| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tree Diagrams |
| Type | Population partition tree diagram |
| Difficulty | Moderate -0.8 This is a straightforward tree diagram question requiring basic probability rules (multiplication along branches, addition of mutually exclusive outcomes) and understanding of independence. The calculations are routine with no conceptual challenges beyond standard S1 material, making it easier than average for A-level. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct 2-branch then 4-branch shape | B1 | First B1: for 2 branch then 4 branch shape |
| Labels showing stitching and 0.03 value correctly placed | B1 | Second B1 (dep. on 1st B1): labels showing stitching (accept letters) and 0.03 value correctly placed |
| Labels showing splitting and 0.7 and 0.02 correctly placed | B1 | Third B1 (dep. on 1st B1): labels showing splitting and 0.7 and 0.02 correctly placed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{Exactly one defect}) = 0.03 \times 0.3 + 0.97 \times 0.02\) or \(P(PS \cup Split) - 2P(PS \cap Split)\) | M1A1ft | M1 for \(0.03 \times p + 0.02 \times q\) where \(p\) and \(q\) follow from their tree diagram |
| \(= [0.009 + 0.0194] = \mathbf{0.0284}\) | A1 cao | 1st A1ft for fully correct expression. Accept \(1-0.7\) for 0.3 and \(1-0.03\) for 0.97. 2nd A1 cao for 0.0284 only (or exact equivalent such as \(\frac{71}{2500}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{No defects}) = (1-0.03)\times(1-0.02)\times(1-0.05)\) (or better) | M1 | for (their \(0.97\))\(\times\)(their \(0.98\))\(\times(1-0.05)\) f.t. values from their tree diagram |
| \(= 0.90307\) awrt \(\mathbf{0.903}\) | A1 cao | for awrt 0.903 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{Exactly one defect}) = (b)\times(1-0.05) + (1-0.03)\times(1-0.02)\times0.05\) | M1 M1 | 1st M1 for one correct triple (or correct ft from their tree): \([0.03\times0.3\times(1-0.05)]+[0.97\times0.02\times(1-0.05)]+[0.97\times0.98\times0.05]\) |
| \(= \text{"0.0284"}\times0.95 + 0.97\times0.98\times0.05\) | A1ft | 2nd M1 for two correct triples or correct ft from their tree and adding or their \((b)\times(1-0.05)\) |
| \(= [0.02698 + 0.04753] = 0.07451\) awrt \(\mathbf{0.0745}\) | A1 cao | 1st A1ft for fully correct expression or f.t. their (b) and 0.2 or 0.3 MR only. 2nd A1 cao for awrt 0.0745 |
## Question 7:
### Part (a) - Tree Diagram
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct 2-branch then 4-branch shape | B1 | First B1: for 2 branch then 4 branch shape |
| Labels showing stitching and 0.03 value correctly placed | B1 | Second B1 (dep. on 1st B1): labels showing stitching (accept letters) and 0.03 value correctly placed |
| Labels showing splitting and 0.7 and 0.02 correctly placed | B1 | Third B1 (dep. on 1st B1): labels showing splitting and 0.7 and 0.02 correctly placed |
*Note: Probabilities shown in brackets are not required and any such values given can be ignored in (a). Allow MR of 0.2 for 0.02 or 0.3 for 0.03 on tree diagram to score all M and A1ft marks only.*
---
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Exactly one defect}) = 0.03 \times 0.3 + 0.97 \times 0.02$ or $P(PS \cup Split) - 2P(PS \cap Split)$ | M1A1ft | M1 for $0.03 \times p + 0.02 \times q$ where $p$ and $q$ follow from their tree diagram |
| $= [0.009 + 0.0194] = \mathbf{0.0284}$ | A1 cao | 1st A1ft for fully correct expression. Accept $1-0.7$ for 0.3 and $1-0.03$ for 0.97. 2nd A1 cao for 0.0284 only (or exact equivalent such as $\frac{71}{2500}$) |
---
### Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{No defects}) = (1-0.03)\times(1-0.02)\times(1-0.05)$ (or better) | M1 | for (their $0.97$)$\times$(their $0.98$)$\times(1-0.05)$ f.t. values from their tree diagram |
| $= 0.90307$ **awrt $\mathbf{0.903}$** | A1 cao | for awrt 0.903 |
*Note: Do not allow 0.5 as MR of 0.05 so no M or A marks in (c) or (d)*
---
### Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Exactly one defect}) = (b)\times(1-0.05) + (1-0.03)\times(1-0.02)\times0.05$ | M1 M1 | 1st M1 for one correct triple (or correct ft from their tree): $[0.03\times0.3\times(1-0.05)]+[0.97\times0.02\times(1-0.05)]+[0.97\times0.98\times0.05]$ |
| $= \text{"0.0284"}\times0.95 + 0.97\times0.98\times0.05$ | A1ft | 2nd M1 for two correct triples or correct ft from their tree and adding or their $(b)\times(1-0.05)$ |
| $= [0.02698 + 0.04753] = 0.07451$ **awrt $\mathbf{0.0745}$** | A1 cao | 1st A1ft for fully correct expression or f.t. their (b) and 0.2 or 0.3 MR only. 2nd A1 cao for awrt 0.0745 |\begin{enumerate}
\item A manufacturer carried out a survey of the defects in their soft toys. It is found that the probability of a toy having poor stitching is 0.03 and that a toy with poor stitching has a probability of 0.7 of splitting open. A toy without poor stitching has a probability of 0.02 of splitting open.\\
(a) Draw a tree diagram to represent this information.\\
(b) Find the probability that a randomly chosen soft toy has exactly one of the two defects, poor stitching or splitting open.\\
(3)
\end{enumerate}
The manufacturer also finds that soft toys can become faded with probability 0.05 and that this defect is independent of poor stitching or splitting open. A soft toy is chosen at random.\\
(c) Find the probability that the soft toy has none of these 3 defects.\\
(d) Find the probability that the soft toy has exactly one of these 3 defects.
\hfill \mbox{\textit{Edexcel S1 2012 Q7 [12]}}