| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2008 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Population partition tree diagram |
| Difficulty | Moderate -0.8 This is a straightforward application of tree diagrams and conditional probability with clearly given percentages. Part (a) is routine construction, part (b) uses the law of total probability and Bayes' theorem with simple arithmetic, and part (c) applies the binomial distribution. All steps are standard S3 techniques with no problem-solving insight required. |
| Spec | 2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a) | ||
| Tree diagram with B, S & D with 3 probabilities | B1 | B, S & D with 3 probabilities |
| \(3 \times\) (E, M & W) each with 3 probabilities | B2 | 3 marks |
| \(\geq 1 \times\) (E, M & W) (each) with 3 probabilities | (B1) | |
| Part (b)(i) | ||
| \(P(E) = (0.25 \times 0.3) + (0.6 \times 0.4) + (0.15 \times 0.55)\) | M1 | \(\geq 1\) term correct |
| \(= 0.075 + 0.24 + 0.0825 = 0.397 \text{ to } 0.398 \text{ or } 159/400\) | A1 | AWFW/CAO (0.3975); 2 marks |
| Part (b)(ii) | ||
| \(P(D \mid E) = \frac{0.0825}{\text{(b)(i)}}\) | M1 | Or equivalent |
| \(0.207 \text{ to } 0.208 \text{ or } 11/53\) | A1 | AWFW/CAO (0.2075); 2 marks |
| Part (c) | ||
| \(X \sim B(10, \text{(b)(ii)})\) | M1 | Used |
| \(P(X = 4) = \binom{10}{4}(0.2075)^4(0.7925)^6 = \) | A1↑ | ft on (b)(ii) |
| \(0.0955 \text{ to } 0.0975\) | A1 | AWFW (0.09645); 3 marks |
| **Part (a)** |
|---|
| Tree diagram with B, S & D with 3 probabilities | B1 | B, S & D with 3 probabilities |
| $3 \times$ (E, M & W) each with 3 probabilities | B2 | 3 marks |
| $\geq 1 \times$ (E, M & W) (each) with 3 probabilities | (B1) | |
| **Part (b)(i)** |
|---|
| $P(E) = (0.25 \times 0.3) + (0.6 \times 0.4) + (0.15 \times 0.55)$ | M1 | $\geq 1$ term correct |
| $= 0.075 + 0.24 + 0.0825 = 0.397 \text{ to } 0.398 \text{ or } 159/400$ | A1 | AWFW/CAO (0.3975); 2 marks |
| **Part (b)(ii)** |
|---|
| $P(D \mid E) = \frac{0.0825}{\text{(b)(i)}}$ | M1 | Or equivalent |
| $0.207 \text{ to } 0.208 \text{ or } 11/53$ | A1 | AWFW/CAO (0.2075); 2 marks |
| **Part (c)** |
|---|
| $X \sim B(10, \text{(b)(ii)})$ | M1 | Used |
| $P(X = 4) = \binom{10}{4}(0.2075)^4(0.7925)^6 = $ | A1↑ | ft on (b)(ii) |
| $0.0955 \text{ to } 0.0975$ | A1 | AWFW (0.09645); 3 marks |
**Total: 10 marks**
---4 A manufacturer produces three models of washing machine: basic, standard and deluxe. An analysis of warranty records shows that $25 \%$ of faults are on basic machines, $60 \%$ are on standard machines and 15\% are on deluxe machines.
For basic machines, 30\% of faults reported during the warranty period are electrical, $50 \%$ are mechanical and $20 \%$ are water-related.
For standard machines, 40\% of faults reported during the warranty period are electrical, $45 \%$ are mechanical and 15\% are water-related.
For deluxe machines, $55 \%$ of faults reported during the warranty period are electrical, $35 \%$ are mechanical and $10 \%$ are water-related.
\begin{enumerate}[label=(\alph*)]
\item Draw a tree diagram to represent the above information.
\item Hence, or otherwise, determine the probability that a fault reported during the warranty period:
\begin{enumerate}[label=(\roman*)]
\item is electrical;
\item is on a deluxe machine, given that it is electrical.
\end{enumerate}\item A random sample of 10 electrical faults reported during the warranty period is selected. Calculate the probability that exactly 4 of them are on deluxe machines.
\end{enumerate}
\hfill \mbox{\textit{AQA S3 2008 Q4 [10]}}