| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Alternating pattern probability |
| Difficulty | Challenging +1.2 Part (a) is a straightforward application of counting arrangements with alternating patterns—students need to recognize there are 2 valid patterns (T-C-T-C... or C-T-C-T...), count arrangements for each, and divide by total arrangements. Part (b) requires understanding hypergeometric distribution and manipulating binomial coefficients, which is more conceptually demanding but still follows standard Further Maths techniques. The 'show that' format and combinatorial manipulation elevate this above routine (0.0), but it doesn't require novel insight beyond applying known formulas. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
| Answer | Marks | Guidance |
|---|---|---|
| \(2\times10!\times10!\) | M1 | Correct method, e.g. MWMW… |
| M1 | \(\times 2\) | |
| \(\div\ 20!\) | M1 | |
| \(\dfrac{1}{92378} = 1.08(25\ldots)\times10^{-5}\) | A1 | Exact or awrt \(1.08\times10^{-5}\ (= 0.000\ 010\ 8)\), www |
| Answer | Marks | Guidance |
|---|---|---|
| \({}^{10}C_0\times{}^{10}C_{10} + \ldots + {}^{10}C_r\times{}^{10}C_{10-r}\) | M1 | Any two pairs of \(^nC_r\) stated or implied |
| \({}^{10}C_1 = {}^{10}C_9\) etc | M1 | At least one such must be clearly stated |
| \(N = {}^{20}C_{10}\ [= 184756]\) | B1 | Clearly stated |
| \(\dfrac{1}{184756}\displaystyle\sum_{i=0}^{r}({}^{10}C_i)^2\) | A1 | Fully correct argument to derive given result, including \(\Sigma\) |
## Question 7:
### Part (a):
$2\times10!\times10!$ | **M1** | Correct method, e.g. MWMW…
| **M1** | $\times 2$
$\div\ 20!$ | **M1** |
$\dfrac{1}{92378} = 1.08(25\ldots)\times10^{-5}$ | **A1** | Exact or awrt $1.08\times10^{-5}\ (= 0.000\ 010\ 8)$, www
[4]
### Part (b):
${}^{10}C_0\times{}^{10}C_{10} + \ldots + {}^{10}C_r\times{}^{10}C_{10-r}$ | **M1** | Any two pairs of $^nC_r$ stated or implied
${}^{10}C_1 = {}^{10}C_9$ etc | **M1** | At least one such must be clearly stated
$N = {}^{20}C_{10}\ [= 184756]$ | **B1** | Clearly stated
$\dfrac{1}{184756}\displaystyle\sum_{i=0}^{r}({}^{10}C_i)^2$ | **A1** | Fully correct argument to derive given result, including $\Sigma$
[4]
---7 The 20 members of a club consist of 10 Town members and 10 Country members.
\begin{enumerate}[label=(\alph*)]
\item All 20 members are arranged randomly in a straight line.
Determine the probability that the Town members and the Country members alternate.
\item Ten members of the club are chosen at random.
Show that the probability that the number of Town members chosen is no more than $r$, where $0 \leqslant r \leqslant 10$, is given by\\
$\frac { 1 } { \mathrm {~N} } \sum _ { \mathrm { i } = 0 } ^ { \mathrm { r } } \left( { } ^ { 10 } \mathrm { C } _ { \mathrm { i } } \right) ^ { 2 }$\\
where $N$ is an integer to be determined.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS 2021 Q7 [8]}}