2023 H2 Maths Paper 2 Q5 — Worked Solution

Part (a)(i)

1 mark

With 36 equally likely slots,

A = \{ 1,3,5,\ldots,35\}\ \left( 18\ \text{outcomes} \right),\quad B = \{ 2,4,6,\ldots,36\}\ \left( 18\ \text{outcomes} \right)

C = \{ 3,6,9,\ldots,36\}\ \left( 12\ \text{outcomes} \right),\quad D = \{ 6,12,18,\ldots,36\}\ \left( 6\ \text{outcomes} \right)

Since every score is either odd or even (never both), $A \cap B = \varnothing$ . Since every multiple of $6$ is even, $D \subseteq B$ , so no multiple of $6$ is odd and $A \cap D = \varnothing$ .

Checking the remaining pairs: $A \cap C = \{ 3,9,15,21,27,33\} \neq \varnothing$ ; $B \cap C = \{ 6,12,\ldots,36\} \neq \varnothing$ ; $B \cap D = D \neq \varnothing$ ; $C \cap D = D \neq \varnothing$ .

To check your answer:

A

and

D

being mutually exclusive is consistent with

D \subseteq B

and

A \cap B = \varnothing

already, since

A \cap D \subseteq A \cap B = \varnothing

✓

$\therefore$ $A$ and $B$ are mutually exclusive, and $A$ and $D$ are mutually exclusive. No other pair is.

Part (a)(ii)

2 marks

P(A) = \frac{18}{36} = \frac{1}{2},\quad\quad P(C) = \frac{12}{36} = \frac{1}{3}

$A \cap C$ is the event that the score is an odd multiple of $3$ : $\{ 3,9,15,21,27,33\}$ , so $P(A \cap C) = \frac{6}{36} = \frac{1}{6}$ .

P(A) \times P(C) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} = P(A \cap C)

Since $P(A) \times P(C) = P(A \cap C)$ , $A$ and $C$ are independent.

$\therefore$ $A$ and $C$ are independent.

Checking $B$ and $C$ : $P(B) = \frac{1}{2}$ , and $B \cap C = \{ 6,12,18,24,30,36\}$ , so $P(B \cap C) = \frac{6}{36} = \frac{1}{6}$ .

P(B) \times P(C) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} = P(B \cap C)

$\therefore$ $B$ and $C$ are another pair of independent events.

Part (b)(i)

1 mark

Now with only $35$ slots (since $36$ is stuck),

A = \{ 1,3,\ldots,35\}\ (18),\quad B = \{ 2,4,\ldots,34\}\ (17),\quad C = \{ 3,6,\ldots,33\}\ (11),\quad D = \{ 6,12,\ldots,30\}\ (5)

Removing slot $36$ does not change the odd/even split, so $A \cap B = \varnothing$ still; and every multiple of $6$ remaining ( $6$ to $30$ ) is still even, so $D \subseteq B$ still and $A \cap D = \varnothing$ still. The other pairs remain non-disjoint — for instance $A \cap C = \{ 3,9,15,21,27,33\} \neq \varnothing$ , unchanged since $36$ was even and was not in $A \cap C$ to begin with.

$\therefore$ $A$ and $B$ , and $A$ and $D$ , are still the only mutually exclusive pairs — the same pairs as in (a)(i).

Part (b)(ii)

1 mark

P(A) = \frac{18}{35},\quad\quad P(C) = \frac{11}{35}

$A \cap C$ is still the set of odd multiples of $3$ up to $35$ : $\{ 3,9,15,21,27,33\}$ (removing slot $36$ , which is even, does not affect this set), so $P(A \cap C) = \frac{6}{35}$ .

P(A) \times P(C) = \frac{18}{35} \times \frac{11}{35} = \frac{198}{1225}

P(A \cap C) = \frac{6}{35} = \frac{210}{1225}

To check your answer: removing slot

36

(a multiple of both

2

and

3

, but not of

A \cap C

) shrinks

P(C)

without changing

P(A \cap C)

, so the two quantities, equal before, no longer match — exactly what breaking independence looks like ✓

Since $\frac{198}{1225} \neq \frac{210}{1225}$ , $P(A) \times P(C) \neq P(A \cap C)$ .

$\therefore$ $A$ and $C$ are no longer independent.

2023 H2 Maths Paper 2, Question 5

Part (a)(i)

Part (a)(ii)

Part (b)(i)

Part (b)(ii)

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