| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Principle of Inclusion/Exclusion |
| Type | Three-Set Venn Diagram Probability Calculation |
| Difficulty | Easy -1.3 This is a straightforward Venn diagram reading exercise requiring only basic probability calculations (counting regions and dividing by total). No problem-solving insight needed—students simply read values from the diagram and apply P(A) = n(A)/n(total). The conditional probability in part (ii) is also routine application of P(A|B) = P(A∩B)/P(B). |
| Spec | 2.03b 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 |
| \(P(\text{Watched cyc but not fb}) = \frac{15}{250} = \frac{3}{50} = 0.06\) | B1 | CAO (aef) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{Watched one or two}) = \frac{33+12+21+14+3+65}{250}\) | M1 | OR: \(\frac{250-(64+38)}{250}\). For M1 terms must be added with no extra terms |
| \(= \frac{148}{250} = \frac{74}{125} = 0.592\) | A1 | CAO (aef) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{Not watched fb} \mid \text{watched cyc}) = \frac{15}{67} = 0.224\) | M1 | For denominator of either 67 or 67/250 or 0.268 |
| A1 | CAO (aef). Allow 0.22 with working |
## Question 2:
### Part (i)(A)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Watched cyc but not fb}) = \frac{15}{250} = \frac{3}{50} = 0.06$ | B1 | CAO (aef) |
### Part (i)(B)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Watched one or two}) = \frac{33+12+21+14+3+65}{250}$ | M1 | OR: $\frac{250-(64+38)}{250}$. For M1 terms must be added with no extra terms |
| $= \frac{148}{250} = \frac{74}{125} = 0.592$ | A1 | CAO (aef) |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Not watched fb} \mid \text{watched cyc}) = \frac{15}{67} = 0.224$ | M1 | For denominator of either 67 or 67/250 or 0.268 |
| | A1 | CAO (aef). Allow 0.22 with working |
---2 A survey is being carried out into the sports viewing habits of people in a particular area. As part of the survey, 250 people are asked which of the following sports they have watched on television in the past month.
\begin{itemize}
\item Football
\item Cycling
\item Rugby
\end{itemize}
The numbers of people who have watched these sports are shown in the Venn diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{6015ae6c-bf76-4a0c-af0f-5c53f9c5ed2a-2_723_917_1183_575}
One of the people is selected at random.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that this person has in the past month\\
(A) watched cycling but not football,\\
(B) watched either one or two of the three sports.
\item Given that this person has watched cycling, find the probability that this person has not watched football.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2015 Q2 [5]}}