Basic Probability: WJEC GCSE Foundation June 2023 Unit 1 Q18

Foundation June 2023 Unit 1 Q18

WJECFoundationCurrent spec5 marksBasic Probability Multiple Events

18. Two fair spinners are shown below.

Spinner A has three equal sections numbered 1, 2, 3 with the pointer on 2; Spinner B has four equal sections numbered 1, 2, 3, 4 with the pointer on 3

In a game, the numbers shown by the two pointers are added together.
In the diagram, the score gained would be 2 + 3 = 5.

A winning score is 6 or more.

How many winning scores would you expect when the game is played 60 times?
You must show all your working. [5]

Mark scheme

Scheme

Marks

Comments

Identifying 12 possible combinations

Award B1 for convincing identification of the 12 combinations, for example:

simply stating 12
(3 × 4 =) 12
showing all combinations 1+1, 1+2, etc.
all 12 ‘totals’
(2,3,3,4,4,4,5,5,5,6,6,7) shown with no extras
completed sample space drawn (3 by 4).

Identifying the 3 correct combinations that give a score of 6 or more (2 and 4, 3 and 3, 3 and 4)

Strict FT only if a list of all possible scores previously stated.

(Probability of ‘6 or more’ =) \(\dfrac{3}{12}\) or equivalent
ISW ( but note comment for M1 below)

A fraction with a denominator of 12 implies the first B1.
Unsupported \(\dfrac{3}{12}\) or equivalent implies previous B1B1.
Probability may be implied in later working
(e.g. 60 ÷ 12 = 5, 5 × 3 = 15).
FT if a clear numerator and denominator can be identified from previous work.

e.g. Possible scores 2, 3, 4 ,5, 6, 7	(B0)
2 scores of 6 or more	(B1 FT)
Probability = \(\dfrac{2}{6}\)	(B1 FT)

(Number of winning scores =) \(\dfrac{3}{12}\) × 60 or equivalent

FT ‘their \(\dfrac{3}{12}\)’
If ‘their \(\dfrac{3}{12}\)’ incorrectly simplified and used then award B0 previously.

= 15

Must not come from incorrect working.
Award M1 A0 for a final answer of ( \(\dfrac{3}{12}\) =) \(\dfrac{15}{60}\)

Note: using ‘a winning score of 6’ instead of ‘a winning score of 6 or more’ can be awarded a maximum of B1B0B1M1A1.