Pre-U Pre-U 9794/3 2016 Specimen — Question 2 11 marks

Exam BoardPre-U
ModulePre-U 9794/3 (Pre-U Mathematics Paper 3)
Year2016
SessionSpecimen
Marks11
TopicCombinations & Selection
TypeCommittee with gender/category constraints
DifficultyModerate -0.8 This question combines basic set theory/Venn diagrams with straightforward combinations. Part (a) requires simple probability calculations from a two-way table, while part (b) involves standard combination formulas with minimal constraints. All techniques are routine A-level material with no novel problem-solving required, making it easier than average.
Spec2.03c Conditional probability: using diagrams/tables5.01a Permutations and combinations: evaluate probabilities

2
  1. A music club has 200 members. 75 members play the piano, 130 members like Elgar, and 30 members do not play the piano, nor do they like Elgar.
    1. Calculate the probability that a member chosen at random plays the piano but does not like Elgar.
    2. Calculate the probability that a member chosen at random plays the piano given that this member likes Elgar.
  2. The music club is organising a concert. The programme is to consist of 7 pieces of music which are to be selected from 9 classical pieces and 6 modern pieces. Find the number of different concert programmes than can be produced if
    1. there are no restrictions,
    2. the programme must consist of 5 classical pieces and 2 modern pieces,
    3. there are to be more modern pieces than classical pieces.

Question 2(a)
(i) \(75 - x + x + 130 - x = 170\)
\(x = 35\) (Finding the intersection) [M1]
State \(75 - 35\) o.e. [A1]
\(\frac{40}{200}\) o.e. [A1]
(ii) Use conditional probability [M1]
\(\frac{\text{their } 35}{\text{their } 130}\)
\(= \frac{35}{130}\) o.e. [A1]
(b)(i) Recognise combination problem [M1]
\(^{15}C_7 = \frac{15!}{8!7!} = 6435\) [A1]
(ii) \(^6C_2 \times {^9C_5}\) correct method [M1]
\(= 1890\) [A1]
(iii) \((6M\ 1C) + (5M\ 2C) + (4M\ 3C)\) correct method [M1]
\(^6C_6 \times {^9C_1} + {^6C_5} \times {^9C_2} + {^6C_4} \times {^9C_3}\) [M1]
\(= 1485\) [A1]
Total: 11 marks
**Question 2(a)**

(i) $75 - x + x + 130 - x = 170$

$x = 35$ (Finding the intersection) [M1]

State $75 - 35$ o.e. [A1]

$\frac{40}{200}$ o.e. [A1]

(ii) Use conditional probability [M1]

$\frac{\text{their } 35}{\text{their } 130}$

$= \frac{35}{130}$ o.e. [A1]

**(b)(i)** Recognise combination problem [M1]

$^{15}C_7 = \frac{15!}{8!7!} = 6435$ [A1]

(ii) $^6C_2 \times {^9C_5}$ correct method [M1]

$= 1890$ [A1]

(iii) $(6M\ 1C) + (5M\ 2C) + (4M\ 3C)$ correct method [M1]

$^6C_6 \times {^9C_1} + {^6C_5} \times {^9C_2} + {^6C_4} \times {^9C_3}$ [M1]

$= 1485$ [A1]

**Total: 11 marks**
2
\begin{enumerate}[label=(\alph*)]
\item A music club has 200 members. 75 members play the piano, 130 members like Elgar, and 30 members do not play the piano, nor do they like Elgar.
\begin{enumerate}[label=(\roman*)]
\item Calculate the probability that a member chosen at random plays the piano but does not like Elgar.
\item Calculate the probability that a member chosen at random plays the piano given that this member likes Elgar.
\end{enumerate}\item The music club is organising a concert. The programme is to consist of 7 pieces of music which are to be selected from 9 classical pieces and 6 modern pieces. Find the number of different concert programmes than can be produced if
\begin{enumerate}[label=(\roman*)]
\item there are no restrictions,
\item the programme must consist of 5 classical pieces and 2 modern pieces,
\item there are to be more modern pieces than classical pieces.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2016 Q2 [11]}}