October 2021 Paper 2 Q9

OCR MEICurrent spec10 marksProbability

9 Labrador puppies may be black, yellow or chocolate in colour. Some information about a litter of 9 puppies is given in the table.

Four puppies are chosen at random to train as guide dogs.

(a) Determine the probability that exactly 3 females are chosen. [3]

(b) Determine the probability that at least 3 black puppies are chosen. [3]

(d) Explain whether the 2 events

‘choosing exactly 3 females’ and ‘choosing at least 3 black puppies’

are independent events. [1]

Mark scheme (a)

Scheme	Marks	AO
\(\dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{3}{7} \times \dfrac{4}{6}\) oe	M1	1.1
\(\times 4\)	M1	1.1
\(\dfrac{40}{126}\) or \(\dfrac{20}{63}\) or 0.317460…isw or rounded to 2 sf or better	A1	1.1
	[3]

M1: M0 if binomial distribution or sampling with replacement used

M1: dependent on award of first M1

Scheme	Marks
\({}_{9}\mathrm{C}_{4}\) [= 126] soi	M1
\({}_{5}\mathrm{C}_{3} \times {}_{4}\mathrm{C}_{1}\) [= 10×4] soi	M1

Mark scheme (b)

Scheme	Marks	AO
\(\dfrac{4}{9} \times \dfrac{3}{8} \times \dfrac{2}{7} \times \dfrac{5}{6} \times 4\) oe or \(\dfrac{4}{9} \times \dfrac{3}{8} \times \dfrac{2}{7} \times \dfrac{1}{6}\) oe	M1 M1	3.1b 1.1
\(\dfrac{1}{6}\) or 0.166666….to 2 sf or better	A1	1.1
	[3]

M1: condone omission of \(\times 4\) in first term;
M0 if binomial distribution or sampling with replacement used

M1: for addition of their terms; dependent on award of first M1

Scheme	Marks
\({}_{4}\mathrm{C}_{4} + {}_{4}\mathrm{C}_{3} \times {}_{5}\mathrm{C}_{1}\)	M1 M1

M1: for \({}_{4}\mathrm{C}_{3} \times {}_{5}\mathrm{C}_{1}\)

M1: for addition of their terms; dependent on award of first M1

NB \(1 + 4 \times 5 = 21\)

Scheme	Marks
\(\dfrac{5}{9} \times \dfrac{4}{8} \times \dfrac{3}{7} \times \dfrac{2}{6} + 4 \times \dfrac{4}{9} \times \dfrac{5}{8} \times \dfrac{4}{7} \times \dfrac{3}{6} + 6 \times \dfrac{4}{9} \times \dfrac{5}{8} \times \dfrac{4}{7} \times \dfrac{3}{6}\) oe	M1
\(1 - \text{their } \dfrac{2520}{3024}\)	M1

M1: for two of these terms

M1: for 1 – the sum of their 3 terms

NB \(\dfrac{2520}{3024} = \dfrac{5}{6}\)

Mark scheme (c)

Scheme	Marks	AO
(3 BF, 1M) + (2BF, 1NBF, 1BM) attempted	M1	3.1b
\(\dfrac{3}{9} \times \dfrac{2}{8} \times \dfrac{1}{7} \times \dfrac{4}{6} \times 4 + \dfrac{3}{9} \times \dfrac{2}{8} \times \dfrac{2}{7} \times \dfrac{1}{6} \times 12 \left[= \dfrac{5}{63}\right]\) oe	A1	1.1
\(\dfrac{10}{21}\) or 0.476190476…to 2 sf or better	A1	1.1
	[3]

M1: M0 if binomial distribution or sampling with replacement used

Scheme	Marks
\({}_{3}\mathrm{C}_{3} \times {}_{4}\mathrm{C}_{1} + {}_{3}\mathrm{C}_{2} \times {}_{2}\mathrm{C}_{1} \times {}_{1}\mathrm{C}_{1}\)	M1
10	A1

M1: for either term

Mark scheme (d)

Scheme	Marks	AO
not independent since \(\dfrac{10}{21} \neq \dfrac{20}{63}\) oe	B1	2.4
	[1]

B1: FT their probabilities
no FT if binomial distribution or sampling with replacement used

Scheme	Marks
not independent since \(\dfrac{20}{63} \times \dfrac{1}{6} \neq \dfrac{5}{63}\)

FT their calculated probabilities from first three parts