2016 H2 Maths Paper 2 Q5 — Worked Solution

Let $R$ , $B$ , $Y$ be the events that the spinner comes to rest over the sections labelled $\mathbf{R}$ , $\mathbf{B}$ , $\mathbf{Y}$ respectively, and let $W$ be the event that the player wins the game.

Since the spinner has $7$ equally likely sections, $1$ labelled $\mathbf{R}$ , $2$ labelled $\mathbf{B}$ and $4$ labelled $\mathbf{Y}$ ,

P(R) = \frac{1}{7},\quad\quad P(B) = \frac{2}{7},\quad\quad P(Y) = \frac{4}{7}.

Since the red die has $3$ faces with $\text{*}$ , the blue die has $2$ faces with $\text{*}$ , and the yellow die has $1$ face with $\text{*}$ , each out of $6$ faces,

P(W \mid R) = \frac{3}{6},\quad\quad P(W \mid B) = \frac{2}{6},\quad\quad P(W \mid Y) = \frac{1}{6}.

Part (i)

2 marks

By the law of total probability, since $R$ , $B$ , $Y$ partition the sample space:

P(W)

= P(R)P(W \mid R) + P(B)P(W \mid B) + P(Y)P(W \mid Y)

= \frac{1}{7} \times \frac{3}{6} + \frac{2}{7} \times \frac{2}{6} + \frac{4}{7} \times \frac{1}{6}

= \frac{11}{42}

$\therefore\ P(W) = \frac{11}{42}$ .

Part (ii)

1 mark

Using Bayes’ theorem,

P(B \mid W)

= \frac{P(B)P(W \mid B)}{P(W)}

= \frac{\frac{2}{7} \times \frac{2}{6}}{\frac{11}{42}}

= \frac{4}{11}

$\therefore\ P(B \mid W) = \frac{4}{11}$ .

Part (iii)

2 marks

Approach

“A different colour die each time” over the $3$ games means the spinner shows $\mathbf{R}$ , $\mathbf{B}$ , $\mathbf{Y}$ in some order, one each, over the $3$ games — there are $3! = 6$ such orders, and by independence of the games each order has the same probability. So rather than listing $6$ cases, we find the probability of one representative order and multiply by $3!$ .

Let $p_{R} = P(R)P(W \mid R)$ , $p_{B} = P(B)P(W \mid B)$ , $p_{Y} = P(Y)P(W \mid Y)$ , the probability that a single game is won with the red, blue, yellow die respectively.

p_{R} = \frac{1}{7} \times \frac{3}{6} = \frac{1}{14},\quad\quad p_{B} = \frac{2}{7} \times \frac{2}{6} = \frac{2}{21},\quad\quad p_{Y} = \frac{4}{7} \times \frac{1}{6} = \frac{2}{21}

Since the $3$ games are independent, and there are $3!$ ways to assign the colours $\mathbf{R}$ , $\mathbf{B}$ , $\mathbf{Y}$ to the $3$ games, all winning games:

P\left( \text{win 3 games, each a different colour die} \right)

= 3! \times p_{R} \times p_{B} \times p_{Y}

= 6 \times \frac{1}{14} \times \frac{2}{21} \times \frac{2}{21}

= \frac{4}{1029}

To check your answer:

\frac{4}{1029} \approx 0.00389

, sensibly small since it requires winning all

3

games with

3

different, specific colours. ✓

$\therefore$ the required probability is $\frac{4}{1029}$ .

2016 H2 Maths Paper 2, Question 5

Part (i)

Part (ii)

Part (iii)

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