2024 H2 Maths Paper 2 Q7 — Worked Solution

Part (a)(i)

1 mark

P\left( A\text{ finds prize} \right) = \frac{2}{n}

Explanation: $A$ opens 2 of the $n$ boxes at random, and the prize is in exactly one of the $n$ boxes, so the probability that it is among the 2 $A$ opens is $\frac{2}{n}$ .

Part (a)(ii)

1 mark

If $A$ fails, the prize is among the remaining $n - 2$ boxes, so

P\left( A\text{ fails} \right) = 1 - \frac{2}{n} = \frac{n - 2}{n}

Given $A$ fails, $B$ opens 2 of the remaining $n - 2$ boxes at random, so

P\left( B\text{ finds prize} \mid A\text{ fails} \right) = \frac{2}{n - 2}

Multiplying,

P\left( B\text{ finds prize} \right) = P\left( A\text{ fails} \right) \times P\left( B\text{ finds prize} \mid A\text{ fails} \right)

= \frac{n - 2}{n} \times \frac{2}{n - 2}

= \frac{2}{n}

$\therefore$ $P\left( B\text{ finds prize} \right) = \frac{2}{n}$ . (shown)

Part (a)(iii)

3 marks

Since $A$ , $B$ , $C$ finding the prize are mutually exclusive and exhaustive,

P\left( C\text{ finds prize} \right) = 1 - P\left( A\text{ finds} \right) - P\left( B\text{ finds} \right)

= 1 - \frac{2}{n} - \frac{2}{n}

= \frac{n - 4}{n}

The probability that either $A$ or $B$ finds the prize (mutually exclusive events) is

P\left( A\text{ finds} \right) + P\left( B\text{ finds} \right) = \frac{2}{n} + \frac{2}{n} = \frac{4}{n}

Requiring $P\left( C\text{ finds prize} \right)$ to be more than 8 times this,

\frac{n - 4}{n} > 8 \times \frac{4}{n}

= \frac{32}{n}

Since $n > 4 > 0$ , we multiply both sides by $n$ :

n - 4 > 32

\Rightarrow n > 36

To check your answer: at

n = 37

,

P(C) = \frac{33}{37} \approx 0.892

and

8 \times \frac{4}{37} = \frac{32}{37} \approx 0.865

; indeed

0.892 > 0.865

✓

$\therefore$ $n > 36$ .

Part (b)

3 marks

Approach

a Venn diagram for $Q$ and $R$ has four regions — $Q \cap R$ , $Q \cap R'$ , $Q'\cap R$ and $(Q \cup R)'$ — whose probabilities sum to $1$ . Writing $P(Q) = q$ and $P(R) = r$ , each given piece of information becomes one linear equation in $q$ , $r$ and $x$ , and solving the resulting system gives $x$ .

Let $P(Q) = q$ and $P(R) = r$ . Since $Q = (Q \cap R) \cup (Q \cap R')$ ,

q = x + 0.1\quad\text{...(1)}

Since $P(Q'\cap R) = P(R) - P(Q \cap R) = r - x$ , and this equals $P(Q) = q$ ,

r - x = q

\Rightarrow r = q + x\quad\text{...(2)}

Since $P\left( (Q \cup R)'\right) = 1 - P(Q \cup R) = 1 - (q + r - x)$ , and this equals $P(Q) + 2P(R) = q + 2r$ ,

1 - q - r + x = q + 2r

\Rightarrow 1 + x = 2q + 3r\quad\text{...(3)}

Substituting (1) into (2),

r = (x + 0.1) + x = 2x + 0.1

Substituting $q = x + 0.1$ and $r = 2x + 0.1$ into (3),

1 + x = 2(x + 0.1) + 3(2x + 0.1)

= 2x + 0.2 + 6x + 0.3

= 8x + 0.5

Solving,

1 + x = 8x + 0.5

\Rightarrow 0.5 = 7x

\Rightarrow x = \frac{1}{14}

To check your answer: with

x = \frac{1}{14}

,

q = x + 0.1 = \frac{6}{35}

and

r = 2x + 0.1 = \frac{17}{70}

, so

1 + x = \frac{15}{14}

and

2q + 3r = \frac{12}{35} + \frac{51}{70} = \frac{15}{14}

— equal ✓

$\therefore$ $x = \frac{1}{14}$ .

2024 H2 Maths Paper 2, Question 7

Part (a)(i)

Part (a)(ii)

Part (a)(iii)

Part (b)

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