| Exam Board | SPS |
|---|---|
| Module | SPS FM Statistics (SPS FM Statistics) |
| Year | 2022 |
| Session | February |
| Marks | 8 |
| Topic | Permutations & Arrangements |
| Type | Arrangements with alternating patterns |
| Difficulty | Standard +0.8 Part (a) requires counting alternating arrangements (2×10!×10! favorable outcomes from 20! total), which is a moderately challenging permutation problem. Part (b) involves proving a summation formula for hypergeometric probabilities, requiring insight into combinatorial identities and careful algebraic manipulation—this elevates the question above routine exercises to require genuine problem-solving and proof skills. |
| Spec | 5.01a Permutations and combinations: evaluate probabilities5.01b Selection/arrangement: probability problems |
6. 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\\
$\frac { 1 } { N } \sum _ { i = 0 } ^ { r } \left( { } ^ { 10 } C _ { i } \right) ^ { 2 }$\\
where $N$ is an integer to be determined.\\[0pt]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Statistics 2022 Q6 [8]}}