2015 H2 Maths Paper 2 Q6 — Worked Solution

Part (i)

2 marks

Let $X$ be the number of red sweets in a small packet of 10 Droppers. Since each sweet is independently red with probability $0.25$ ,

X \sim B(10,\ 0.25).

We want $P(X \geq 4)$ . Converting to cumulative form,

P(X \geq 4)

\Rightarrow 1 - P(X \leq 3)

= 1 - 0.776

= 0.224\ \text{(3 s.f., by GC).}

$\therefore$ $P(X \geq 4) = 0.224$ .

Part (ii)

3 marks

Let $Y$ be the number of red sweets in a large packet of 100 Droppers, so $Y \sim B(100,\ 0.25)$ .

Approach

With $n = 100$ trials, evaluating the binomial probability term by term is impractical, so we approximate $Y$ by a normal distribution. Since $n$ is large and $np = 25 > 5$ and $nq = 75 > 5$ , the approximation is valid.

Y \sim N(25,\ 18.75)\ \text{approximately,}

since $E(Y) = np = 25$ and $Var(Y) = \text{npq} = 100(0.25)(0.75) = 18.75$ .

Note: $Y$ is discrete and we are approximating it by a continuous distribution, so a continuity correction is needed.

P(Y \geq 30)

\approx P(Y > 29.5)

Standardising, with $Z = \frac{Y - 25}{\sqrt{18.75}}$ ,

= P\left( Z > \frac{29.5 - 25}{\sqrt{18.75}} \right)

= P(Z > 1.039)

= 0.149\ \text{(3 s.f., by GC).}

To check your answer: the exact binomial value is

P(Y \geq 30) = 0.1495

(by GC), close to the normal approximation

0.1493

. ✓

$\therefore$ $P(Y \geq 30) = 0.149$ .

Part (iii)

2 marks

Let $W$ be the number of large packets, out of the 15 Yip buys, that contain at least 30 red sweets. Using the unrounded probability from part (ii), $p = P(Y \geq 30) = 0.14935$ , and since the packets are independent of each other,

W \sim B(15,\ 0.14935).

We want $P(W \leq 3)$ :

P(W \leq 3) = 0.825\ \text{(3 s.f., by GC).}

$\therefore$ $P(W \leq 3) = 0.825$ .

2015 H2 Maths Paper 2, Question 6

Part (i)

Part (ii)

Part (iii)

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