| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2011 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Basic two-way table probability |
| Difficulty | Easy -1.3 This is a straightforward S1 question testing basic probability calculations from a two-way table. Parts (a)(i)-(iii) are simple division, (a)(iv)-(v) are standard conditional probability, and parts (b)-(c) involve routine combinations. All techniques are direct applications of formulas with no problem-solving or insight required—easier than average A-level content. |
| Spec | 2.03d Calculate conditional probability: from first principles |
| \multirow{2}{*}{} | Political Party | |||||
| Labour | Conservative | Liberal Democrat | Other | Total | ||
| \multirow{2}{*}{Gender} | Male | 255 | 175 | 54 | 35 | 519 |
| Female | 94 | 18 | 9 | 5 | 126 | |
| Total | 349 | 193 | 63 | 40 | 645 | |
| Answer | Marks |
|---|---|
| \(P(\text{male Conservative}) = \frac{175}{645} = 0.271\) | B1 |
| Answer | Marks |
|---|---|
| \(P(\text{male}) = \frac{519}{645} = 0.805\) | B1 |
| Answer | Marks |
|---|---|
| \(P(\text{Liberal Democrat}) = \frac{63}{645} = 0.0977\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{Labour} \mid \text{female}) = \frac{94}{126} = 0.746\) | M1 A1 | M1 for \(\frac{94}{126}\) seen |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{not Labour}) = \frac{296}{645}\) | M1 | |
| \(P(\text{male and not Labour}) = \frac{264}{645}\) | M1 | |
| \(P(\text{male} \mid \text{not Labour}) = \frac{264}{296} = 0.892\) | A1 | awrt \(0.892\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{both Labour female}) = \frac{94}{645} \times \frac{93}{644}\) | M1 | |
| \(= 0.0211\) | A1 | awrt \(0.0211\) |
| Answer | Marks | Guidance |
|---|---|---|
| Ways to select exactly 1 Labour and 1 Conservative from 3: need 1 Labour, 1 Conservative, 1 other | M1 | |
| \(P = \frac{\binom{349}{1}\binom{193}{1}\binom{103}{1}}{\binom{645}{3}}\) | M1 | |
| \(= \frac{349 \times 193 \times 103}{645 \times 644 \times 643 / 6}\) | M1 | |
| \(= 0.0977\) | A1 | awrt \(0.0977\) |
# Question 2:
## Part (a)(i)
| $P(\text{male Conservative}) = \frac{175}{645} = 0.271$ | B1 | |
## Part (a)(ii)
| $P(\text{male}) = \frac{519}{645} = 0.805$ | B1 | |
## Part (a)(iii)
| $P(\text{Liberal Democrat}) = \frac{63}{645} = 0.0977$ | B1 | |
## Part (a)(iv)
| $P(\text{Labour} \mid \text{female}) = \frac{94}{126} = 0.746$ | M1 A1 | M1 for $\frac{94}{126}$ seen |
## Part (a)(v)
| $P(\text{not Labour}) = \frac{296}{645}$ | M1 | |
| $P(\text{male and not Labour}) = \frac{264}{645}$ | M1 | |
| $P(\text{male} \mid \text{not Labour}) = \frac{264}{296} = 0.892$ | A1 | awrt $0.892$ |
## Part (b)
| $P(\text{both Labour female}) = \frac{94}{645} \times \frac{93}{644}$ | M1 | |
| $= 0.0211$ | A1 | awrt $0.0211$ |
## Part (c)
| Ways to select exactly 1 Labour and 1 Conservative from 3: need 1 Labour, 1 Conservative, 1 other | M1 | |
| $P = \frac{\binom{349}{1}\binom{193}{1}\binom{103}{1}}{\binom{645}{3}}$ | M1 | |
| $= \frac{349 \times 193 \times 103}{645 \times 644 \times 643 / 6}$ | M1 | |
| $= 0.0977$ | A1 | awrt $0.0977$ |2 The number of MPs in the House of Commons was 645 at the beginning of August 2009. The genders of these MPs and the political parties to which they belonged are shown in the table.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{4}{|c|}{Political Party} & \\
\hline
& & Labour & Conservative & Liberal Democrat & Other & Total \\
\hline
\multirow{2}{*}{Gender} & Male & 255 & 175 & 54 & 35 & 519 \\
\hline
& Female & 94 & 18 & 9 & 5 & 126 \\
\hline
& Total & 349 & 193 & 63 & 40 & 645 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item One MP was selected at random for an interview. Calculate, to three decimal places, the probability that the MP was:
\begin{enumerate}[label=(\roman*)]
\item a male Conservative;
\item a male;
\item a Liberal Democrat;
\item Labour, given that the MP was female;
\item male, given that the MP was not Labour.
\end{enumerate}\item Two female MPs were selected at random for an enquiry. Calculate, to three decimal places, the probability that both MPs were Labour.
\item Three MPs were selected at random for a committee. Calculate, to three decimal places, the probability that exactly one MP was Labour and exactly one MP was Conservative.
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2011 Q2 [14]}}