Year: 2023
Paper: 3
Question Number: 12
Course: LFM Stats And Pure
Section: Discrete Probability Distributions
No solution available for this problem.
The total entry was a marginal increase on that of 2022 (by just over 1%). Two questions were attempted by more than 90% of candidates, another two by 80%, and another two by about two thirds. The least popular questions were attempted by more than a sixth of candidates. All the questions were perfectly answered by at least three candidates (but mostly more than this), with one being perfectly answered by eighty candidates. Very nearly 90% of candidates attempted no more than 7 questions. One general comment regarding all the questions is that candidates need to make sure that they read the question carefully, paying particular attention to command words such as "hence" and "show that".
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A drawer contains $n$ pairs of socks. The two socks in each pair are indistinguishable, but each pair of socks is a different colour from all the others. A set of $2k$ socks, where $k$ is an integer with $2k \leqslant n$, is selected at random from this drawer: that is, every possible set of $2k$ socks is equally likely to be selected.
\begin{questionparts}
\item Find the probability that, among the socks selected, there is no pair of socks.
\item Let $X_{n,k}$ be the random variable whose value is the number of pairs of socks found amongst those selected. Show that
\[\mathrm{P}(X_{n,k} = r) = \frac{\dbinom{n}{r}\dbinom{n-r}{2(k-r)}\, 2^{2(k-r)}}{\dbinom{2n}{2k}}\]
for $0 \leqslant r \leqslant k$.
\item Show that
\[r\,\mathrm{P}(X_{n,k} = r) = \frac{k(2k-1)}{2n-1}\,\mathrm{P}(X_{n-1,k-1} = r-1)\,,\]
for $1 \leqslant r \leqslant k$, and hence find $\mathrm{E}(X_{n,k})$.
\end{questionparts}
This was only a little more popular than questions 9 and 11, and it was the seventh most successfully attempted with a mean score of 7.4/20. While there were a number of attempts that did not manage to make any significant progress, those who were able to analyse the situation were able to make very good progress. When finding the probability in part (i), the most common methods employed were to count the number of ways in which the outcome could be achieved or to create a product of individual probabilities by considering the socks being taken one at a time. Answers to part (ii) were generally well done, with most candidates providing a good explanation of the role of each part of the required formula. For the final part, most candidates chose to use algebra to show the given result, and this was generally done successfully, although a small number of candidates did not show fully the steps that were being taken. Almost all candidates who reached this point were able to apply the result shown to the formula for the expectation and most realised that several factors could be moved outside the sum. Only a small number did not realise that the summation that remained was the sum of the probabilities of all possible outcomes for that random variable.