| Exam Board | OCR |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Digit arrangements forming numbers |
| Difficulty | Moderate -0.3 This is a straightforward permutations and combinations question with standard techniques. Part (i) uses basic permutation counting (9P4 and case analysis for odd numbers), while part (ii) applies simple probability calculations with enumeration of favorable outcomes. All methods are routine for S1 level with no novel insight required, making it slightly easier than average. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \({}^9P_4\) or \(\frac{9!}{5!}\) or \({}^9C_4 \times 4!\) | M1 | alone; oe e.g. \({}^9C_1 \times {}^8C_1 \times {}^7C_1 \times {}^6C_1\) or \(9{\times}8{\times}7{\times}6\) |
| \(= 3024\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \({}^8P_3\) or \(8 \times 7 \times 6\) oe or \({}^8C_3 \times 3!\) | M1 | Allow \(\times \ldots\) or \(\div \ldots\) |
| \(\times 5\) (or \({}^5C_1\)) | M1 | Correct \(\times 5\) or \({}^8C_3 \times 5\) (or \({}^5C_1\)); or \(({}^9P_4\) or "3024") \(\times \frac{5}{9}\) M2 |
| \(= 1680\) | A1 | Not ISW, e.g. \(\frac{1680}{3024}\): M1M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \({}^5C_3 \times {}^4C_1\) or \({}^5C_4\) oe | M1 | \({}^5C_3 \times {}^4C_1 \times 4!\) (or \({}^5P_3 \times 4 \times 4\)) or \(5!\) (or \({}^5P_4\)) |
| \({}^5C_3 \times {}^4C_1 + {}^5C_4\) oe correct method so far \((= 45)\) | M1 | \(960 + 120\) oe correct method so far |
| \(\div {}^9C_4\) — Allow anything \(\div {}^9C_4\) | M1 | \(\div {}^9P_4\) [must involve any P or any !] \(\div {}^9P_4\) |
| \(= \frac{5}{14}\) or \(0.357\) (3 sfs) oe, e.g. \(\frac{35}{98}\) or \(\frac{45}{126}\) | A1 | Marks must come from one method, not mixture of two methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(9, 8, 7, 4\) or \(9, 8, 6, 5\) — No mark yet | ||
| \(2\) | M1 | \(\frac{1}{9}{\times}\frac{1}{8}{\times}\frac{1}{7}{\times}\frac{1}{6} \cdot \frac{4}{9}{\times}\frac{3}{8}{\times}\frac{2}{7}{\times}\frac{1}{6}\); Allow \(\times\) or \(+\ldots\); \(4! + 4!\) or \(2 \times 4!\) oe |
| \(\div {}^9C_4\) oe — Must be (1 or 2 or 4) \(\div {}^9C_4\) | M1 | \(\times 4! \times 2\ \mid \times 2\) fully correct method; \(\div {}^9P_4\) or \(\div \text{(i)(a)}\) oe |
| \(= \frac{1}{63}\) oe or \(0.0159\) (3 sfs) | A1 |
## Question 9:
### Part (i)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| ${}^9P_4$ or $\frac{9!}{5!}$ or ${}^9C_4 \times 4!$ | M1 | alone; oe e.g. ${}^9C_1 \times {}^8C_1 \times {}^7C_1 \times {}^6C_1$ or $9{\times}8{\times}7{\times}6$ |
| $= 3024$ | A1 | |
### Part (i)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| ${}^8P_3$ or $8 \times 7 \times 6$ oe or ${}^8C_3 \times 3!$ | M1 | Allow $\times \ldots$ or $\div \ldots$ |
| $\times 5$ (or ${}^5C_1$) | M1 | Correct $\times 5$ or ${}^8C_3 \times 5$ (or ${}^5C_1$); or $({}^9P_4$ or "3024") $\times \frac{5}{9}$ M2 |
| $= 1680$ | A1 | Not ISW, e.g. $\frac{1680}{3024}$: M1M1A0 |
### Part (ii)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| ${}^5C_3 \times {}^4C_1$ or ${}^5C_4$ oe | M1 | ${}^5C_3 \times {}^4C_1 \times 4!$ (or ${}^5P_3 \times 4 \times 4$) or $5!$ (or ${}^5P_4$) |
| ${}^5C_3 \times {}^4C_1 + {}^5C_4$ oe correct method so far $(= 45)$ | M1 | $960 + 120$ oe correct method so far |
| $\div {}^9C_4$ — Allow anything $\div {}^9C_4$ | M1 | $\div {}^9P_4$ [must involve any P or any !] $\div {}^9P_4$ |
| $= \frac{5}{14}$ or $0.357$ (3 sfs) oe, e.g. $\frac{35}{98}$ or $\frac{45}{126}$ | A1 | Marks must come from one method, not mixture of two methods |
### Part (ii)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $9, 8, 7, 4$ or $9, 8, 6, 5$ — No mark yet | | |
| $2$ | M1 | $\frac{1}{9}{\times}\frac{1}{8}{\times}\frac{1}{7}{\times}\frac{1}{6} \cdot \frac{4}{9}{\times}\frac{3}{8}{\times}\frac{2}{7}{\times}\frac{1}{6}$; Allow $\times$ or $+\ldots$; $4! + 4!$ or $2 \times 4!$ oe |
| $\div {}^9C_4$ oe — Must be (1 or 2 or 4) $\div {}^9C_4$ | M1 | $\times 4! \times 2\ \mid \times 2$ fully correct method; $\div {}^9P_4$ or $\div \text{(i)(a)}$ oe |
| $= \frac{1}{63}$ oe or $0.0159$ (3 sfs) | A1 | |9 A bag contains 9 discs numbered 1, 2, 3, 4, 5, 6, 7, 8, 9 .\\
(i) Andrea chooses 4 discs at random, without replacement, and places them in a row.
\begin{enumerate}[label=(\alph*)]
\item How many different 4 -digit numbers can be made?
\item How many different odd 4-digit numbers can be made?\\
(ii) Andrea's 4 discs are put back in the bag. Martin then chooses 4 discs at random, without replacement. Find the probability that\\
(a) the 4 digits include at least 3 odd digits,\\
(b) the 4 digits add up to 28 .
\end{enumerate}
\hfill \mbox{\textit{OCR S1 2012 Q9 [12]}}