2018 H2 Maths Paper 2 Q8 — Worked Solution

Part (i)

4 marks

Approach

The bag has $n + 5$ balls in total: $2$ numbered $3$ , $3$ numbered $4$ , and $n$ numbered $5$ . Drawing $2$ balls without replacement, the possible values of $S$ come from pairing these three labels: $3 + 3 = 6$ , $3 + 4 = 7$ , $3 + 5 = 8$ or $4 + 4 = 8$ , $4 + 5 = 9$ , $5 + 5 = 10$ . We count the number of ways to get each sum and divide by the total number of ways to choose $2$ balls from $n + 5$ .

The total number of ways to choose $2$ balls from the bag is

\binom{n + 5}{2} = \frac{(n + 5)(n + 4)}{2}.

For each value of $S$ :

$S = 6$ : both balls numbered $3$ , giving $\binom{2}{2} = 1$ way.
$S = 7$ : one numbered $3$ and one numbered $4$ , giving $2 \times 3 = 6$ ways.
$S = 8$ : either both numbered $4$ ( $\binom{3}{2} = 3$ ways) or one numbered $3$ and one numbered $5$ ( $2n$ ways), giving $2n + 3$ ways.
$S = 9$ : one numbered $4$ and one numbered $5$ , giving $3n$ ways.
$S = 10$ : both numbered $5$ , giving $\binom{n}{2} = \frac{n(n - 1)}{2}$ ways.

Dividing each by $\binom{n + 5}{2}$ and simplifying to a common denominator $(n + 5)(n + 4)$ ,

$s$	$6$	$7$	$8$	$9$	$10$
$P(S = s)$	$\frac{2}{(n + 4)(n + 5)}$	$\frac{12}{(n + 4)(n + 5)}$	$\frac{4n + 6}{(n + 4)(n + 5)}$	$\frac{6n}{(n + 4)(n + 5)}$	$\frac{n(n - 1)}{(n + 4)(n + 5)}$

To check your answer: summing the five numerators gives

2 + 12 + (4n + 6) + 6n + n(n - 1) = n^{2} + 9n + 20 = (n + 4)(n + 5)

, matching the common denominator exactly, so the probabilities sum to

1

for every

n

. ✓

Part (ii)

1 mark

With $n = 1$ there are $6$ balls in the bag, with only one ball numbered $5$ . Substituting $n = 1$ into the formula for $P(S = 10)$ from (i),

P(S = 10) = \frac{1(1 - 1)}{(1 + 4)(1 + 5)} = \frac{0}{30} = 0.

Explanation: $S = 10$ can only arise from drawing two balls both numbered $5$ . Since there is only one ball numbered $5$ in the bag when $n = 1$ , it is impossible to draw two of them, so this outcome cannot occur.

$\therefore P(S = 10) = 0$ , since the bag contains only one $5$ -numbered ball when $n = 1$ , so two $5$ s cannot both be drawn.

Part (iii)

6 marks

Approach

We compute $E(S)$ and $E\left( S^{2} \right)$ directly from the distribution in (i), then use $Var(S) = E\left( S^{2} \right) - \left\lbrack E(S) \right\rbrack^{2}$ and combine everything over the common denominator $(n + 4)(n + 5)$ (squared, where needed) to match the required form.

Using the distribution from (i),

E(S) = \sum s\, P(S = s)

= \frac{6(2) + 7(12) + 8(4n + 6) + 9(6n) + 10\, n(n - 1)}{(n + 4)(n + 5)}

= \frac{10n^{2} + 76n + 144}{(n + 4)(n + 5)}

The numerator has $(n + 4)$ as a factor:

10n^{2} + 76n + 144 = (n + 4)(10n + 36)

E(S) = \frac{(n + 4)(10n + 36)}{(n + 4)(n + 5)} = \frac{10n + 36}{n + 5}.\ \ \left( \text{shown} \right)

For the variance, we first find $E\left( S^{2} \right)$ :

E\left( S^{2} \right) = \sum s^{2}\, P(S = s)

= \frac{36(2) + 49(12) + 64(4n + 6) + 81(6n) + 100\, n(n - 1)}{(n + 4)(n + 5)}

= \frac{100n^{2} + 642n + 1044}{(n + 4)(n + 5)}

Var(S) = E\left( S^{2} \right) - \left\lbrack E(S) \right\rbrack^{2}

= \frac{100n^{2} + 642n + 1044}{(n + 4)(n + 5)} - \left( \frac{10n + 36}{n + 5} \right)^{2}

Writing both terms over the common denominator $(n + 4)(n + 5)^{2}$ ,

Var(S) = \frac{\left( 100n^{2} + 642n + 1044 \right)(n + 5) - (10n + 36)^{2}(n + 4)}{(n + 4)(n + 5)^{2}}

By GC, the numerator expands and simplifies to

\left( 100n^{2} + 642n + 1044 \right)(n + 5) - (10n + 36)^{2}(n + 4) = 22n^{2} + 78n + 36 = 2(n + 3)(11n + 6)

Var(S) = \frac{22n^{2} + 78n + 36}{(n + 5)^{2}(n + 4)}.\ \ \left( \text{shown} \right)

To check your answer: at

n = 1

(the case in part (ii)),

E(S) = \frac{46}{6} = \frac{23}{3}

and

Var(S) = \frac{136}{180} = \frac{34}{45}

; computing these directly from the six-ball distribution in (ii) by brute force gives the same values. ✓

$\therefore E(S) = \frac{10n + 36}{n + 5}$ and $Var(S) = \frac{g(n)}{(n + 5)^{2}(n + 4)}$ , where $g(n) = 22n^{2} + 78n + 36$ .

2018 H2 Maths Paper 2, Question 8

Part (i)

Part (ii)

Part (iii)

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