| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | January |
| 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 the independence test formula P(R∩L) = P(R)×P(L), followed by routine Venn diagram completion and conditional probability calculation P(L|R) = P(R∩L)/P(R). All parts require only direct substitution into standard formulas with no problem-solving or insight needed, making it 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/Working | Marks | Guidance |
| \(P(R) \times P(L) = 0.36 \times 0.25 = 0.09 \neq P(R \cap L)\) | M1 | For \(0.36 \times 0.25\) or 0.09 seen |
| Not equal so not independent. (Allow \(0.36 \times 0.25 \neq 0.2\) or \(0.09 \neq 0.2\) or \(\neq P(R \cap L)\) so not independent) | A1 | Numerical justification needed |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Two overlapping circles labelled R and L | G1 | For two overlapping circles labelled |
| 0.2 and either 0.16 or 0.05 in correct places | G1 | For 0.2 and either 0.16 or 0.05 in correct places |
| All 4 correct probs: 0.16, 0.2, 0.05, 0.59 in correct places | G1 | All 4 correct probs in correct places (including 0.59); last two G marks independent of labels |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(L \mid R) = \frac{P(L \cap R)}{P(R)} = \frac{0.2}{0.36} = \frac{5}{9} = 0.556\) (awrt 0.56) | M1 | For 0.2/0.36 o.e. |
| A1 cao | ||
| This is the probability that Anna is late given that it is raining (must be in context). Condone 'if' or 'when' or 'on a rainy day' for 'given that' but not 'and' or 'because' or 'due to' | E1 | indep of M1A1; order/structure must be correct i.e. no reverse statement |
| TOTAL | 8 |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(R) \times P(L) = 0.36 \times 0.25 = 0.09 \neq P(R \cap L)$ | M1 | For $0.36 \times 0.25$ or 0.09 seen |
| Not equal so not independent. (Allow $0.36 \times 0.25 \neq 0.2$ or $0.09 \neq 0.2$ or $\neq P(R \cap L)$ so not independent) | A1 | Numerical justification needed |
| **Total** | **2** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Two overlapping circles labelled R and L | G1 | For two overlapping circles labelled |
| 0.2 and either 0.16 or 0.05 in correct places | G1 | For 0.2 and either 0.16 or 0.05 in correct places |
| All 4 correct probs: 0.16, 0.2, 0.05, 0.59 in correct places | G1 | All 4 correct probs in correct places (including 0.59); last two G marks independent of labels |
| **Total** | **3** | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(L \mid R) = \frac{P(L \cap R)}{P(R)} = \frac{0.2}{0.36} = \frac{5}{9} = 0.556$ (awrt 0.56) | M1 | For 0.2/0.36 o.e. |
| | A1 cao | |
| This is the probability that Anna is late given that it is raining (must be in context). Condone 'if' or 'when' or 'on a rainy day' for 'given that' but not 'and' or 'because' or 'due to' | E1 | indep of M1A1; order/structure must be correct i.e. no reverse statement |
| **TOTAL** | **8** | |
---5 Each day Anna drives to work.
\begin{itemize}
\item $R$ is the event that it is raining.
\item $L$ is the event that Anna arrives at work late.
\end{itemize}
You are given that $\mathrm { P } ( R ) = 0.36 , \mathrm { P } ( L ) = 0.25$ and $\mathrm { P } ( R \cap L ) = 0.2$.\\
(i) Determine whether the events $R$ and $L$ are independent.\\
(ii) Draw a Venn diagram showing the events $R$ and $L$. Fill in the probability corresponding to each of the four regions of your diagram.\\
(iii) Find $\mathrm { P } ( L \mid R )$. State what this probability represents.
\hfill \mbox{\textit{OCR MEI S1 2009 Q5 [8]}}