| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conditional Probability |
| Type | Sampling without replacement from bags/boxes |
| Difficulty | Moderate -0.8 This is a straightforward conditional probability question from S1. Part (i) requires simple division (25/37) using the definition of conditional probability. Part (ii) involves basic probability without replacement but with clear structure: P(both Jaguars|first female) = (15/23) × (14/59). Both parts are routine applications of standard probability formulas with no conceptual challenges or problem-solving required. |
| Spec | 2.03c Conditional probability: using diagrams/tables2.03d Calculate conditional probability: from first principles |
| Male | Female | |
| Jaguar | 25 | 15 |
| Bentley | 12 | 8 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{25}{37p}\) | B2 2 | B1 num, B1 denom 25/(37×p) B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{15}{23}\) seen or implied | M1 | |
| \(\times\frac{39}{59}\) seen or implied | M2 | M1 num, M1 denom Allow M1 for 39/59x or + wrong p |
| \(= \frac{585}{1357}\) or \(0.431\) (3 sfs) oe | A1 4 |
### (i)
$\frac{25}{37p}$ | B2 2 | B1 num, B1 denom 25/(37×p) B1
### (ii)
$\frac{15}{23}$ seen or implied | M1 |
$\times\frac{39}{59}$ seen or implied | M2 | M1 num, M1 denom Allow M1 for 39/59x or + wrong p
$= \frac{585}{1357}$ or $0.431$ (3 sfs) oe | A1 4 |
**Total [6]**
---The table shows the numbers of male and female members of a vintage car club who own either a Jaguar or a Bentley. No member owns both makes of car.
\begin{tabular}{|c|c|c|}
\hline
& Male & Female \\
\hline
Jaguar & 25 & 15 \\
\hline
Bentley & 12 & 8 \\
\hline
\end{tabular}
One member is chosen at random from these 60 members.
\begin{enumerate}[label=(\roman*)]
\item Given that this member is male, find the probability that he owns a Jaguar. [2]
\end{enumerate}
Now two members are chosen at random from the 60 members. They are chosen one at a time, without replacement.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Given that the first one of these members is female, find the probability that both own Jaguars. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2010 Q7 [6]}}