Higher June 2025 Paper 2R Q14

EdexcelCurrent spec7 marksMultiple Events Tree Diagrams

14 Sara has two bags, A and B

In bag A, there are only 5 red beads and 4 green beads.
In bag B, there are only 7 red beads and 3 green beads.

Sara takes at random a bead from bag A
She then takes at random a bead from bag B

(a) Use this information to complete the probability tree diagram. (2)

Probability tree diagram: Bag A branches red and green, each followed by Bag B branches red and green, with blank probabilities

(b) Work out the probability that Sara takes two red beads. (2)

Sara puts the beads back into their original bags.

Sara also has a box of beads.
In the box, there are only red beads and green beads.

When a bead is taken at random from the box, the probability that it is a green bead is \(\dfrac{2}{11}\)

Sara takes at random a bead from bag A
She then takes at random a bead from bag B
She then takes at random a bead from the box.

Mark scheme (a)

Scheme	Marks
\(\dfrac{5}{9}\) and \(\dfrac{4}{9}\) \(\dfrac{7}{10}\) and \(\dfrac{3}{10}\) \(\dfrac{7}{10}\) and \(\dfrac{3}{10}\)	B2
	(2)

Notes

B2: for all 3 correct pairs of probabilities on the correct branches

(B1 for 1 or 2 correct pairs of probabilities on the correct branches)

Accept decimals or percentages rounded or truncated to at least 2sf.

NB \(\dfrac{5}{9} = 0.55(55\ldots)\) and \(\dfrac{4}{9} = 0.44(44\ldots)\)

Mark scheme (b)

Scheme	Marks
\(\dfrac{5}{9} \times \dfrac{7}{10}\) or \(1 - \left(\dfrac{5}{9} \times \dfrac{3}{10} + \dfrac{4}{9} \times \dfrac{7}{10} + \dfrac{4}{9} \times \dfrac{3}{10}\right)\)	M1ft
Correct answer only scores full marks (unless from obviously incorrect working) Answer: \(\dfrac{7}{18}\)	A1ft
	(2)

Notes

M1ft: ft diagram, oe

A1ft: ft diagram, oe fraction, decimal or percentage.

NB \(\dfrac{7}{18} = \dfrac{35}{90} = 0.38(88\ldots)\)

Allow ft their tree diagram provided the relevant probabilities are less than 1 in each case

For A1, allow decimals or percentages that round or truncate correctly to at least 2sf. ISW any attempt to convert to other form once correct probability seen

Mark scheme (c)

Scheme	Marks
\((RRR =)\;\dfrac{5}{9} \times \dfrac{7}{10} \times \left(1 - \dfrac{2}{11}\right)\;\left(= \dfrac{315}{990} = \dfrac{7}{22}\right)\) or \((RRG =)\;\dfrac{5}{9} \times \dfrac{7}{10} \times \dfrac{2}{11}\;\left(= \dfrac{70}{990} = \dfrac{7}{99}\right)\) or \((RGR =)\;\dfrac{5}{9} \times \dfrac{3}{10} \times \left(1 - \dfrac{2}{11}\right)\;\left(= \dfrac{135}{990} = \dfrac{3}{22}\right)\) or \((GRR =)\;\dfrac{4}{9} \times \dfrac{7}{10} \times \left(1 - \dfrac{2}{11}\right)\;\left(= \dfrac{252}{990} = \dfrac{14}{55}\right)\) OR \((GGG =)\;\dfrac{4}{9} \times \dfrac{3}{10} \times \dfrac{2}{11}\;\left(= \dfrac{24}{990} = \dfrac{4}{165}\right)\) or \((GGR =)\;\dfrac{4}{9} \times \dfrac{3}{10} \times \left(1 - \dfrac{2}{11}\right)\;\left(= \dfrac{108}{990} = \dfrac{6}{55}\right)\) or \((GRG =)\;\dfrac{4}{9} \times \dfrac{7}{10} \times \dfrac{2}{11}\;\left(= \dfrac{56}{990} = \dfrac{28}{495}\right)\) or \((RGG =)\;\dfrac{5}{9} \times \dfrac{3}{10} \times \dfrac{2}{11}\;\left(= \dfrac{30}{990} = \dfrac{1}{33}\right)\)	M1ft
\(\text{``}{\dfrac{7}{22}}\text{''} + \text{``}{\dfrac{7}{99}}\text{''} + \text{``}{\dfrac{3}{22}}\text{''} + \text{``}{\dfrac{14}{55}}\text{''}\) oe OR \(1 - \left(\text{``}{\dfrac{4}{165}}\text{''} + \text{``}{\dfrac{6}{55}}\text{''} + \text{``}{\dfrac{28}{495}}\text{''} + \text{``}{\dfrac{1}{33}}\text{''}\right)\)	M1ft
Correct answer only scores full marks (unless from obviously incorrect working) Answer: \(\dfrac{386}{495}\)	A1ft
	(3)
	(7 marks)

Notes

M1ft: ft diagram

for a correct calculation to find the probability of one relevant outcome,

eg \(RRR\) or \(RRG\) or \(RGR\) or \(GRR\)

eg \(GGG\) or \(GGR\) or \(GRG\) or \(RGG\)

May see \(RRR\) or \(RRG\) found using their answer to part (b):

\((RRR =)\left[\dfrac{7}{18}\right] \times \left(1 - \dfrac{2}{11}\right)\)

\((RRG =)\left[\dfrac{7}{18}\right] \times \dfrac{2}{11}\)

where \(\left[\dfrac{7}{18}\right]\) is their answer to part (b) and must be less than 1

M1ft: ft diagram, for a method to find the probability required. Condone one error in one of the four relevant outcomes or omission of one outcome

Note P(\(RR\)) = P(\(RRR\)) + P(\(RRG\)), so may see \(\dfrac{5}{9} \times \dfrac{7}{10}\left(= \dfrac{7}{18}\right)\) in place of \(\text{``}{\dfrac{7}{22}}\text{''} + \text{``}{\dfrac{7}{99}}\text{''}\) for this mark (similar with P(\(GG\)))

A1ft: ft diagram, correct probability oe fraction, decimal or percentage

NB \(\dfrac{386}{495} = \dfrac{772}{990} = 0.77(979\ldots)\)

Allow ft their tree diagram provided the relevant probabilities are less than 1 in each case

For A1, allow decimals or percentages that round or truncate correctly to at least 2sf. ISW any attempt to convert to other form once correct probability seen