2022 H2 Mathematics Paper 2 Question 7

Permutations and Combinations (P&C)
Sampling Theory

Answers

(a)

They form a sample as these 53{53} employees form a subset of the population (all the 75{75} employees).

(b)

She should select a sample (e.g. of size 30{30}) of employees to survey their views and decide on which 30{30} employees are chosen into the sample using a random number generator.

Using a random number generator ensures that the sample selected is random: i.e. each employee has an equal chance to be selected into the sample with all employees in the sample independently chosen. This will give a fairer survey of the employees' views that is free from bias.

(c)

33642.{33642.}

Full solutions

(a)

They form a sample as these 53{53} employees form a subset of the population (all the 75{75} employees). ■{\blacksquare}

(b)

She should select a sample (e.g. of size 30{30}) of employees to survey their views and decide on which 30{30} employees are chosen into the sample using a random number generator.

Using a random number generator ensures that the sample selected is random: i.e. each employee has an equal chance to be selected into the sample with all employees in the sample independently chosen. This will give a fairer survey of the employees' views that is free from bias. ■{\blacksquare}

(c)

Case 1: 5{5} from Administration
Number of ways=(75)(61)(41)(31)=1512\begin{align*} &\textrm{Number of ways} \\ &= { 7 \choose 5 } { 6 \choose 1 } { 4 \choose 1 } { 3 \choose 1 } \\ &= 1512 \end{align*}
Case 2: 4{4} from Administration
Number of ways=(74)(62)(41)(31)+(74)(61)(42)(31)+(74)(61)(41)(32)=12600\begin{align*} &\textrm{Number of ways} \\ &= { 7 \choose 4 } { 6 \choose 2 } { 4 \choose 1 } { 3 \choose 1 } + { 7 \choose 4 } { 6 \choose 1 } { 4 \choose 2 } { 3 \choose 1 } + { 7 \choose 4 } { 6 \choose 1 } { 4 \choose 1 } { 3 \choose 2 } \\ &= 12600 \end{align*}
Case 3: 3{3} from Administration
Number of ways=(73)(62)(42)(31)+(73)(62)(41)(32)+(73)(61)(42)(32)=19530\begin{align*} &\textrm{Number of ways} \\ &= { 7 \choose 3 } { 6 \choose 2 } { 4 \choose 2 } { 3 \choose 1 } + { 7 \choose 3 } { 6 \choose 2 } { 4 \choose 1 } { 3 \choose 2 } + { 7 \choose 3 } { 6 \choose 1 } { 4 \choose 2 } { 3 \choose 2 } \\ &= 19530 \end{align*}
Required number of ways=1512+12600+19530=33642  ■\begin{align*} &\textrm{Required number of ways} \\ &= 1512 + 12600 + 19530 \\ &= 33642 \; \blacksquare \end{align*}

Question Commentary

Theoretical question about sampling have been pretty in trend recently, with similar questions the past two years (2021 paper 2 question 11 and 2020 paper 2 question 9).

In part (c), we have a 5 mark questions on permutations and combinations. Identify the cases (in our approach we had 3 main cases with sub-cases for two of them), and ensuring we don't take shortcuts leading to over-counting will be key to scoring here.