| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Permutations & Arrangements |
| Type | Alternating pattern probability |
| Difficulty | Challenging +1.2 This is a Further Maths combinatorics problem requiring careful counting of arrangements. Part (i) needs recognition of alternating gender patterns (2 cases: MWMW... or WMWM...) with probability 2×(4!)²/(8!). Part (ii) requires treating the Adams as a single unit, giving (2x-1)!×2 favorable outcomes over (2x)! total, simplifying to 2/(2x) = 1/x. While systematic, this demands more sophisticated counting techniques than standard A-level and involves algebraic manipulation in part (ii), placing it moderately above average difficulty. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (i) | 2(cid:117)4!(cid:117)4!(cid:62)(cid:32)1152(cid:64) |
| Answer | Marks |
|---|---|
| 35 | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| i | Method (allow 2 omitted) |
| Answer | Marks |
|---|---|
| 70 | OR |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (ii) | (2x(cid:16)1)!(cid:117)2 |
| Answer | Marks |
|---|---|
| 2x x | c |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| Answer | Marks |
|---|---|
| 2.1 | Taking Mr and Mrs Adam as one |
| Answer | Marks |
|---|---|
| 4(ii) | i |
Question 4:
4 | (i) | 2(cid:117)4!(cid:117)4!(cid:62)(cid:32)1152(cid:64)
(cid:121)8!
(cid:32) 1 or 0.0286
35 | M1
M1
A1
[3] | 1.1
1.1
2.1
i | Method (allow 2 omitted)
n
e
awrt 0.0286
mClear working or justification
must be given
SC2 for one number
1
extra/omitted, e.g.
70 | OR
M1 Attempt at alternating male
female
M1 denominator decreasing
4 3 3 2 2 1
1(cid:117) (cid:117) (cid:117) (cid:117) (cid:117) (cid:117) (cid:117)1
7 6 5 4 3 2
A1 (cid:32) 1 or 0.0286
35
4 | (ii) | (2x(cid:16)1)!(cid:117)2
p
(2x)!
S
2 1
(cid:32) (cid:32)
2x x | c
e
M1
M1
E1
[2] | 3.1a
3.1a
2.1 | Taking Mr and Mrs Adam as one
item, ie 2x seen
Forming the probability and
cancelling the factorial
expressions
AG An intermediate step must be
shown
--- 4(i) ---
4(i)
n
e
m
--- 4(ii) ---
4(ii) | i
c
e
p
S
5(i)(a)
5(i)(b)
n
e
5(i)(c)
m
i
c
e\begin{enumerate}[label=(\roman*)]
\item Four men and four women stand in a random order in a straight line. Determine the probability that no one is standing next to a person of the same gender. [3]
\item $x$ men, including Mr Adam, and $x$ women, including Mrs Adam, are arranged at random in a straight line. Show that the probability that Mr Adam is standing next to Mrs Adam is $\frac{1}{x}$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS Q4 [6]}}