Tree Diagrams

Topic

Higher June 2025 Paper 1 Q14

EdexcelCurrent spec4 marksTree Diagrams

14 Ricardo is going to play one game of badminton and one game of tennis.

He can either win or lose each game.

The probability that he will win the game of badminton is 0.7
The probability that he will win the game of tennis is 0.4

(a) Complete the probability tree diagram. (2)

Probability tree diagram with blank branches: badminton win or lose, then tennis win or lose

(b) Work out the probability that Ricardo loses both games. (2)

Completed tree: badminton win 0.7, lose 0.3; tennis win 0.4, lose 0.6 on both pairs of branches — Mark scheme (a)

Scheme	Marks
Answer: Correct probabilities	B2
	(2)

Mark scheme (b)

Scheme	Marks
“0.3” × “0.6”	M1ft
Correct answer scores full marks (unless from obvious incorrect working) Answer: 0.18	A1ft
	(2)
	(4 marks)

Notes

M1ft: (Both probabilities must be less than 1)

A1ft: oe eg \(\dfrac{18}{100}\) or \(\dfrac{9}{50}\) or \(\dfrac{0.18}{1}\) or 18%

The printed mark scheme gives the total for this question as 3 marks; the parts are worth 2 + 2 = 4 marks (corrected from the printed mark scheme: “Total 3 marks”).

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

Higher November 2024 Paper 1 Q13

EdexcelCurrent spec4 marksMultiple Events Tree Diagrams

13 The probability that Thomas gets to work early on Saturday is 0.7

If Thomas gets to work early on Saturday, the probability that he will get to work early on Sunday is 0.9

If Thomas does not get to work early on Saturday, the probability that he will get to work early on Sunday is 0.6

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

Probability tree diagram: Saturday branches early (0.7) and not early (blank); each followed by Sunday branches early and not early with blank probabilities

(b) Work out the probability that Thomas gets to work early on both Saturday and Sunday. (2)

Completed tree: Saturday early 0.7, not early 0.3; after early, Sunday early 0.9, not early 0.1; after not early, Sunday early 0.6, not early 0.4 — Mark scheme (a)

Scheme	Marks
Answer: Correct probabilities	B2
	(2)

Mark scheme (b)

Scheme	Marks
0.7 × “0.9” oe	M1ft
Correct answer scores full marks (unless from obvious incorrect working) Answer: 0.63	A1
	(2)
	(4 marks)

Notes

M1ft: (probabilities < 1) for a complete method involving one product

A1: ft oe eg \(\dfrac{63}{100}\) or 63%