| Exam Board | SPS |
|---|---|
| Module | SPS FM Statistics (SPS FM Statistics) |
| Year | 2025 |
| Session | April |
| Marks | 7 |
| Topic | Permutations & Arrangements |
| Type | Specific items together |
| Difficulty | Standard +0.3 This is a straightforward combinatorics problem involving arrangements and basic probability calculations. Part (i) requires counting adjacent pairs in a linear arrangement (standard technique). Parts (ii) and (iii) involve conditional arrangements with questions split into two groups, requiring careful case analysis but using only A-level counting methods. The problem is slightly easier than average as it's methodical application of permutations without requiring novel insight. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
An examination paper consists of 8 questions, of which one is on geometric distributions and one is on binomial distributions.
\begin{enumerate}[label=(\roman*)]
\item If the 8 questions are arranged in a random order, find the probability that the question on geometric distributions is next to the question on binomial distributions. [2]
\end{enumerate}
Four of the questions, including the one on geometric distributions, are worth 7 marks each, and the remaining four questions, including the one on binomial distributions, are worth 9 marks each. The 7-mark questions are the first four questions on the paper, but are arranged in random order. The 9-mark questions are the last four questions, but are arranged in random order. Find the probability that
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item the questions on geometric distributions and on binomial distributions are next to one another, [2]
\item the questions on geometric distributions and on binomial distributions are separated by at least 2 other questions. [3]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Statistics 2025 Q5 [7]}}