| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Independent Events |
| Type | Test independence using definition |
| Difficulty | Moderate -0.8 This is a straightforward application of the independence definition P(R∩L) = P(R)×P(L), requiring only basic probability calculations and a standard Venn diagram. The conditional probability in part (iii) is also routine. All parts involve direct recall and application of formulas with minimal problem-solving. |
| 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 |
|---|---|---|
| \(P(R) \times P(L) = 0.36 \times 0.25 = 0.09 \neq P(R \cap L)\); Not equal so not independent | M1 for \(0.36 \times 0.25\) or \(0.09\) seen; A1 (numerical justification needed) | Allow \(0.36 \times 0.25 \neq 0.2\) or \(0.09 \neq 0.2\) or \(\neq p(R \cap L)\) so not independent |
| Answer | Marks | Guidance |
|---|---|---|
| Two overlapping circles labelled \(R\) and \(L\) with values \(0.166\), \(0.2\), \(0.05\), \(0.59\) | G1 for two overlapping circles labelled; G1 for \(0.2\) and either \(0.16\) or \(0.05\) in correct places; G1 for all 4 correct probs in correct places (including \(0.59\)) | The last two G marks are independent of the labels |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(L \mid R) = \dfrac{P(L \cap R)}{P(R)} = \dfrac{0.2}{0.36} = \dfrac{5}{9} = 0.556\) (awrt 0.56); This is the probability that Anna is late given that it is raining | M1 for \(0.2/0.36\) o.e.; A1 cao; E1 (independent of M1A1) | Order/structure must be correct i.e. no reverse statement; Condone 'if'/'when'/'on a rainy day' for 'given that' but not 'and'/'because'/'due to' |
## Question 2:
### Part (i)
| $P(R) \times P(L) = 0.36 \times 0.25 = 0.09 \neq P(R \cap L)$; Not equal so not independent | M1 for $0.36 \times 0.25$ or $0.09$ seen; A1 (numerical justification needed) | Allow $0.36 \times 0.25 \neq 0.2$ or $0.09 \neq 0.2$ or $\neq p(R \cap L)$ so not independent |
### Part (ii)
| Two overlapping circles labelled $R$ and $L$ with values $0.166$, $0.2$, $0.05$, $0.59$ | G1 for two overlapping circles labelled; G1 for $0.2$ and either $0.16$ or $0.05$ in correct places; G1 for all 4 correct probs in correct places (including $0.59$) | The last two G marks are independent of the labels |
### Part (iii)
| $P(L \mid R) = \dfrac{P(L \cap R)}{P(R)} = \dfrac{0.2}{0.36} = \dfrac{5}{9} = 0.556$ (awrt 0.56); This is the probability that Anna is late given that it is raining | M1 for $0.2/0.36$ o.e.; A1 cao; E1 (independent of M1A1) | Order/structure must be correct i.e. no reverse statement; Condone 'if'/'when'/'on a rainy day' for 'given that' but not 'and'/'because'/'due to' |
---2 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 Q2 [8]}}