| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Combinations & Selection |
| Type | Multi-stage selection problems |
| Difficulty | Standard +0.8 This is a multi-part Further Maths statistics question requiring careful combinatorial reasoning. Part (a) is straightforward selection, but parts (b) and (c) involve complex arrangement problems with constraints. Part (c) is particularly challenging as it requires treating two books as a block while ensuring the third is separated, demanding careful case analysis and systematic counting—well above typical A-level but standard for Further Maths. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \({}^5C_3 \times {}^5C_2 \div {}^{10}C_5\) | M1 | Two correct terms: M1A0. Numerator \((= 100)\) can be \({}^5C_3 \times {}^3C_2 + {}^5C_3 \times {}^3C_2 \times {}^2C_1 + {}^5C_3 \times {}^2C_2\) |
| \(= \frac{25}{63}\) or \(0.397\ (0.396825\ldots)\) | A1, A1 [3] | Completely correct expression; Answer, exact or awrt \(0.397\) |
| OR: \(\frac{5}{10}\times\frac{4}{9}\times\frac{3}{8}\times\frac{5}{7}\times\frac{4}{6}\ \left[=\frac{5}{126}\right]\) | B1 | |
| \(\times {}^5C_3\ [=10]\) | M1 | |
| \(= \frac{25}{63}\) or \(0.397\) | A1 | Answer, exact or awrt \(0.397\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(7! \times 3! \times 2! \div 10!\) | M1 | Two correct terms: M1A0. E.g. \((6\times3!) \times (6\times2!) \div 10!\): M1A0 |
| \(= \frac{1}{60}\) or \(0.0167\ (0.01666\ldots)\) | A1, A1 [3] | Correct expression; Answer, exact or awrt \(0.0167\) |
| OR: \(\frac{3}{10}\times\frac{2}{9}\times\frac{1}{8}\times\frac{2}{7}\times\frac{1}{6}\ \left[=\frac{1}{2520}\right]\) | B1 | Five equivalent probs multiplied |
| \(\times {}^7P_2\ [=42]\) | M1 | Needs at least 3 probs |
| \(= \frac{1}{60}\) or \(0.0167\) | A1 | Answer, exact or awrt \(0.0167\) |
| Answer | Marks | Guidance |
|---|---|---|
| DR \( | x\ | x\ |
| Pair can go in 8 places, single in 7, and \(3\times2\) ways of choosing the pair in order | A1 | Correct application of strategy. AAAAATT arranged in \({}^7C_2\) ways M1. This gives \(21\times56 = 1176\) ways A1 |
| Total \(7! \times 8 \times 7 \times (3\times2) \div 10!\) | M1 | \(\times 7! \div 10!\). Total arrangements \(10!/(5!3!2!) = 2520\) |
| \(= \frac{7}{15}\) or \(0.466\ldots\) | A1 [4] | Final answer, exact or awrt \(0.467\). SR: If M0, \(7!\) or \(8!\times\)integers\(\div 10!\): B1. Therefore answer is \(\frac{1176}{2520} = \frac{7}{15}\) A1 |
| OR: \(P(\text{all 3 Calculus together}) = \frac{8!\cdot 3!}{10!}\ \left[=\frac{1}{15}\right]\) | B1 | Or other valid strategy. Two calculus books at end: \(\left[\frac{3}{10}\times\frac{2}{9}\times\frac{2}{8}\right]\times 2\) M1 |
| \(P(\text{no Calculus book is next to another}) = \frac{(8'\cdot 7'\cdot 6')\cdot 7!}{10!}\ \left[=\frac{7}{15}\right]\) | B1 | Two calcs books together, not at end: \(\left[\frac{7}{10}\times\frac{3}{8}\times\frac{2}{8}\times\frac{6}{7}\right]\times 7\) M1 |
| \(1 - P(\text{all 3 together}) - P(\text{all separate})\) | M1 | Total \(= \frac{7}{60} + \frac{7}{20}\) A1. (A1 for both expressions correct) |
| \(= 1 - \frac{1}{15} - \frac{7}{15} = \frac{7}{15}\) or \(0.466\ldots\) | A1 | \(= \frac{7}{15}\) or \(0.466\ldots\) A1 |
# Question 6:
## Part (a):
${}^5C_3 \times {}^5C_2 \div {}^{10}C_5$ | **M1** | Two correct terms: M1A0. Numerator $(= 100)$ can be ${}^5C_3 \times {}^3C_2 + {}^5C_3 \times {}^3C_2 \times {}^2C_1 + {}^5C_3 \times {}^2C_2$
$= \frac{25}{63}$ or $0.397\ (0.396825\ldots)$ | **A1, A1 [3]** | Completely correct expression; Answer, exact or awrt $0.397$
**OR:** $\frac{5}{10}\times\frac{4}{9}\times\frac{3}{8}\times\frac{5}{7}\times\frac{4}{6}\ \left[=\frac{5}{126}\right]$ | **B1** |
$\times {}^5C_3\ [=10]$ | **M1** |
$= \frac{25}{63}$ or $0.397$ | **A1** | Answer, exact or awrt $0.397$
## Part (b):
$7! \times 3! \times 2! \div 10!$ | **M1** | Two correct terms: M1A0. E.g. $(6\times3!) \times (6\times2!) \div 10!$: M1A0
$= \frac{1}{60}$ or $0.0167\ (0.01666\ldots)$ | **A1, A1 [3]** | Correct expression; Answer, exact or awrt $0.0167$
**OR:** $\frac{3}{10}\times\frac{2}{9}\times\frac{1}{8}\times\frac{2}{7}\times\frac{1}{6}\ \left[=\frac{1}{2520}\right]$ | **B1** | Five equivalent probs multiplied
$\times {}^7P_2\ [=42]$ | **M1** | Needs at least 3 probs
$= \frac{1}{60}$ or $0.0167$ | **A1** | Answer, exact or awrt $0.0167$
## Part (c):
**DR** $|x\ |x\ |x\ |x\ |x\ |x\ |x\ |$ ($x$ = non-Calculus) | **M1** | *(Decide first whether candidate is using direct calc or the complement)* Use of barriers clearly indicated. Equivalent: (same barrier idea M1). Pair can go in 8 places, single in 7
Pair can go in 8 places, single in 7, and $3\times2$ ways of choosing the pair in order | **A1** | Correct application of strategy. AAAAATT arranged in ${}^7C_2$ ways M1. This gives $21\times56 = 1176$ ways A1
Total $7! \times 8 \times 7 \times (3\times2) \div 10!$ | **M1** | $\times 7! \div 10!$. Total arrangements $10!/(5!3!2!) = 2520$
$= \frac{7}{15}$ or $0.466\ldots$ | **A1 [4]** | Final answer, exact or awrt $0.467$. SR: If M0, $7!$ or $8!\times$integers$\div 10!$: B1. Therefore answer is $\frac{1176}{2520} = \frac{7}{15}$ A1
**OR:** $P(\text{all 3 Calculus together}) = \frac{8!\cdot 3!}{10!}\ \left[=\frac{1}{15}\right]$ | **B1** | Or other valid strategy. Two calculus books at end: $\left[\frac{3}{10}\times\frac{2}{9}\times\frac{2}{8}\right]\times 2$ M1
$P(\text{no Calculus book is next to another}) = \frac{(8'\cdot 7'\cdot 6')\cdot 7!}{10!}\ \left[=\frac{7}{15}\right]$ | **B1** | Two calcs books together, not at end: $\left[\frac{7}{10}\times\frac{3}{8}\times\frac{2}{8}\times\frac{6}{7}\right]\times 7$ M1
$1 - P(\text{all 3 together}) - P(\text{all separate})$ | **M1** | Total $= \frac{7}{60} + \frac{7}{20}$ A1. (A1 for both expressions correct)
$= 1 - \frac{1}{15} - \frac{7}{15} = \frac{7}{15}$ or $0.466\ldots$ | **A1** | $= \frac{7}{15}$ or $0.466\ldots$ A1
---6 A teacher has 10 different mathematics books. Of these books, 5 are on Algebra, 3 are on Calculus and 2 are on Trigonometry.
The teacher chooses 5 of the books at random.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that 3 of the books are on Algebra.
The teacher now arranges all 10 books in random order on a shelf.
\item Find the probability that the Calculus books are next to each other and the Trigonometry books are next to each other.
\section*{In this question you must show detailed reasoning.}
\item Find the probability that 2 of the Calculus books are next to each other but the third Calculus book is separated from the other 2 by at least 1 other book.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS 2022 Q6 [10]}}