| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Population partition tree diagram |
| Difficulty | Moderate -0.3 This is a straightforward tree diagram question requiring standard conditional probability calculations. Part (a) is routine setup, parts (b)(i-iii) involve basic probability rules (addition, Bayes' theorem), and part (c) requires permutations with given probabilities. All techniques are standard S3 material with no novel insight required, making it slightly easier than average. |
| 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| First branches: C (0.65), V (0.20), L (0.15) | B1 | Correct first-level branches with labels and probabilities |
| Second branches for C: F (0.30), M (0.55), A (0.15) | B1 | Correct second-level branches for cars |
| Second branches for V: F (0.35), M (0.45), A (0.20) and for L: F (0.10), M (0.65), A (0.25) | B1 | Correct second-level branches for vans and lorries |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{car} \cap M) + P(\text{lorry} \cap M) = 0.65 \times 0.55 + 0.15 \times 0.65\) | M1 | Correct products identified |
| \(= 0.3575 + 0.0975 = 0.455\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(A) = 0.65 \times 0.15 + 0.20 \times 0.20 + 0.15 \times 0.25 = 0.0975 + 0.04 + 0.0375 = 0.175\) | M1 | Correct total \(P(A)\) |
| \(P(L \mid A) = \frac{0.15 \times 0.25}{0.175} = \frac{0.0375}{0.175} = \frac{3}{14} \approx 0.214\) | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(\text{not } F \mid \text{not } C) = \frac{P(\text{not } C \cap \text{not } F)}{P(\text{not } C)}\) | M1 | Correct conditional probability structure |
| \(P(\text{not } C) = 0.35\); \(P(\text{not } C \cap \text{not } F) = 0.20 \times 0.65 + 0.15 \times 0.90 = 0.13 + 0.135 = 0.265\) | M1 A1 | Correct numerator and denominator |
| \(= \frac{0.265}{0.35} = \frac{53}{70} \approx 0.757\) | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(F) = 0.65 \times 0.30 + 0.20 \times 0.35 + 0.15 \times 0.10 = 0.195 + 0.07 + 0.015 = 0.28\) | B1 | Correct total \(P(F)\) |
| \(P(\text{C}\mid F) = \frac{0.195}{0.28}\), \(P(\text{V}\mid F) = \frac{0.07}{0.28}\), \(P(\text{L}\mid F) = \frac{0.015}{0.28}\) | M1 | Correct conditional probabilities identified |
| \(3! \times \frac{0.195}{0.28} \times \frac{0.07}{0.28} \times \frac{0.015}{0.28}\) | M1 | Multiply by \(3!\) for arrangements |
| \(= 6 \times \frac{0.195 \times 0.07 \times 0.015}{0.28^3} = \frac{6 \times 0.0002048}{0.021952} \approx 0.0560\) | A1 | Correct final answer |
# Question 3:
## Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| First branches: C (0.65), V (0.20), L (0.15) | B1 | Correct first-level branches with labels and probabilities |
| Second branches for C: F (0.30), M (0.55), A (0.15) | B1 | Correct second-level branches for cars |
| Second branches for V: F (0.35), M (0.45), A (0.20) and for L: F (0.10), M (0.65), A (0.25) | B1 | Correct second-level branches for vans and lorries |
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{car} \cap M) + P(\text{lorry} \cap M) = 0.65 \times 0.55 + 0.15 \times 0.65$ | M1 | Correct products identified |
| $= 0.3575 + 0.0975 = 0.455$ | A1 | Correct answer |
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(A) = 0.65 \times 0.15 + 0.20 \times 0.20 + 0.15 \times 0.25 = 0.0975 + 0.04 + 0.0375 = 0.175$ | M1 | Correct total $P(A)$ |
| $P(L \mid A) = \frac{0.15 \times 0.25}{0.175} = \frac{0.0375}{0.175} = \frac{3}{14} \approx 0.214$ | A1 | Correct final answer |
## Part (b)(iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(\text{not } F \mid \text{not } C) = \frac{P(\text{not } C \cap \text{not } F)}{P(\text{not } C)}$ | M1 | Correct conditional probability structure |
| $P(\text{not } C) = 0.35$; $P(\text{not } C \cap \text{not } F) = 0.20 \times 0.65 + 0.15 \times 0.90 = 0.13 + 0.135 = 0.265$ | M1 A1 | Correct numerator and denominator |
| $= \frac{0.265}{0.35} = \frac{53}{70} \approx 0.757$ | A1 | Correct final answer |
## Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(F) = 0.65 \times 0.30 + 0.20 \times 0.35 + 0.15 \times 0.10 = 0.195 + 0.07 + 0.015 = 0.28$ | B1 | Correct total $P(F)$ |
| $P(\text{C}\mid F) = \frac{0.195}{0.28}$, $P(\text{V}\mid F) = \frac{0.07}{0.28}$, $P(\text{L}\mid F) = \frac{0.015}{0.28}$ | M1 | Correct conditional probabilities identified |
| $3! \times \frac{0.195}{0.28} \times \frac{0.07}{0.28} \times \frac{0.015}{0.28}$ | M1 | Multiply by $3!$ for arrangements |
| $= 6 \times \frac{0.195 \times 0.07 \times 0.015}{0.28^3} = \frac{6 \times 0.0002048}{0.021952} \approx 0.0560$ | A1 | Correct final answer |
I can see these are answer space pages (blank lined pages for student responses) from what appears to be an AQA Statistics exam paper (P71758/Jun14/MS03). The pages shown (7-11) are blank answer spaces for Questions 3 and 4, with no mark scheme content visible.
The actual question text visible is for **Question 4** on page 10:
---3 An investigation was carried out into the type of vehicle being driven when its driver was caught speeding. The investigation was restricted to drivers who were caught speeding when driving vehicles with at least 4 wheels.
An analysis of the results showed that $65 \%$ were driving cars ( C ), $20 \%$ were driving vans (V) and 15\% were driving lorries (L).
Of those driving cars, $30 \%$ were caught by fixed speed cameras (F), 55\% were caught by mobile speed cameras (M) and 15\% were caught by average speed cameras (A).
Of those driving vans, $35 \%$ were caught by fixed speed cameras (F), $45 \%$ were caught by mobile speed cameras (M) and 20\% were caught by average speed cameras (A).
Of those driving lorries, $10 \%$ were caught by fixed speed cameras $( \mathrm { F } )$, $65 \%$ were caught by mobile speed cameras (M) and $25 \%$ were caught by average speed cameras (A).
\begin{enumerate}[label=(\alph*)]
\item Represent this information by a tree diagram on which are shown labels and percentages or probabilities.
\item Hence, or otherwise, calculate the probability that a driver, selected at random from those caught speeding:
\begin{enumerate}[label=(\roman*)]
\item was driving either a car or a lorry and was caught by a mobile speed camera;
\item was driving a lorry, given that the driver was caught by an average speed camera;
\item was not caught by a fixed speed camera, given that the driver was not driving a car.\\[0pt]
[8 marks]
\end{enumerate}\item Three drivers were selected at random from those caught speeding by fixed speed cameras.
Calculate the probability that they were driving three different types of vehicle.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2014 Q3 [12]}}