Basic Probability: WJEC GCSE Foundation June 2024 Unit 1 Q18

Foundation June 2024 Unit 1 Q18

WJECFoundationCurrent spec7 marksBasic Probability Multiple Events

18. Ahmed organises a game using two fair spinners, as shown below.
The first spinner shows the values 10, 20, 30 and 40.
The second spinner shows the values 1, 2, 3, 4 and 5.

Two spinners: the first has four equal sections 10, 20, 30, 40 with the arrow on 30; the second has five equal sections 1, 2, 3, 4, 5 with the arrow on 2

In the game, the two spinners are spun and the values shown are added to give a score.
For example, the spinners above score 32.

Ahmed charges £1 for each attempt at the game.
Any player who scores 43 or more wins £5.

Calculate Ahmed’s expected profit when this game is played 100 times. [7]

Mark scheme

Scheme	Marks	Comments
(Probability of winning score =) \(\dfrac{3}{20}\) or equivalent	B2	Award B2 for sight of \(\frac{1}{4} \times \frac{3}{5}\) ‘3 winners out of 20’ the 3 winning combinations/scores clearly identified in a list/table of the 20 possible combinations/scores. B2 may be implied in later workings. Award B1 for one of the following: for convincing identification of the 20 combinations/scores, for example: ✓ sight of 20 ✓ 4 × 5 ✓ showing all 20 correct combinations 10 + 1,10 + 2...with no extras ✓ all 20 correct scores listed with no extras ✓ completed sample space drawn (4 by 5) ✓ sight of \(\frac{1}{4}\) AND \(\frac{3}{5}\) or equivalent. identifying the three correct possible winning scores (43, 44, 45) with no extras identifying the three correct winning combinations (40 + 3, 40 + 4, 40 + 5) with no extras \(\dfrac{3}{x}\) provided \(x\) > 3 and correct winning combinations/scores identified \(\dfrac{y}{20}\) provided with \(y\) < 20 \(\frac{3}{20}\) from incorrect winning combinations or scores identified strict FT from ‘their list’ provided clearly stated \(\dfrac{\textit{'their number of winning scores'}}{\textit{'their total number of possible scores'}}\)
(Number of winners = ) \(\dfrac{3}{20} \times 100\) or equivalent	M1	Award M1 for \(\frac{1}{4} \times \frac{3}{5} \times 100\). May be implied e.g. 100 ÷ 20 = 5, 5 × 3 = 15. FT ‘their probability of winning score’ × 100, provided ‘their probability of winning score’ < 1, or ≠ \(\frac{x}{100}\). M0 awarded if ‘their probability of winning score’ is simplified incorrectly.
= 15	A1	May be implied by ‘15 out of 100’ or equivalent. If 15 is not seen but final answer of £15 is given (i.e. ‘people’ confused with ‘money’) then allow only M1A0. Answer must be whole number.
(Profit =) (£)100 – 15 × (£)5 OR (£)85 – 15 × (£)4	M2	FT ‘their number of winners’, provided ≠ 3 and <100. Award M1 for one of the following: 15 × (£)5 an appropriate sight of (£)75 ‘their number of winners’× (£)5 ‘their number of winners’× (£)5 evaluated correctly (£)100 – (£)15 AND 15 × (£)4 (£)100 – ‘their number of winners’ × (£)1 AND ‘their number of winners’ × (£)4.
= (£)25	A1	FT provided M2 (not M1M1) previously awarded. Unsupported answer of (£)25 is awarded B2 M1A1M2A1.

Alternative

18. Alternative method for the final 5 marks

Scheme	Marks	Comments
Must clearly be working with groups of 20 throughout
20 × (£)1 – 3 × (£)5	M2	Method must be seen for M2. FT ‘their 20’ × (£)1 – ‘their 3’ × (£)5.
(£)5	A1	May be implied in later working.
× 5	m1	FT 100 ÷ ‘their 20’
=(£)25	A1

Corrected by Westie's Workshop

The board's mark scheme printed “‘their total number of possibles scores’”. Changed to “‘their total number of possible scores’” because “possibles” should be “possible”.

Original: see the board's mark scheme PDF (link above).