2020 H2 Maths Paper 2 Q9 — Worked Solution

Part (i)

1 mark

A random sample in this context means that each of the $10$ pens tested is selected from the box of $100$ pens in such a way that every pen has an equal chance of being chosen, and the pens are chosen independently of one another (so that whether one pen is faulty does not affect the chance that another is chosen).

Part (ii)

1 mark

Let $X$ be the number of faulty pens in a random sample of $10$ pens from a box. Since each pen is independently faulty with probability $0.06$ ,

X \sim B(10,\ 0.06).

The box is accepted if $2$ or fewer faulty pens are found, so

P\left( \text{box accepted} \right)

= P(X \leq 2)

= 0.981\ \ \text{(by GC).}

$\therefore$ $P\left( \text{box accepted} \right) = 0.981$ .

Part (iii)

4 marks

Approach

“More than $5\%$ of these boxes are rejected” is a statement about the number of boxes out of $75$ , so we need a second binomial variable, $Y$ , counting rejected boxes among the $75$ , with success probability $P\left( \text{reject} \right) = 1 - P\left( \text{accept} \right)$ found in part (ii).

From part (ii), the probability that a box is rejected is

P\left( \text{reject} \right) = 1 - 0.9811621... = 0.0188378...\ \ \text{(by GC, unrounded value kept).}

Let $Y$ be the number of rejected boxes among the $75$ boxes tested. Since the boxes are tested independently,

Y \sim B(75,\ 0.0188378...).

$5\%$ of $75$ boxes is $3.75$ boxes, so “more than $5\%$ of these boxes are rejected” means $Y > 3.75$ , i.e. $Y \geq 4$ . Converting the tail to cumulative form,

P(Y \geq 4)

\Rightarrow 1 - P(Y \leq 3)

= 1 - 0.9465

= 0.0535\ \ \text{(by GC).}

$\therefore$ $P\left( \text{more than }5\%\text{ of boxes rejected} \right) = 0.0535$ .

Part (iv)

5 marks

Approach

The alternative procedure has three outcomes after the first sample of $5$ (accept outright, reject outright, or take a second sample), so $P\left( \text{accept} \right)$ is built up as a sum over these cases rather than as a single binomial tail.

Let $W$ be the number of faulty pens in the first sample of $5$ pens, so $W \sim B(5,\ 0.06)$ . The box is accepted immediately if $W = 0$ . If $W = 1$ or $W = 2$ , a second sample of $5$ pens is taken; let $V$ be the number of faulty pens in this second sample, so $V \sim B(5,\ 0.06)$ independently of $W$ , and the box is accepted if $W + V \leq 2$ .

If $W = 1$ , the box is accepted when $V \leq 1$ .
If $W = 2$ , the box is accepted when $V \leq 0$ .

By GC,

P(W = 0) = 0.733904,\ \ P(W = 1) = 0.234225,\ \ P(W = 2) = 0.0299010,

P(V \leq 1) = 0.968129,\ \ P(V \leq 0) = 0.733904.

Therefore

P\left( \text{box accepted} \right)

= P(W = 0) + P(W = 1)P(V \leq 1) + P(W = 2)P(V \leq 0)

= 0.733904 + (0.234225)(0.968129) + (0.0299010)(0.733904)

= 0.983\ \ \text{(by GC).}

$\therefore$ $P\left( \text{box accepted under alternative procedure} \right) = 0.983$ .

Part (v)

1 mark

Explanation: Under the alternative procedure a second sample of $5$ pens is only needed when the first sample gives $1$ or $2$ faulty pens; otherwise testing stops after just $5$ pens. The expected number of pens tested per box is therefore

5 + 5 \times P\left( W = 1\text{ or }W = 2 \right) = 5 + 5(0.264126) \approx 6.32,

which is fewer than the $10$ pens always tested under the original procedure, while the probability of accepting a good box ( $0.983$ ) is about the same as before ( $0.981$ ). The factory manager might prefer the alternative procedure because it tests fewer pens on average, reducing the cost, time and product wasted in testing, without materially changing the chance that a box is correctly accepted.

2020 H2 Maths Paper 2, Question 9

Part (i)

Part (ii)

Part (iii)

Part (iv)

Part (v)

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