

Part (a)(iii)

Part (a)(iv)

Part (a)(v)

Part (b)

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