2022 H2 Maths Paper 2 Q9 — Worked Solution

Part (a)

2 marks

Approach

From the diagram, the counter makes exactly $5$ moves in total from $S$ to a top endpoint: the first move (from $S$ to $L$ or $R$ ) is equally likely either way, and each of the remaining $4$ moves independently goes left with probability $p$ or right with probability $q$ . Tracking net horizontal displacement (each left move $- 1$ , each right move $\text{+}1$ ), the six endpoints $A,\ldots,F$ correspond to the six possible totals $- 5, - 3, - 1,1,3,5$ , with $B$ (second from the left) corresponding to a total of $- 3$ . There are exactly two ways to reach a total of $- 3$ , and we find the probability of each.

Route 1: first move left, then exactly $1$ right move among the remaining $4$ (net $- 2$ over those $4$ , combined with the first move’s $- 1$ , giving $- 3$ ). There are $\binom{4}{1} = 4$ ways to place the right move among the $4$ moves:

P\left( \text{Route 1} \right) = \frac{1}{2} \times \binom{4}{1}p^{3}q = 2p^{3}q

Route 2: first move right, then all $4$ remaining moves left (net $\text{+}1 - 4 = - 3$ ):

P\left( \text{Route 2} \right) = \frac{1}{2} \times p^{4} = \frac{1}{2}p^{4}

Since these are the only two ways to reach $B$ , and they are mutually exclusive,

P\left( \text{arrives at }B \right) = 2p^{3}q + \frac{1}{2}p^{4}\ \ \left( \text{shown} \right)

To check your answer: summing

P\left( \text{arrives at each endpoint} \right)

by the same method for every one of

A,\ldots,F

gives

p^{4} + 4p^{3}q + 6p^{2}q^{2} + 4pq^{3} + q^{4} = (p + q)^{4} = 1

, since

p + q = 1

, confirming all six probabilities are correctly accounted for. ✓

$\therefore$ $P\left( \text{arrives at }B \right) = 2p^{3}q + \frac{1}{2}p^{4}$ (shown).

Part (b)

4 marks

Approach

There are $5$ distinct routes from $S$ to $B$ : the $4$ routes of “Route 1” type from part (a), each with probability $\frac{1}{2}p^{3}q$ , and the $1$ route of “Route 2” type, with probability $\frac{1}{2}p^{4}$ . Since Jon’s and Kath’s counters move independently, we sum, over each of the $5$ routes, the probability that both take that specific route, then divide by $P\left( \text{both at }B \right)$ .

P\left( \text{same route} \mid \text{both at }B \right) = \frac{P\left( \text{same route and both at }B \right)}{P\left( \text{both at }B \right)}

Since each player’s route is chosen independently, the probability both take one specific route is the square of that route’s probability:

P\left( \text{same route and both at }B \right)

= 4 \times \left( \frac{1}{2}p^{3}q \right)^{2} + \left( \frac{1}{2}p^{4} \right)^{2}

= p^{6}q^{2} + \frac{1}{4}p^{8}

Since Jon’s and Kath’s counters are independent, $P\left( \text{both at }B \right) = P(B)^{2}$ :

P\left( \text{both at }B \right) = \left( 2p^{3}q + \frac{1}{2}p^{4} \right)^{2}

= \left\lbrack \frac{1}{2}p^{3}(4q + p) \right\rbrack^{2}

= \frac{1}{4}p^{6}(4q + p)^{2}

So:

P\left( \text{same route} \mid \text{both at }B \right) = \frac{p^{6}q^{2} + \frac{1}{4}p^{8}}{\frac{1}{4}p^{6}(4q + p)^{2}}

= \frac{4q^{2} + p^{2}}{(4q + p)^{2}}

Substituting $q = 1 - p$ :

4q + p = 4(1 - p) + p = 4 - 3p

4q^{2} + p^{2} = 4(1 - p)^{2} + p^{2} = 5p^{2} - 8p + 4

To check your answer: at

p = q = \frac{1}{2}

,

\frac{4q^{2} + p^{2}}{(4q + p)^{2}} = \frac{1 + \frac{1}{4}}{\left( 2 + \frac{1}{2} \right)^{2}} = \frac{5/4}{25/4} = \frac{1}{5}

, a valid probability strictly between

0

and

1

. ✓

$\therefore$ $P\left( \text{same route} \mid \text{both at }B \right) = \frac{5p^{2} - 8p + 4}{(4 - 3p)^{2}}$ .

Part (c)

4 marks

By the same method as (a), $C$ (third from the left) corresponds to a net displacement of $- 1$ , reached either by first move left with $2$ right moves among the remaining $4$ , or first move right with $1$ right move among the remaining $4$ :

P\left( \text{arrives at }C \right) = \frac{1}{2}\binom{4}{2}p^{2}q^{2} + \frac{1}{2}\binom{4}{1}p^{3}q = 3p^{2}q^{2} + 2p^{3}q

Setting $P(B) = P(C)$ :

2p^{3}q + \frac{1}{2}p^{4} = 3p^{2}q^{2} + 2p^{3}q

\Rightarrow \frac{1}{2}p^{4} = 3p^{2}q^{2}

Dividing both sides by $p^{2}$ (since $p \neq 0$ ),

\frac{1}{2}p^{2} = 3q^{2}

\Rightarrow p^{2} = 6q^{2}

Since $p,q > 0$ , taking the positive square root,

p = \sqrt{6}\, q

Substituting into $p + q = 1$ :

\sqrt{6}\, q + q = 1

\Rightarrow q\left( \sqrt{6} + 1 \right) = 1

\Rightarrow q = \frac{1}{\sqrt{6} + 1} = \frac{\sqrt{6} - 1}{5}

p = \sqrt{6}\, q = \frac{6 - \sqrt{6}}{5}

To check your answer:

p = \frac{6 - \sqrt{6}}{5} \approx 0.7101

and

q = \frac{\sqrt{6} - 1}{5} \approx 0.2899

, and

p + q \approx 1

. ✓ Substituting back,

P(B) \approx 0.3347

and

P(C) \approx 0.3347

, which match. ✓

$\therefore$ $p = \frac{6 - \sqrt{6}}{5}$ .

2022 H2 Maths Paper 2, Question 9

Part (a)

Part (b)

Part (c)

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