2023 H2 Maths Paper 2 Q6 — Worked Solution

Part (a)

8 marks

Let $X$ be the number of red counters among Mei’s $12$ counters. Since the counters are drawn without replacement from $r$ red and $b$ blue counters ( $r + b$ in total), $X$ follows a hypergeometric distribution:

P(X = k) = \frac{\binom{r}{k}\binom{b}{12 - k}}{\binom{r + b}{12}}

Using $P(X = 4) = P(X = 3)$ ,

\binom{r}{4}\binom{b}{8} = \binom{r}{3}\binom{b}{9}

\Rightarrow \frac{\binom{r}{4}}{\binom{r}{3}} = \frac{\binom{b}{9}}{\binom{b}{8}}

Using $\binom{r}{4} = \binom{r}{3} \cdot \frac{r - 3}{4}$ and $\binom{b}{9} = \binom{b}{8} \cdot \frac{b - 8}{9}$ ,

\frac{r - 3}{4} = \frac{b - 8}{9}

\Rightarrow 9(r - 3) = 4(b - 8)

\Rightarrow 9r - 27 = 4b - 32

\Rightarrow 9r + 5 = 4b

$\therefore\ 9r + 5 = 4b$ (shown)

Part (b)

Approach

We relate $P(X = 3)$ to $P(X = 2)$ in exactly the same way as in part (a), giving a second linear equation in $r$ and $b$ . Solving this simultaneously with $9r + 5 = 4b$ then fixes $r$ and $b$ , from which $P(X = 1)$ follows directly.

Using $P(X = 3) = \frac{5}{3}P(X = 2)$ ,

\binom{r}{3}\binom{b}{9} = \frac{5}{3}\binom{r}{2}\binom{b}{10}

Using $\binom{r}{3} = \binom{r}{2} \cdot \frac{r - 2}{3}$ and $\binom{b}{10} = \binom{b}{9} \cdot \frac{b - 9}{10}$ , so that $\frac{\binom{b}{9}}{\binom{b}{10}} = \frac{10}{b - 9}$ ,

\frac{r - 2}{3} \times \frac{10}{b - 9} = \frac{5}{3}

\Rightarrow 10(r - 2) = 5(b - 9)

\Rightarrow 2(r - 2) = b - 9

\Rightarrow b = 2r + 5

Solving simultaneously with $9r + 5 = 4b$ from part (a):

9r + 5 = 4(2r + 5)

\Rightarrow 9r + 5 = 8r + 20

\Rightarrow r = 15

b = 2(15) + 5

= 35

To check your answer: with

r = 15

b = 35

(

50

counters in total), direct computation gives

P(X = 4) = P(X = 3) \approx 0.265

and

P(X = 3)/P(X = 2) = 5/3

exactly ✓

Both values satisfy $r > 12$ and $b > 12$ , as required by the question.

We now find the probability that exactly $1$ of the $12$ counters removed is red:

P(X = 1) = \frac{\binom{15}{1}\binom{35}{11}}{\binom{50}{12}}

= \frac{687\, 735}{13\, 340\, 621}

\approx 0.0516\ \left( 3\text{ s.f.} \right)

$\therefore\ 2r + 5 = b$ , and $r = 15$ , $b = 35$ ; $P(X = 1) = \frac{687\, 735}{13\, 340\, 621} \approx 0.0516$ .

2023 H2 Maths Paper 2, Question 6

Part (a)

Part (b)

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