| Exam Board | AQA |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Basic two-way table probability |
| Difficulty | Easy -1.8 This is a straightforward S1 question requiring only basic probability calculations from a two-way table: simple division for parts (a)(i)-(iii), elementary conditional probability for (a)(iv)-(v), and a standard combinations calculation for (b). No problem-solving or conceptual insight needed—purely mechanical application of formulas. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| \multirow{2}{*}{} | Nationality | ||||
| English | Welsh | Scottish | Irish | ||
| \multirow[b]{2}{*}{Playing position} | Forward | 14 | 5 | 2 | 6 |
| Back | 8 | 7 | 2 | 6 | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(\text{Welsh back}) = \frac{7}{50}\) or \(0.14\) | B1 | 1; CAO; OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(\text{English}) = \frac{14+8}{50}\) | B1 | Correct expression; PI |
| \(\frac{22}{50}\) or \(\frac{11}{25}\) or \(0.44\) | B1 | 2; CAO; OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(\text{not English}) = 1 - \text{(ii)}\) | ||
| \(\frac{28}{50}\) or \(\frac{14}{25}\) or \(0.56\) | \(\text{B1}\sqrt{}\) | 1; \(\sqrt{}\) on (ii) if used; \(0 < p < 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(\text{Irish} \mid \text{back}) = \frac{P(\text{Irish} \cap \text{back})}{P(\text{back})} = \frac{6}{\sum(\text{back})}\) | M1 | Used; may be implied by values or answer |
| \(\frac{6}{23}\) or \(0.26\) to \(0.261\) | A1 | 2; CAO/AWFW \((6/50 \Rightarrow 0)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(\text{forward} \mid \text{not Scottish}) = \frac{P(\text{forward} \cap \text{not Scottish})}{P(\text{not Scottish})} = \frac{14+5+6}{50-4} = \frac{27-2}{50-4}\) | M1 | Used; OE. May be implied by values or answer |
| \(\frac{25}{46}\) or \(0.54\) to \(0.544\) | A1 | 2; CAO/AWFW \((25/50 \Rightarrow 0)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(4 \times \text{English}) = \left(\frac{22}{50}\right)\times\left(\frac{21}{49}\right)\times\left(\frac{20}{48}\right)\times\left(\frac{19}{47}\right)\) | M1 M1 | Reducing non-tabulated value 4 times; Reducing 50 and multiplying 4 terms (ignore multipliers) |
| \(\frac{175560}{5527200}\) or \(\frac{209}{6580}\) or \(0.0317\) to \(0.032\) | A1 | 3; CAO/AWFW |
## Question 2:
**Part (a)(i):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{Welsh back}) = \frac{7}{50}$ or $0.14$ | B1 | 1; CAO; OE |
**Part (a)(ii):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{English}) = \frac{14+8}{50}$ | B1 | Correct expression; PI |
| $\frac{22}{50}$ or $\frac{11}{25}$ or $0.44$ | B1 | 2; CAO; OE |
**Part (a)(iii):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{not English}) = 1 - \text{(ii)}$ | | |
| $\frac{28}{50}$ or $\frac{14}{25}$ or $0.56$ | $\text{B1}\sqrt{}$ | 1; $\sqrt{}$ on (ii) if used; $0 < p < 1$ |
**Part (a)(iv):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{Irish} \mid \text{back}) = \frac{P(\text{Irish} \cap \text{back})}{P(\text{back})} = \frac{6}{\sum(\text{back})}$ | M1 | Used; may be implied by values or answer |
| $\frac{6}{23}$ or $0.26$ to $0.261$ | A1 | 2; CAO/AWFW $(6/50 \Rightarrow 0)$ |
**Part (a)(v):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\text{forward} \mid \text{not Scottish}) = \frac{P(\text{forward} \cap \text{not Scottish})}{P(\text{not Scottish})} = \frac{14+5+6}{50-4} = \frac{27-2}{50-4}$ | M1 | Used; OE. May be implied by values or answer |
| $\frac{25}{46}$ or $0.54$ to $0.544$ | A1 | 2; CAO/AWFW $(25/50 \Rightarrow 0)$ |
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(4 \times \text{English}) = \left(\frac{22}{50}\right)\times\left(\frac{21}{49}\right)\times\left(\frac{20}{48}\right)\times\left(\frac{19}{47}\right)$ | M1 M1 | Reducing non-tabulated value 4 times; Reducing 50 and multiplying 4 terms (ignore multipliers) |
| $\frac{175560}{5527200}$ or $\frac{209}{6580}$ or $0.0317$ to $0.032$ | A1 | 3; CAO/AWFW |
---2 The British and Irish Lions 2005 rugby squad contained 50 players. The nationalities and playing positions of these players are shown in the table.
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{4}{|c|}{Nationality} \\
\hline
& & English & Welsh & Scottish & Irish \\
\hline
\multirow[b]{2}{*}{Playing position} & Forward & 14 & 5 & 2 & 6 \\
\hline
& Back & 8 & 7 & 2 & 6 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item A player was selected at random from the squad for a radio interview. Calculate the probability that the player was:
\begin{enumerate}[label=(\roman*)]
\item a Welsh back;
\item English;
\item not English;
\item Irish, given that the player was a back;
\item a forward, given that the player was not Scottish.
\end{enumerate}\item Four players were selected at random from the squad to visit a school. Calculate the probability that all four players were English.
\end{enumerate}
\hfill \mbox{\textit{AQA S1 2007 Q2 [11]}}