| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Arrangements with adjacency constraints |
| Difficulty | Standard +0.8 This is a Further Maths statistics question requiring sophisticated counting techniques. Part (a) involves treating adjacent items as a single unit (a standard but non-trivial technique), while part (b) requires careful hypergeometric probability calculation with 'at least' conditions. Both parts demand precise combinatorial reasoning beyond typical A-level, placing it moderately above average difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Class Baroque CDs as single unit | M1 | e.g. \(23! \times 7!\) seen, with or without other terms, *or* 24! (with 7! omitted) |
| \(24! \times 7!\ (= 6.2 \times 10^{23} \times 5040)\) | A1 | These, and no other terms, in numerator (allow even if no denominator) |
| \(\div 30! = 1.18 \times 10^{-5} = 0.000\,011\,8\) | A1[3] | Awrt \(1.2 \times 10^{-5}\), or \(\frac{1}{84825}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(6:\ ^7C_6 \times\ ^{23}C_4\quad (= 7 \times 8855\ \text{or}\ 61\,985)\) | M1* | Clear attempt at one (allow for \(^7C_6 \times\ ^{23}C_4 \times\) other things), allow \(^{10}C_6 \times \ldots\) |
| \(7: 1 \times\ ^{23}C_3\quad (= 1771)\) | A1 | Both expressions correct |
| Add, and divide by \(^{30}C_{10}\quad (= 30\,045\,015)\) | depM1 | \(\left[= \frac{7}{3393} + \frac{1}{16965}\right]\) Needs two terms, allow dividing by \(^{30}P_{10}\) if consistent |
| \(= \frac{4}{1885}\) or \(0.002\,12\ (0.002\,122\ldots)\) | A1[4] | Any equivalent exact fraction, or 0.00212 or better. SC: \(B(10, \frac{7}{30})\), 0.014(0): B1 max. SC: \((^7P_6 \times\ ^{23}P_4) + (^7P_7 \times\ ^{23}P_3)\), M1; \(\times 20!/30!\), M1 (same as \(\div\ ^{30}P_{10}\)) |
## Question 2:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Class Baroque CDs as single unit | M1 | e.g. $23! \times 7!$ seen, with or without other terms, *or* 24! (with 7! omitted) |
| $24! \times 7!\ (= 6.2 \times 10^{23} \times 5040)$ | A1 | These, and no other terms, in numerator (allow even if no denominator) |
| $\div 30! = 1.18 \times 10^{-5} = 0.000\,011\,8$ | A1[3] | Awrt $1.2 \times 10^{-5}$, or $\frac{1}{84825}$ |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $6:\ ^7C_6 \times\ ^{23}C_4\quad (= 7 \times 8855\ \text{or}\ 61\,985)$ | M1* | Clear attempt at one (allow for $^7C_6 \times\ ^{23}C_4 \times$ other things), allow $^{10}C_6 \times \ldots$ |
| $7: 1 \times\ ^{23}C_3\quad (= 1771)$ | A1 | Both expressions correct |
| Add, and divide by $^{30}C_{10}\quad (= 30\,045\,015)$ | depM1 | $\left[= \frac{7}{3393} + \frac{1}{16965}\right]$ Needs two terms, allow dividing by $^{30}P_{10}$ if consistent |
| $= \frac{4}{1885}$ or $0.002\,12\ (0.002\,122\ldots)$ | A1[4] | Any equivalent exact fraction, or 0.00212 or better. SC: $B(10, \frac{7}{30})$, 0.014(0): B1 max. SC: $(^7P_6 \times\ ^{23}P_4) + (^7P_7 \times\ ^{23}P_3)$, M1; $\times 20!/30!$, M1 (same as $\div\ ^{30}P_{10}$) |
---2 A music lover has 30 CDs arranged in a random order in a line on a shelf. Of these CDs, 7 are classed as Baroque, 10 as Classical and 13 as Romantic.
\begin{enumerate}[label=(\alph*)]
\item Determine the probability that all 7 Baroque CDs are next to each other.
\item Determine the probability that, of the 10 CDs furthest to the left on the shelf, at least 6 are Baroque.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS 2023 Q2 [7]}}