2020 H2 Maths Paper 2 Q8 — Worked Solution

Part (i)

2 marks

Approach

A direct comparison of $P(R = 1)$ and $P(R = 2)$ as decimals hides the structure. Since both probabilities share the same denominator $\binom{28}{12}$ , it is quicker to form the ratio $P(R = 2) \div P(R = 1)$ and show it exceeds $1$ .

Let $R$ be the number of rectangles chosen. Since $12$ shapes are chosen from $11$ circles and $17$ rectangles, $R$ has a hypergeometric distribution, and

P(R = 1) = \frac{\binom{17}{1}\binom{11}{11}}{\binom{28}{12}},\ \ \ P(R = 2) = \frac{\binom{17}{2}\binom{11}{10}}{\binom{28}{12}}.

The two probabilities have the same denominator, so we compare numerators. Simplifying:

\frac{P(R = 2)}{P(R = 1)}

= \frac{\binom{17}{2}\binom{11}{10}}{\binom{17}{1}\binom{11}{11}}

= \frac{\binom{17}{2}}{\binom{17}{1}} \times \frac{\binom{11}{10}}{\binom{11}{11}}

= \frac{136}{17} \times \frac{11}{1}

= 8 \times 11

= 88.

Since $\frac{P(R = 2)}{P(R = 1)} = 88 > 1$ and $P(R = 1) > 0$ , we have $P(R = 2) > P(R = 1)$ , i.e. $P(R = 1) < P(R = 2)$ . (shown)

To check your answer: By GC,

P(R = 1) = \frac{17}{30421755} \approx 5.59 \times 10^{- 7}

and

P(R = 2) = \frac{1496}{30421755} \approx 4.92 \times 10^{- 5}

, and indeed

88 \times 5.59 \times 10^{- 7} \approx 4.92 \times 10^{- 5}

. ✓

Part (ii)

5 marks

Approach

With $r$ extra rectangles, the total number of shapes and the size of the choice both change, but $\binom{28 + r}{12}$ again appears in every probability and cancels out of the given ratio, leaving an equation in $r$ alone.

With the number of rectangles increased by $r$ , there are $(17 + r)$ rectangles and $11$ circles, so

P(R = 3) = \frac{\binom{17 + r}{3}\binom{11}{9}}{\binom{28 + r}{12}},\ \ \ P(R = 4) = \frac{\binom{17 + r}{4}\binom{11}{8}}{\binom{28 + r}{12}}.

Using the given condition $P(R = 4) = 15\, P(R = 3)$ , the common denominator $\binom{28 + r}{12}$ cancels:

\binom{17 + r}{4}\binom{11}{8} = 15\binom{17 + r}{3}\binom{11}{9}

\Rightarrow \frac{\binom{17 + r}{4}}{\binom{17 + r}{3}} = 15 \times \frac{\binom{11}{9}}{\binom{11}{8}}.

Now $\frac{\binom{17 + r}{4}}{\binom{17 + r}{3}} = \frac{14 + r}{4}$ , and $\binom{11}{9} = 55$ , $\binom{11}{8} = 165$ , so $\frac{\binom{11}{9}}{\binom{11}{8}} = \frac{1}{3}$ . Substituting:

\frac{14 + r}{4} = 15 \times \frac{1}{3}

= 5.

14 + r = 20

r = 6.

$\therefore$ $r = 6$ .

2020 H2 Maths Paper 2, Question 8

Part (i)

Part (ii)

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