4MA0 Higher January 2018 Paper 4 Q19

EdexcelHigherOld spec8 marksMultiple Events Tree Diagrams

19 Mary and Naveed play two games of table tennis against each other.
For each game they play, the probability that Mary wins is 0.32

(a) Complete the probability tree diagram.

Probability tree diagram for the first and second games, with branches Mary wins and Naveed wins; 0.32 is given on the first-game Mary wins branch and the other branches are blank

(2)

(b) Work out the probability that Mary wins exactly one game of table tennis. (3)

In a fridge there are only
  3 bottles of water
  5 bottles of orange juice
  2 bottles of cola

After their games of table tennis, Mary takes at random a bottle from the fridge and then Naveed takes at random a bottle from the fridge.

Scheme	Marks
0.68 0.32, 0.68, 0.32, 0.68	B2
	(2)

Mark scheme (b)

Scheme	Marks
0.32 × 0.68 (=0.2176)	M1
2 × “0.2176” oe	M1
0.4352	A1
	(3)

Notes

M1: ft from tree diagram (allow 0.218)

M1: ft from tree diagram

A1: [Allow 0.435 from correct working] oe eg. \(\dfrac{272}{625}\) 43.5(2)%

Mark scheme (c)

Scheme	Marks
\(\dfrac{3}{10} \times \dfrac{2}{9}\) or \(\dfrac{5}{10} \times \dfrac{4}{9}\) or \(\dfrac{2}{10} \times \dfrac{1}{9}\)	M1
\(\dfrac{3}{10} \times \dfrac{2}{9} + \dfrac{5}{10} \times \dfrac{4}{9} + \dfrac{2}{10} \times \dfrac{1}{9}\)	M1
\(\dfrac{28}{90}\) oe	A1
	(3)

Notes

A1: eg \(\dfrac{14}{45}\) or 0.311(1…..)

Method with replacement (maximum 2 marks)

M1: \(\dfrac{3}{10} \times \dfrac{3}{10}\) or \(\dfrac{5}{10} \times \dfrac{5}{10}\) or \(\dfrac{2}{10} \times \dfrac{2}{10}\)

M1: \(\dfrac{3}{10} \times \dfrac{3}{10} + \dfrac{5}{10} \times \dfrac{5}{10} + \dfrac{2}{10} \times \dfrac{2}{10}\left(= \dfrac{38}{100}\right)\)