Basic Probability: WJEC GCSE Foundation June 2022 Unit 2 Q14

Foundation June 2022 Unit 2 Q14

WJECFoundationCurrent spec8 marksBasic Probability Multiple Events

14.

A triangular spinner divided into three equal parts numbered 1, 3, 5 with its arrow on 3, and a square spinner divided into four equal parts numbered 2, 4, 6, 8 with its arrow on 2

Two fair spinners are shown in the diagram above.
In a game, the two spinners are spun.

The two numbers obtained are multiplied together to get a score.
For example, in the diagram above, the score is 6 because \(3 \times 2 = 6\).

Some of the scores are shown in the table below.

		Square spinner
		2	4	6	8
Triangular spinner	1		4		8
	3	6		18	24
	5		20

(a) Complete the table to show all the possible scores. [1]

(b) Explain why all the scores are even numbers. [1]

(d) Players are charged £2.50 to play the game once.
Each player who gets a score of 10 or more wins £3.50.
How much profit would you expect to make when 228 people each play the game once?
You must show all your working. [4]

Mark scheme (a)

Scheme

Marks

Comments

		Square spinner
		2	4	6	8
Triangular Spinner	1	2	(4)	6	(8)
	3	(6)	12	(18)	(24)
	5	10	(20)	30	40

All six entries correct.

Mark scheme (b)

Scheme	Marks	Comments
Valid explanation given e.g. “odd × even = even” "because it’s odd times even” “even times any whole number is always even”	E1	Do not accept “because all the numbers on the square spinner are even” Allow “as they are multiplied by even numbers which make even numbers” "because it’s multiplied with an even number"

Mark scheme (c)

Scheme	Marks	Comments
\(\dfrac{7}{12}\) ISW	B2	FT ‘their fully completed table’. Award B2 for unsupported 58·3(333…)%. Penalise −1 for only words (7 out of 12) or only ratio (7:12). B1 for \(x\)/12 if \(x\) < 12. B1 for 7/\(y\) if \(y\) > 7 (FT ‘their 7’). B1 for unsupported 58%.

Mark scheme (d)

Scheme	Marks	Comments
(Amount taken = 228 × £2.50 =) (£)570	B1
(Expected number of winners = 7/12 × 228) 133 (winners)	B1	If 7/12 or correct % or decimal seen in part (c), it must be used for this B1. FT 228 × ‘their 7/12’ provided less than 1 Allow 133/228 or ‘133 out of 228’. Must be whole number. Award B0 for 7/12 × 228 = 0·58(333…) × 228 = 132 winners. Award B0 for 7/12 × 228 = 0·6 × 228 = 136 or 137 winners.
(Expected prize money = 133 × £3.50 = ) (£)465.5(0)	B1	FT £3.50 ×‘their 133’ (provided < 228).
(Expected profit = 228 × £2.50 – 133 × £3.50 =) (£)104.5(0)	B1	(£)570 – (£)465.5(0) FT ‘their (£)570’ – ‘their (£)465.5(0)’ Award B1B1B1B0 for sight of 228 × £2.50 – 133 × £3.50 with an incorrect final answer. If the FT results in a loss, the ‘Loss’ must be stated, or the answer left as a negative.

Alternative Method 1

Scheme	Marks	Comments
(Expected number of winners = 7/12 × 228 ) 133 (winners)	B1	If 7/12 or correct % or decimal seen in part (c), it must be used for this B1. FT ‘their 7/12’ if less than 1 × 228 Allow 133/228 or ‘133 out of 228’ Must be whole number Award B0 for 7/12 × 228 = 0·58(333…) × 228 = 132 winners. Award B0 for 7/12 × 228 = 0·6 × 228 = 136 or 137 winners.
(Expected number that don’t win = 228 – 133) 95 (non-winners)	B1	FT 228 – ‘their 133’ (provided < 228)
(Amount taken = 95 × £2.50 = ) (£)237.5(0)	B1	FT £2.50 × ‘their 95’ provided < 133
(Expected profit = 95 × £2.50 – 133 × £1 = ) (£)104.5(0)	B1	(£)237.5(0) – (£)133 FT ‘their (£)237.5(0)’ – ‘their (£)133’ Award B1B1B1B0 for sight of 95 × £2.50 – 133 × £1 with an incorrect final answer. If the FT results in a loss, the ‘Loss’ must be stated, or the answer left as a negative.

Alternative Method 2

Scheme	Marks	Comments
Working with 12 players (Amount taken = 12 × £2.50 = ) (£)30(.00)	B1
(Expected prize money = 7 × £3.50 = ) (£)24.5(0)	B1	FT ‘their 7’ (provided < 12)
(Expected profit for 12 players = (£)30(.00) - (£)24.5(0) =) (£)5.5(0)	B1	FT ‘their (£)30(.00)’ – ‘their (£)24.5(0)’
(Expected profit for 228 players = \(\dfrac{\underline{228}}{12}\) × (£)5.5(0) = ) (£)104.5(0)	B1	FT 19 × ‘their (£)5.5(0)’ If the FT results in a loss, the ‘Loss’ must be stated, or the answer left as a negative.