| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Independence test with conditional probability |
| Difficulty | Easy -1.2 This is a straightforward S1 probability question testing basic concepts: independence check (comparing P(L|R) to P(L)), conditional probability formula P(L∩R) = P(L|R)×P(R), and constructing a Venn diagram. All parts are routine applications of standard formulas 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 | Marks | Guidance |
| Because \(P(L \mid R) \neq P(L)\) | E1 | If two or more methods given and only one correct, do not award the mark. Allow \(0.45 \neq 0.15\). Look out for complement methods, etc |
| Allow 45 ÷ 15 | ||
| Total | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(L \cap R) = P(L \mid R) \times P(R) = 0.45 \times 0.22 = 0.099\) | M1 | For product. Allow if done correctly in part(i) |
| A1 | CAO. Allow 99/1000 | |
| Total | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Venn diagram with two labelled intersecting circles, provided no incorrect labelling | G1 | Condone labels such as \(P(L)\) etc. Allow other shapes in place of circles. No need for 'box' |
| With regions labelled: 0.051, 0.099, 0.121 | G1 | FT from 0.033 in (ii) gives 0.117, 0.033, 0.187, 0.663. In general \(0.15 - x\), \(x\), \(0.22 - x\), \(0.63 + x\). May also see 0.0825, 0.0675, 0.1525, 0.6975 |
| 0.729 outside the circles | G1 | For at least 2 correct probabilities. FT their \(P(L \cap R)\) from part (ii) provided \(\leq 0.15\) |
| For remaining probabilities. FT their \(P(L \cap R)\) providing probabilities between 0 and 1. | ||
| Total | [3] |
## (i)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| Because $P(L \mid R) \neq P(L)$ | E1 | If two or more methods given and only one correct, do not award the mark. Allow $0.45 \neq 0.15$. Look out for complement methods, etc |
| | | Allow 45 ÷ 15 |
| **Total** | **[1]** | |
## (ii)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| $P(L \cap R) = P(L \mid R) \times P(R) = 0.45 \times 0.22 = 0.099$ | M1 | For product. Allow if done correctly in part(i) |
| | A1 | CAO. Allow 99/1000 |
| **Total** | **[2]** | |
## (iii)
| **Answer** | **Marks** | **Guidance** |
|---|---|---|
| Venn diagram with two labelled intersecting circles, provided no incorrect labelling | G1 | Condone labels such as $P(L)$ etc. Allow other shapes in place of circles. No need for 'box' |
| With regions labelled: 0.051, 0.099, 0.121 | G1 | FT from 0.033 in (ii) gives 0.117, 0.033, 0.187, 0.663. In general $0.15 - x$, $x$, $0.22 - x$, $0.63 + x$. May also see 0.0825, 0.0675, 0.1525, 0.6975 |
| 0.729 outside the circles | G1 | For at least 2 correct probabilities. FT their $P(L \cap R)$ from part (ii) provided $\leq 0.15$ |
| | | For remaining probabilities. FT their $P(L \cap R)$ providing probabilities between 0 and 1. |
| **Total** | **[3]** | |
---Each weekday, Marta travels to school by bus. Sometimes she arrives late.
• $L$ is the event that Marta arrives late.
• $R$ is the event that it is raining.
You are given that $\mathrm{P}(L) = 0.15$, $\mathrm{P}(R) = 0.22$ and $\mathrm{P}(L \mid R) = 0.45$.
\begin{enumerate}[label=(\roman*)]
\item Use this information to show that the events $L$ and $R$ are not independent. [1]
\item Find $\mathrm{P}(L \cap R)$. [2]
\item Draw a Venn diagram showing the events $L$ and $R$, and fill in the probability corresponding to each of the four regions of your diagram. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2014 Q3 [6]}}