4MA0 Higher June 2015 Paper 4 Q15

EdexcelHigherOld spec6 marksTree Diagrams

15 There are 6 milk chocolates and 4 plain chocolates in a box.
Rob takes at random a chocolate from the box and eats it.
Then Alison takes at random a chocolate from the box and eats it.

(a) Complete the probability tree diagram.

Probability tree: Rob milk chocolate or plain chocolate, then Alison milk chocolate or plain chocolate; all branches blank

(3)

(b) Work out the probability that there are now exactly 3 plain chocolates in the box. (3)

Mark scheme (a)

Scheme	Marks
Probabilities on branches correct. Answer: \(\dfrac{6}{10}, \dfrac{4}{10}, \dfrac{5}{9}, \dfrac{4}{9}, \dfrac{6}{9}, \dfrac{3}{9}\)	B1 B1 B1
	(3)

Notes

B1 B1 B1: B1 for \(\dfrac{6}{10}, \dfrac{4}{10}\) oe on LH branches

B1 for \(\dfrac{5}{9}, \dfrac{4}{9}\) oe on top RH branches

B1 for \(\dfrac{6}{9}, \dfrac{3}{9}\) oe on bottom RH branches

Decimals given on the 2^nd set of branches to be to at least 2dp (truncated or rounded).

Mark scheme (b)

Scheme	Marks
\(\dfrac{6}{10} \times \dfrac{4}{9}\) or \(\dfrac{4}{10} \times \dfrac{6}{9}\) or ft from their tree diagram	M1
\(\dfrac{6}{10} \times \dfrac{4}{9} + \dfrac{4}{10} \times \dfrac{6}{9}\) or ft from their tree diagram	M1dep
\(\dfrac{48}{90}\) oe	A1
	(3)

Notes

M1: or \(\dfrac{6}{10} \times \dfrac{5}{9} + \dfrac{4}{10} \times \dfrac{3}{9}\) \(\left(= \dfrac{42}{90}\right)\)

M1dep: or \(1 - \text{"}\dfrac{42}{90}\text{"}\)

A1: Allow 0.53(33…)
Note: If all 4 probability products are seen at the ends of the branches on the tree diagram or in lists in the working space for (b), marks can only be awarded in (b) if it is clear which product(s) they are intending to use.