2018 H2 Maths Paper 2 Q6 — Worked Solution

Part (i)

2 marks

Since there are $9$ endpoints A–I, every journey from S to an endpoint passes through exactly $8$ junctions, and at each junction the bug independently takes the left fork (probability $p$ ) or the right fork (probability $q$ ). Finishing at the $k$ -th endpoint from the top (A being $k = 1$ , I being $k = 9$ ) requires exactly $(k - 1)$ right forks and $(9 - k)$ left forks among these $8$ junctions — moving one letter further down the row for each extra right fork taken.

D is the $4$ th endpoint, so reaching D requires exactly $3$ right forks and $5$ left forks. By independence of the junctions, any one such route has probability $p^{5}q^{3}$ , and the number of such routes is the number of ways to choose which $3$ of the $8$ junctions are the right forks, $\binom{8}{3}$ .

P\left( \text{bug finishes at D} \right)

= \binom{8}{3}p^{5}q^{3}

= 56p^{5}q^{3}.\ \ \left( \text{shown} \right)

Part (ii)

4 marks

Let $D_{k}$ be the probability that the bug finishes at the $k$ -th endpoint from the top, so

P\left( D_{k} \right) = \binom{8}{k - 1}p^{9 - k}q^{k - 1},\quad\quad k = 1,\ldots,9.

D is $k = 4$ ; its immediate neighbours are C ( $k = 3$ ) and E ( $k = 5$ ).

Approach

Comparing $P\left( D_{k + 1} \right)$ with $P\left( D_{k} \right)$ gives the ratio $\frac{P\left( D_{k + 1} \right)}{P\left( D_{k} \right)} = \frac{9 - k}{k} \cdot \frac{q}{p}$ , which is strictly decreasing in $k$ . This means the sequence $P\left( D_{1} \right),\ldots,P\left( D_{9} \right)$ increases and then decreases (it has a single peak), so it is enough to check that $P(D)$ exceeds its two neighbours C and E — this then guarantees $P(D)$ exceeds every other endpoint’s probability as well, without checking each one individually.

Comparing D with C:

\frac{P(D)}{P(C)} = \frac{56p^{5}q^{3}}{28p^{6}q^{2}} = \frac{2q}{p}.

We need $P(D) > P(C)$ , i.e. $\frac{2q}{p} > 1$ :

2q > p

\Rightarrow 2(1 - p) > p

\Rightarrow 2 > 3p

\Rightarrow p < \frac{2}{3}.

Comparing D with E:

\frac{P(D)}{P(E)} = \frac{56p^{5}q^{3}}{70p^{4}q^{4}} = \frac{4p}{5q}.

We need $P(D) > P(E)$ , i.e. $\frac{4p}{5q} > 1$ :

4p > 5q

\Rightarrow 4p > 5(1 - p)

\Rightarrow 9p > 5

\Rightarrow p > \frac{5}{9}.

To check your answer: at

p = \frac{5}{9}

,

P(D) = P(E)

exactly, and at

p = \frac{2}{3}

,

P(D) = P(C)

exactly — the two boundary values recovered directly by setting each ratio equal to

1

. ✓

$\therefore$ the possible range of values of $p$ is $\frac{5}{9} < p < \frac{2}{3}$ .

Part (iii)

1 mark

At each of the $8$ junctions, independently of the other junctions, the bug either continues along a fork (probability $0.9p + 0.9q = 0.9(p + q) = 0.9$ ) or is swallowed by the black hole (probability $0.1$ ). The bug reaches one of the endpoints A–I if and only if it survives all $8$ junctions.

P\left( \text{reaches an endpoint} \right) = {0.9}^{8} = 0.430\ \text{(3 s.f., by GC).}

Explanation: reaching an endpoint at all does not depend on which endpoint, only on never being swallowed, so the fork probabilities $p$ and $q$ do not appear in the answer — only their combined survival probability $0.9$ at each junction matters.

$\therefore$ $P\left( \text{bug reaches an endpoint without being swallowed} \right) = 0.430$ .

2018 H2 Maths Paper 2, Question 6

Part (i)

Part (ii)

Part (iii)

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