2020 H2 Maths Paper 2 Q5 — Worked Solution

Part (i)

1 mark

The bag has one green disc (worth $0$ ), $r$ red discs (worth $5$ each) and $2r$ blue discs (worth $2$ each). Tina’s score is the product of the two discs’ point values, so the possible pairs and their scores are: green with anything gives $0$ ; red with red gives $5 \times 5 = 25$ ; red with blue gives $5 \times 2 = 10$ ; blue with blue gives $2 \times 2 = 4$ .

$\therefore$ Tina’s possible scores are $0,\ 4,\ 10,\ 25$ .

Part (ii)

7 marks

Approach

There are $3r + 1$ discs in total, so we find the probability of each score by counting unordered pairs of discs, using combinations. Writing every probability over the common denominator $3(3r + 1)$ makes the sum $E(X)$ come out as a single fraction that matches the given expression.

Let $X$ be Tina’s score. The total number of ways to take $2$ discs from the $3r + 1$ discs is $\binom{3r + 1}{2} = \frac{(3r + 1)(3r)}{2}$ .

Counting the ways to get each score:

P(X = 25) = \frac{\binom{r}{2}}{\binom{3r + 1}{2}} = \frac{r(r - 1)}{(3r + 1)(3r)} = \frac{r - 1}{3(3r + 1)}

P(X = 10) = \frac{r \cdot 2r}{\binom{3r + 1}{2}} = \frac{2r^{2}}{\frac{1}{2}(3r + 1)(3r)} = \frac{4r}{3(3r + 1)}

P(X = 4) = \frac{\binom{2r}{2}}{\binom{3r + 1}{2}} = \frac{2r(2r - 1)}{(3r + 1)(3r)} = \frac{2(2r - 1)}{3(3r + 1)}

P(X = 0) = \frac{1 \cdot 3r}{\binom{3r + 1}{2}} = \frac{3r}{\frac{1}{2}(3r + 1)(3r)} = \frac{2}{3r + 1}

To check your answer: the four probabilities should sum to

1

. Writing

P(X = 0) = \frac{6}{3(3r + 1)}

, the numerators are

(r - 1) + 4r + 2(2r - 1) + 6 = 9r + 3 = 3(3r + 1)

, so the sum is

\frac{3(3r + 1)}{3(3r + 1)} = 1

. ✓

Then

E(X)

= 0 \cdot P(X = 0) + 4 \cdot P(X = 4) + 10 \cdot P(X = 10) + 25 \cdot P(X = 25)

= \frac{4 \cdot 2(2r - 1) + 10 \cdot 4r + 25(r - 1)}{3(3r + 1)}

= \frac{(16r - 8) + 40r + (25r - 25)}{3(3r + 1)}

= \frac{81r - 33}{3(3r + 1)}

= \frac{27r - 11}{3r + 1}\ \ \text{(shown)}

For the variance we also need $E\left( X^{2} \right)$ :

E\left( X^{2} \right)

= 16 \cdot P(X = 4) + 100 \cdot P(X = 10) + 625 \cdot P(X = 25)

= \frac{16 \cdot 2(2r - 1) + 100 \cdot 4r + 625(r - 1)}{3(3r + 1)}

= \frac{(64r - 32) + 400r + (625r - 625)}{3(3r + 1)}

= \frac{1089r - 657}{3(3r + 1)}

= \frac{3(121r - 73)}{3r + 1}

Then $Var(X) = E\left( X^{2} \right) - \left\lbrack E(X) \right\rbrack^{2}$ :

Var(X) = \frac{3(121r - 73)}{3r + 1} - \left( \frac{27r - 11}{3r + 1} \right)^{2}

= \frac{3(121r - 73)(3r + 1) - (27r - 11)^{2}}{(3r + 1)^{2}}

= \frac{\left( 1089r^{2} - 294r - 219 \right) - \left( 729r^{2} - 594r + 121 \right)}{(3r + 1)^{2}}

= \frac{360r^{2} + 300r - 340}{(3r + 1)^{2}}

\therefore\ Var(X) = \frac{20\left( 18r^{2} + 15r - 17 \right)}{(3r + 1)^{2}}

Part (iii)

2 marks

Approach

setting the variance expression from (ii) equal to $38$ gives a quadratic in $r$ ; since $r > 1$ , only one root can be accepted.

\frac{20\left( 18r^{2} + 15r - 17 \right)}{(3r + 1)^{2}} = 38

360r^{2} + 300r - 340 = 38(3r + 1)^{2}

360r^{2} + 300r - 340 = 342r^{2} + 228r + 38

18r^{2} + 72r - 378 = 0

r^{2} + 4r - 21 = 0

(r + 7)(r - 3) = 0

r = - 7\ \left( \text{rej. as}\ r > 1 \right),\ \ r = 3

To check your answer: with

r = 3

there are

1 + 3 + 6 = 10

discs; enumerating all

\binom{10}{2} = 45

equally likely pairs of discs directly gives

E(X) = 7

(matching

\frac{27(3) - 11}{3(3) + 1} = \frac{70}{10} = 7

) and

Var(X) = 38

. ✓

$\therefore r = 3$ .

2020 H2 Maths Paper 2, Question 5

Part (i)

Part (ii)

Part (iii)

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