Question 6 - A-Level Maths

AQA S1 2005 January — Question 6 14 marks

Exam Board	AQA
Module	S1 (Statistics 1)
Year	2005
Session	January
Marks	14
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Probability Definitions
Type	Two-way table probabilities
Difficulty	Easy -1.2 This is a straightforward two-way table probability question requiring basic probability calculations (simple fractions from table totals), conditional probability, sampling without replacement, and set notation interpretation. All parts follow standard S1 procedures with no novel problem-solving required—purely routine application of definitions and formulas.
Spec	2.03a Mutually exclusive and independent events 2.03c Conditional probability: using diagrams/tables

6 The table below shows the numbers of males and females in each of three employment categories at a university on 31 July 2003.

\cline { 2 - 4 } \multicolumn{1}{c\|}{}	Employment category
\cline { 2 - 4 } \multicolumn{1}{c\|}{}	Managerial	Academic	Support
Male	38	369	303
Female	26	275	643

An employee is selected at random. Determine the probability that the employee is:
1. female;
2. a female academic;
3. either female or academic or both;
4. female, given that the employee is academic.
Three employees are selected at random, without replacement. Determine the probability that:
1. all three employees are male;
2. exactly one employee is male.
The event "employee selected is academic" is denoted by \(A\). The event "employee selected is female" is denoted by \(F\). Describe in context, as simply as possible, the events denoted by:
1. \(F \cap A\);
2. \(F ^ { \prime } \cup A\).
  Surname Other Names
  Centre Number Candidate Number
  Candidate Signature
  General Certificate of Education
  January 2005
  Advanced Subsidiary Examination MS/SS1B AQA
  459:5EMLM
  : 11 P וPII " 1 : : ר
  ALLI.ub c \section*{STATISTICS} Unit Statistics 1B Insert for use in Question 3.
  Fill in the boxes at the top of this page.
  Fasten this insert securely to your answer book. \begin{figure}[h]
  \captionsetup{labelformat=empty} \caption{Scatter diagram for parcel deliveries by a van} \includegraphics[alt={},max width=\textwidth]{7faa4a2d-f5cc-4cc3-a3a9-5d8290ceabdc-8_2420_1664_349_175}
  \end{figure} Figure 1 (for Question 3)

Show mark scheme Show mark scheme source

Question 6:

Part (a)(i):

Answer	Marks	Guidance
Working	Mark	Guidance
\(P(F) = 944/1654\ (= 0.571)\)	M1	Use of

Part (a)(ii):

Answer	Marks	Guidance
Working	Mark	Guidance
\(P(F \cap A) = 275/1654\ (= 0.166)\)	M1	Use of

Part (a)(iii):

Answer	Marks	Guidance
Working	Mark	Guidance
\(P(F \cup A) = \frac{944 + 369}{1654}\)	M1	Use of; OE
\(= 1313/1654\) or \(0.793\) to \(0.795\)	A1	CAO/AWFW (0.7938)

Part (a)(iv):

Answer	Marks	Guidance
Working	Mark	Guidance
\(P(F \mid A) = \frac{\text{their (ii)}}{644/1654}\)	M1	Use of
\(= 275/644\) or \(0.426\) to \(0.428\)	A1	CAO/AWFW (0.4270)

Part (b)(i):

Answer	Marks	Guidance
Working	Mark	Guidance
\(P(MMM) = \frac{710 \times 709 \times 708}{1654 \times 1653 \times 1652}\)	M1	Use of (without replacement)
\(= 0.0788\) to \(0.0791\)	A1	AWFW (no fraction) (0.07891)

Part (b)(ii):

Answer	Marks	Guidance
Working	Mark	Guidance
\(P(MFF) = \frac{710 \times 944 \times 943 \times 3}{1654 \times 1653 \times 1652}\)	M1 M1	Use of one combination of \(MFF\) (without replacement); use of multiplier of 3
\(= 0.419\) to \(0.421\)	A1	AWFW (no fraction) (0.4198)

Part (c)(i):

Answer	Marks	Guidance
Working	Mark	Guidance
Female (and) Academic	B1	CAO

Part (c)(ii):

Answer	Marks	Guidance
Working	Mark	Guidance
Male	B1	Not female \(\Rightarrow\) B0
OR		'OR' must be clearly stated or implied
Academic (or both)	B1	Addition of 'not both' \(\Rightarrow\) B0

# Question 6:

## Part (a)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(F) = 944/1654\ (= 0.571)$ | M1 | Use of |

## Part (a)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(F \cap A) = 275/1654\ (= 0.166)$ | M1 | Use of |

## Part (a)(iii):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(F \cup A) = \frac{944 + 369}{1654}$ | M1 | Use of; OE |
| $= 1313/1654$ or $0.793$ to $0.795$ | A1 | CAO/AWFW (0.7938) |

## Part (a)(iv):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(F \mid A) = \frac{\text{their (ii)}}{644/1654}$ | M1 | Use of |
| $= 275/644$ or $0.426$ to $0.428$ | A1 | CAO/AWFW (0.4270) |

## Part (b)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(MMM) = \frac{710 \times 709 \times 708}{1654 \times 1653 \times 1652}$ | M1 | Use of (without replacement) |
| $= 0.0788$ to $0.0791$ | A1 | AWFW (no fraction) (0.07891) |

## Part (b)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $P(MFF) = \frac{710 \times 944 \times 943 \times 3}{1654 \times 1653 \times 1652}$ | M1 M1 | Use of one combination of $MFF$ (without replacement); use of multiplier of 3 |
| $= 0.419$ to $0.421$ | A1 | AWFW (no fraction) (0.4198) |

## Part (c)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| Female (and) Academic | B1 | CAO |

## Part (c)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| Male | B1 | Not female $\Rightarrow$ B0 |
| OR | | 'OR' must be clearly stated or implied |
| Academic (or both) | B1 | Addition of 'not both' $\Rightarrow$ B0 |

Show LaTeX source

6 The table below shows the numbers of males and females in each of three employment categories at a university on 31 July 2003.

\begin{center}
\begin{tabular}{ | l | c | c | c | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & \multicolumn{3}{c|}{Employment category} \\
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & Managerial & Academic & Support \\
\hline
Male & 38 & 369 & 303 \\
\hline
Female & 26 & 275 & 643 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item An employee is selected at random. Determine the probability that the employee is:
\begin{enumerate}[label=(\roman*)]
\item female;
\item a female academic;
\item either female or academic or both;
\item female, given that the employee is academic.
\end{enumerate}\item Three employees are selected at random, without replacement. Determine the probability that:
\begin{enumerate}[label=(\roman*)]
\item all three employees are male;
\item exactly one employee is male.
\end{enumerate}\item The event "employee selected is academic" is denoted by $A$. The event "employee selected is female" is denoted by $F$.

Describe in context, as simply as possible, the events denoted by:
\begin{enumerate}[label=(\roman*)]
\item $F \cap A$;
\item $F ^ { \prime } \cup A$.

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|}
\hline
Surname & \multicolumn{9}{|c|}{Other Names} \\
\hline
\multicolumn{2}{|c|}{Centre Number} &  &  & \multicolumn{2}{|l|}{Candidate Number} &  &  &  &  \\
\hline
\multicolumn{3}{|c|}{Candidate Signature} &  &  &  &  &  &  &  \\
\hline
\end{tabular}
\end{center}

General Certificate of Education\\
January 2005\\
Advanced Subsidiary Examination

MS/SS1B

AQA\\
459:5EMLM\\
: 11 P וPII " 1 : : ר\\
ALLI.ub c

\section*{STATISTICS}
Unit Statistics 1B

Insert for use in Question 3.\\
Fill in the boxes at the top of this page.\\
Fasten this insert securely to your answer book.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Scatter diagram for parcel deliveries by a van}
  \includegraphics[alt={},max width=\textwidth]{7faa4a2d-f5cc-4cc3-a3a9-5d8290ceabdc-8_2420_1664_349_175}
\end{center}
\end{figure}

Figure 1 (for Question 3)
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA S1 2005 Q6 [14]}}

This paper (6 questions)

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Q1 7 Q2 9 Q3 12 Q4 15 Q5 15 Q6 14

Surname	Other Names
Centre Number		Candidate Number
Candidate Signature