| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Given conditional, find joint or marginal |
| Difficulty | Moderate -0.8 This is a straightforward conditional probability question requiring standard formula manipulation: P(L∩W) = P(L|W)×P(W), then using P(L∪W) to complete a Venn diagram, and checking independence via P(L∩W) = P(L)×P(W). All steps are routine applications of basic probability rules with no problem-solving insight required, 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: \(P(L \cap W) = P(L | W) \times P(W) = 0.4 \times 0.07 = 0.028\) | |
| Marks | Guidance | |
| M1 | For \(P(L\ | W) \times P(W)\) |
| A1 | cao | |
| [2] |
| Answer | Marks |
|---|---|
| Marks | Guidance |
| B1 | For two labelled intersecting circles |
| B1 | For at least 2 correct probabilities. FT their 0.028 provided \(< 0.038\) |
| B1 | For remaining probabilities |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Marks | Guidance | |
| M1 | For correct use of \(P(L) \times P(W)\). If \(P(L)\) wrong, max M1A0E0. No marks if \(P(W)\) wrong | |
| A1 | For 0.00266 | |
| E1* | For 0.00266. Allow 'they are dependent'. Do not award E1 if \(P(L \cap W)\) wrong | |
| [3] | dep on M1. Or EG \(P(L\ | W) = 0.4, P(L) = 0.038\). Not equal so not independent. M1 is for comparing with some attempt at numbers. If \(P(L)\) wrong, max M1A0E0 |
## Part (i)
**Answer:** $P(L \cap W) = P(L|W) \times P(W) = 0.4 \times 0.07 = 0.028$
| **Marks** | **Guidance** |
|-----------|--------------|
| M1 | For $P(L\|W) \times P(W)$ |
| A1 | cao |
| **[2]** | |
## Part (ii)
**Answer:** Two labelled intersecting circles with:
- L circle: 0.01
- Intersection: 0.028
- W circle: 0.042
- Outside both: 0.92
| **Marks** | **Guidance** |
|-----------|--------------|
| B1 | For two labelled intersecting circles |
| B1 | For at least 2 correct probabilities. FT their 0.028 provided $< 0.038$ |
| B1 | For remaining probabilities |
| **[3]** | |
## Part (iii)
**Answer:** $P(L \cap W) = 0.028$, $P(L) \times P(W) = 0.038 \times 0.07 = 0.00266$
Not equal so not independent
| **Marks** | **Guidance** |
|-----------|--------------|
| M1 | For correct use of $P(L) \times P(W)$. If $P(L)$ wrong, max M1A0E0. No marks if $P(W)$ wrong |
| A1 | For 0.00266 |
| E1* | For 0.00266. Allow 'they are dependent'. Do not award E1 if $P(L \cap W)$ wrong |
| **[3]** | dep on M1. Or EG $P(L\|W) = 0.4, P(L) = 0.038$. Not equal so not independent. M1 is for comparing with some attempt at numbers. If $P(L)$ wrong, max M1A0E0 |
---3 Each weekday Alan drives to work. On his journey, he goes over a level crossing. Sometimes he has to wait at the level crossing for a train to pass.
\begin{itemize}
\item $W$ is the event that Alan has to wait at the level crossing.
\item $L$ is the event that Alan is late for work.
\end{itemize}
You are given that $\mathrm { P } ( L \mid W ) = 0.4 , \mathrm { P } ( W ) = 0.07$ and $\mathrm { P } ( L \cup W ) = 0.08$.\\
(i) Calculate $\mathrm { P } ( L \cap W )$.\\
(ii) Draw a Venn diagram, showing the events $L$ and $W$. Fill in the probability corresponding to each of the four regions of your diagram.\\
(iii) Determine whether the events $L$ and $W$ are independent, explaining your method clearly.
\hfill \mbox{\textit{OCR MEI S1 2013 Q3 [8]}}