| Exam Board | OCR |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2017 |
| Session | Specimen |
| Marks | 6 |
| Topic | Permutations & Arrangements |
| Type | Alternating pattern probability |
| Difficulty | Standard +0.8 Part (i) requires counting alternating gender arrangements (derangement-like problem) with careful casework, while part (ii) involves a clever combinatorial argument treating a couple as a single unit. Both require non-routine problem-solving beyond standard permutation formulas, typical of Further Statistics questions, but are accessible with systematic counting. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities |
\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 FS1 AS 2017 Q4 [6]}}