4400 Higher November 2009 Paper 4 Q16

EdexcelHigherOld spec5 marksTree Diagrams

16 There are 10 chocolates in a box.
7 of the chocolates have soft centres and 3 of the chocolates have hard centres.
Kyla takes at random a chocolate from the box and eats it.
She then takes at random another chocolate from the box and eats it.

(a) Complete the probability tree diagram.

Probability tree: first chocolate branches soft centre 7/10 and hard centre blank; second chocolate branches to be drawn

(2)

(b) Calculate the probability that at least one of the chocolates Kyla eats has a hard centre. (3)

Completed tree: first soft 7/10, hard 3/10; after soft: soft 6/9, hard 3/9; after hard: soft 7/9, hard 2/9 — Mark scheme (a)

Scheme	Marks
Correct tree diagram (as above)	B2
	(2)

Mark scheme (b)

Scheme	Marks
\(\dfrac{7}{10} \times \dfrac{3}{9} + \dfrac{3}{10} \times \dfrac{7}{9} + \dfrac{3}{10} \times \dfrac{2}{9}\) \(\left(= \dfrac{21}{90} + \dfrac{21}{90} + \dfrac{6}{90}\right)\) or \(\dfrac{7}{10} \times \dfrac{3}{9} + \dfrac{3}{10}\) \(\left(= \dfrac{21}{90} + \dfrac{3}{10}\right)\)	M1
	M1
\(\dfrac{48}{90}\)	A1
	(3)
	(5 marks)

Notes

M1: for one correct product

M1: for completely correct expression

A1: for \(\frac{48}{90}\) oe inc \(\frac{8}{15}\) or for 0.53 or for answer rounding to 0.53

Alternative: M2 for \(1 - \frac{7}{10} \times \frac{6}{9}\)
SC M2 for \(1 - \frac{7}{10} \times \frac{7}{10}\)
SC M1 for \(\frac{7}{10} \times \frac{3}{10}\) or \(\frac{3}{10} \times \frac{7}{10}\) or \(\frac{3}{10} \times \frac{3}{10}\); SC M1 (dep) for sum of above products or for \(\frac{7}{10} \times \frac{3}{10} + \frac{3}{10}\)
ft for method marks from their tree diagram, provided probabilities \(\lt 1\)