| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Definitions |
| Type | Listing outcomes and counting |
| Difficulty | Moderate -0.8 This is a straightforward S1 question requiring completion of an LCM table (basic arithmetic), counting favorable outcomes from the table, and applying the independence formula P(A∩B)=P(A)P(B). All steps are routine with no conceptual challenges beyond basic probability definitions. |
| Spec | 2.03a Mutually exclusive and independent events2.03b Probability diagrams: tree, Venn, sample space |
| \multirow{2}{*}{} | Second integer | ||||||
| 1 | 2 | 3 | 4 | 5 | 6 | ||
| \multirow{6}{*}{First integer} | 1 | 1 | 2 | 3 | 4 | 5 | 6 |
| 2 | 2 | 2 | 6 | 4 | 10 | 6 | |
| 3 | 3 | 6 | 3 | 12 | 15 | 6 | |
| 4 | 4 | 4 | 12 | 12 | |||
| 5 | 5 | 10 | 15 | ||||
| 6 | 6 | 6 | 6 | 12 | |||
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Complete correct LCM table (6×6 grid) | B1 | All correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((A)\) \(P(\text{LCM} > 6) = \frac{1}{3}\) | B1 | |
| \((B)\) \(P(\text{LCM} = 5n) = \frac{11}{36}\) | B1 | |
| \((C)\) \(P(\text{LCM} > 6 \cap \text{LCM} = 5n) = \frac{2}{9}\) | M1 A1 cao | Use of diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{3} \times \frac{11}{36} \neq \frac{2}{9}\) | M1 | Use of definition |
| Hence events are not independent | E1 |
## Question 5:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Complete correct LCM table (6×6 grid) | B1 | All correct |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $(A)$ $P(\text{LCM} > 6) = \frac{1}{3}$ | B1 | |
| $(B)$ $P(\text{LCM} = 5n) = \frac{11}{36}$ | B1 | |
| $(C)$ $P(\text{LCM} > 6 \cap \text{LCM} = 5n) = \frac{2}{9}$ | M1 A1 cao | Use of diagram |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{3} \times \frac{11}{36} \neq \frac{2}{9}$ | M1 | Use of definition |
| Hence events are not independent | E1 | |
---\begin{enumerate}[label=(\roman*)]
\item On the insert, complete the table giving the lowest common multiples of all pairs of integers between 1 and 6 .\\[0pt]
[1]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
\multicolumn{2}{|c|}{\multirow{2}{*}{}} & \multicolumn{6}{|c|}{Second integer} \\
\hline
& & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
\multirow{6}{*}{First integer} & 1 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
& 2 & 2 & 2 & 6 & 4 & 10 & 6 \\
\hline
& 3 & 3 & 6 & 3 & 12 & 15 & 6 \\
\hline
& 4 & 4 & 4 & 12 & & & 12 \\
\hline
& 5 & 5 & 10 & 15 & & & \\
\hline
& 6 & 6 & 6 & 6 & 12 & & \\
\hline
\end{tabular}
\end{center}
Two fair dice are thrown and the lowest common multiple of the two scores is found.
\item Use the table to find the probabilities of the following events.\\
(A) The lowest common multiple is greater than 6 .\\
(B) The lowest common multiple is a multiple of 5 .\\
(C) The lowest common multiple is both greater than 6 and a multiple of 5 .
\item Use your answers to part (ii) to show that the events "the lowest common multiple is greater than 6 " and "the lowest common multiple is a multiple of 5 " are not independent.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2005 Q5 [6]}}