| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Dice/random device selects population |
| Difficulty | Moderate -0.8 This is a straightforward conditional probability question requiring application of the law of total probability and Bayes' theorem to a clearly structured table. All parts involve direct calculation from given percentages with no conceptual challenges—part (a) is simple complement, parts (b)-(c) use basic probability rules, and part (d) applies standard conditional probability formula. The table organization makes it easier than average A-level probability questions. |
| Spec | 2.03a Mutually exclusive and independent events2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| \multirow{2}{*}{} | Percentage of visitors using | |||
| Road | Funicular railway | Cable car | ||
| \multirow{3}{*}{Age (years)} | Under 18 | 15 | 25 | 10 |
| 18 to 64 | 80 | 60 | 55 | |
| Over 64 | 5 | 15 | 35 | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(\geq 18 \mid \text{Road}) = 0.85\) | B1 | CAO; OE; not 85 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(18 \text{ to } 64) = P(\text{Route}) \times P(18 \text{ to } 64 \mid \text{Route})\) | M1 | Use of 3 possibilities, each the product of 2 probabilities |
| \((0.25 \times 0.80) + (0.60 \times 0.35) + (0.55 \times 0.40)\) | A1 | At least 1 term correct |
| \(= 0.20 + 0.21 + 0.22 = 0.63\) | A1 | CAO; OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(FR \cap {>}64) = P(FR) \times P({>}64 \mid FR) = 0.35 \times 0.15\) | B1 | Correct expression |
| \(= 0.052\) to \(0.053\) | B1 | AWFW \((0.0525)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(FR \mid {>}64) = \dfrac{\text{(c)}}{P({>}64)}\) | M1, M1 | \(\dfrac{\text{answer(c)}}{\sum(3\times 2) \text{ probabilities}}\) |
| \(\dfrac{0.0525}{(0.25\times0.05)+(0.35\times0.15)+(0.40\times0.35)}\) | A1 | At least 2 terms correct |
| \(= \dfrac{0.0525}{0.0125+0.0525+0.1400} = \dfrac{0.0525}{0.205}\) | A1 | CAO |
| \(= 0.256\) or \(\dfrac{21}{82}\) | A1 | AWRT/CAO; OE |
## Question 2:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(\geq 18 \mid \text{Road}) = 0.85$ | B1 | CAO; OE; **not** 85 |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(18 \text{ to } 64) = P(\text{Route}) \times P(18 \text{ to } 64 \mid \text{Route})$ | M1 | Use of 3 possibilities, each the product of 2 probabilities |
| $(0.25 \times 0.80) + (0.60 \times 0.35) + (0.55 \times 0.40)$ | A1 | At least 1 term correct |
| $= 0.20 + 0.21 + 0.22 = 0.63$ | A1 | CAO; OE |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(FR \cap {>}64) = P(FR) \times P({>}64 \mid FR) = 0.35 \times 0.15$ | B1 | Correct expression |
| $= 0.052$ to $0.053$ | B1 | AWFW $(0.0525)$ |
### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(FR \mid {>}64) = \dfrac{\text{(c)}}{P({>}64)}$ | M1, M1 | $\dfrac{\text{answer(c)}}{\sum(3\times 2) \text{ probabilities}}$ |
| $\dfrac{0.0525}{(0.25\times0.05)+(0.35\times0.15)+(0.40\times0.35)}$ | A1 | At least 2 terms correct |
| $= \dfrac{0.0525}{0.0125+0.0525+0.1400} = \dfrac{0.0525}{0.205}$ | A1 | CAO |
| $= 0.256$ or $\dfrac{21}{82}$ | A1 | AWRT/CAO; OE |
---2 A hill-top monument can be visited by one of three routes: road, funicular railway or cable car. The percentages of visitors using these routes are 25, 35 and 40 respectively.
The age distribution, in percentages, of visitors using each route is shown in the table. For example, 15 per cent of visitors using the road were under 18 .
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{3}{|c|}{Percentage of visitors using} \\
\hline
& & Road & Funicular railway & Cable car \\
\hline
\multirow{3}{*}{Age (years)} & Under 18 & 15 & 25 & 10 \\
\hline
& 18 to 64 & 80 & 60 & 55 \\
\hline
& Over 64 & 5 & 15 & 35 \\
\hline
\end{tabular}
\end{center}
Calculate the probability that a randomly selected visitor:
\begin{enumerate}[label=(\alph*)]
\item who used the road is aged 18 or over;
\item is aged between 18 and 64;
\item used the funicular railway and is aged over 64;
\item used the funicular railway, given that the visitor is aged over 64.
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2007 Q2 [11]}}