| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Independence test requiring preliminary calculations |
| Difficulty | Moderate -0.8 This is a straightforward application of standard independence test and conditional probability formulas with all values given. Parts (i)-(iii) require only direct substitution into P(W)P(C) vs P(W∩C) and P(W|C) = P(W∩C)/P(C). Part (iv) asks for a simple comparison. No problem-solving or insight needed—purely routine S1 material. |
| 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 |
|---|---|---|
| \(P(W) \times P(C) = 0.20 \times 0.17 = 0.034\); \(P(W \cap C) = 0.06\) (given); Not equal so not independent | M1 for multiplying or \(0.034\) seen; A1 (numerical justification needed) | Allow \(0.20 \times 0.17 \neq 0.06\) or \(\neq p(W \cap C)\) so not independent |
| Answer | Marks | Guidance |
|---|---|---|
| Two overlapping circles labelled \(W\) and \(C\) with values \(0.14\), \(0.06\), \(0.11\), \(0.69\) | G1 for two overlapping circles labelled; G1 for \(0.06\) and either \(0.14\) or \(0.11\) in correct places; G1 for all 4 correct probs in correct places (including \(0.69\)) | NB No credit for Karnaugh maps here; Last two G marks independent of labels |
| Answer | Marks |
|---|---|
| \(P(W \mid C) = \dfrac{P(W \cap C)}{P(C)} = \dfrac{0.06}{0.17} = \dfrac{6}{17} = 0.353\) (awrt 0.35) | M1 for \(0.06/0.17\); A1 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Children are more likely than adults to be able to speak Welsh; or 'proportionally more children speak Welsh than adults' | E1 FT once correct idea seen, apply ISW | Do not accept: 'more Welsh children speak Welsh than adults' |
## Question 3:
### Part (i)
| $P(W) \times P(C) = 0.20 \times 0.17 = 0.034$; $P(W \cap C) = 0.06$ (given); Not equal so not independent | M1 for multiplying or $0.034$ seen; A1 (numerical justification needed) | Allow $0.20 \times 0.17 \neq 0.06$ or $\neq p(W \cap C)$ so not independent |
### Part (ii)
| Two overlapping circles labelled $W$ and $C$ with values $0.14$, $0.06$, $0.11$, $0.69$ | G1 for two overlapping circles labelled; G1 for $0.06$ and either $0.14$ or $0.11$ in correct places; G1 for all 4 correct probs in correct places (including $0.69$) | NB No credit for Karnaugh maps here; Last two G marks independent of labels |
### Part (iii)
| $P(W \mid C) = \dfrac{P(W \cap C)}{P(C)} = \dfrac{0.06}{0.17} = \dfrac{6}{17} = 0.353$ (awrt 0.35) | M1 for $0.06/0.17$; A1 cao | |
### Part (iv)
| Children are more likely than adults to be able to speak Welsh; or 'proportionally more children speak Welsh than adults' | E1 FT once correct idea seen, apply ISW | Do not accept: 'more Welsh children speak Welsh than adults' |
---3 In the 2001 census, people living in Wales were asked whether or not they could speak Welsh. A resident of Wales is selected at random.
\begin{itemize}
\item $W$ is the event that this person speaks Welsh.
\item $C$ is the event that this person is a child.
\end{itemize}
You are given that $\mathrm { P } ( W ) = 0.20 , \mathrm { P } ( C ) = 0.17$ and $\mathrm { P } ( W \cap C ) = 0.06$.\\
(i) Determine whether the events $W$ and $C$ are independent.\\
(ii) Draw a Venn diagram, showing the events $W$ and $C$, and fill in the probability corresponding to each region of your diagram.\\
(iii) Find $\mathrm { P } ( W \mid C )$.\\
(iv) Given that $\mathrm { P } \left( W \mid C ^ { \prime } \right) = 0.169$, use this information and your answer to part (iii) to comment very briefly on how the ability to speak Welsh differs between children and adults. [1]
\hfill \mbox{\textit{OCR MEI S1 Q3 [8]}}