Tree Diagrams: WJEC GCSE Higher June 2025 Unit 2 Q8

Higher June 2025 Unit 2 Q8

WJECHigherCurrent spec5 marksTree Diagrams

8. A bag contains 5 red balls and 4 blue balls.
Megan chooses a ball from the bag at random.
She replaces the ball in the bag.
Then she chooses a second ball from the bag at random.

(a) Complete the tree diagram below. [2]

Tree diagram: Ball 1 Red with probability 5/9, Blue blank; from Red, Ball 2 Red 5/9 and Blue blank; from Blue, Ball 2 Red and Blue both blank

(b) Find the probability that Megan chooses two balls of the same colour. [3]

Completed tree diagram: Ball 1 Red 5/9, Blue 4/9; each Ball 2 branch Red 5/9, Blue 4/9 — Mark scheme (a)

Scheme	Marks	Comments
Complete diagram	B2	Award B1 for any one of the following: \(\dfrac{4}{9}\) or equivalent on ‘Blue’ Ball 1 branch 2 out of the 3 correct Ball 2 branches.

Mark scheme (b)

Scheme	Marks	Comments
Sight of \(\dfrac{5}{9}\) × \(\dfrac{5}{9}\) (= 0.308(6..)) OR \(\dfrac{4}{9}\) × \(\dfrac{4}{9}\) (= 0.197(5..)) or equivalent	B1	Check diagram for answers. 0·55(5…) × 0·55(5…) OR 0·44(4…) × 0·44(4…) Award B1 for an appropriate sight of any one of the following: \(\dfrac{25}{81}\) or 0·308(6..) or 0·31 or 0·3 \(\dfrac{16}{81}\) or 0·197(5..) or 0·2 FT ‘their \(\dfrac{4}{9}\)’ from Blue Ball 1 branch, only if between 0 and 1. FT ‘their \(\dfrac{4}{9}\)’ from Blue Ball 2 branch, only if between 0 and 1.
\(\dfrac{5}{9}\) × \(\dfrac{5}{9}\) + \(\dfrac{4}{9}\) × \(\dfrac{4}{9}\) or equivalent	M1	0·55(5…) × 0·55(5…) + 0·44(4…) × 0·44(4…) Award M1 for sight of \(\dfrac{25}{81}\) + \(\dfrac{16}{81}\). FT ‘their \(\dfrac{5}{9}\) × \(\dfrac{5}{9}\)’ + ‘their \(\dfrac{4}{9}\) × \(\dfrac{4}{9}\)’, provided both less than 1. The addition of both single fractions (or products) can be implied by an appropriate sight of \(\dfrac{\textit{sum of numerators}}{\textit{sum of denominators}}\).
\(\dfrac{41}{81}\) ISW or 0·506(…) or equivalent	A1	CAO for ‘their tree diagram’. Allow answers in the range 0·505 to 0·507 inclusive or equivalent. Allow an answer of 0·51 or equivalent. Only allow 0·5 provided a more accurate answer has been seen (e.g. \(\dfrac{41}{81}\), 0·506(…)). Only FT provided answer is less than 1.

Alternative method

Scheme	Marks	Comments
1 – [\(\dfrac{5}{9}\) × \(\dfrac{4}{9}\) + \(\dfrac{4}{9}\) × \(\dfrac{5}{9}\)] or 1 – [2 × \(\dfrac{5}{9}\) × \(\dfrac{4}{9}\)] or equivalent	M2	Check diagram for answers. (= 1 – [\(\dfrac{20}{81}\) + \(\dfrac{20}{81}\)] = 1 – \(\dfrac{40}{81}\) or equivalent) FT ‘their \(\dfrac{4}{9}\)’ from Blue Ball 1 branch, only if between 0 and 1. FT ‘their \(\dfrac{4}{9}\)’ from Blue Ball 2 branch, only if between 0 and 1. Award M1 for any one of the following: 1 – [\(\dfrac{5}{9}\) × \(\dfrac{4}{9}\)] (= 1 – \(\dfrac{20}{81}\)) 1 – [\(\dfrac{4}{9}\) × \(\dfrac{5}{9}\)] (= 1 – \(\dfrac{20}{81}\)) \(\dfrac{5}{9}\) × \(\dfrac{4}{9}\) + \(\dfrac{4}{9}\) × \(\dfrac{5}{9}\) (= \(\dfrac{20}{81}\) + \(\dfrac{20}{81}\) = \(\dfrac{40}{81}\)) 2 × \(\dfrac{5}{9}\) × \(\dfrac{4}{9}\) (= 2 × \(\dfrac{20}{81}\) = \(\dfrac{40}{81}\)) or equivalent Allow use of 0·55(5…) and 0·44(4…)
\(\dfrac{41}{81}\) ISW or 0·506(…) or equivalent	A1	CAO for ‘their tree diagram’. Allow answers in the range 0·505 to 0·507 inclusive or equivalent. Allow an answer of 0·51 or equivalent. Only allow 0·5 provided a more accurate answer has been seen (e.g. \(\dfrac{41}{81}\), 0·506(…)). Only FT provided answer is less than 1.