Year: 2008
Paper: 2
Question Number: 13
Course: LFM Stats And Pure
Section: Tree Diagrams
There were around 850 candidates for this paper – a slight increase on the 800 of the past two years – and the scripts received covered the full range of marks (and beyond!). The questions on this paper in recent years have been designed to be a little more accessible to all top A-level students, and this has been reflected in the numbers of candidates making good attempts at more than just a couple of questions, in the numbers making decent stabs at the six questions required by the rubric, and in the total scores achieved by candidates. Most candidates made attempts at five or more questions, and most genuinely able mathematicians would have found the experience a positive one in some measure at least. With this greater emphasis on accessibility, it is more important than ever that candidates produce really strong, essentially-complete efforts to at least four questions. Around half marks are required in order to be competing for a grade 2, and around 70 for a grade 1. The range of abilities on show was still quite wide. Just over 100 candidates failed to score a total mark of at least 30, with a further 100 failing to reach a total of 40. At the other end of the scale, more than 70 candidates scored a mark in excess of 100, and there were several who produced completely (or nearly so) successful attempts at more than six questions; if more than six questions had been permitted to contribute towards their paper totals, they would have comfortably exceeded the maximum mark of 120. While on the issue of the "best-six question-scores count" rubric, almost a third of candidates produced efforts at more than six questions, and this is generally a policy not to be encouraged. In most such cases, the seventh, eighth, or even ninth, question-efforts were very low scoring and little more than a waste of time for the candidates concerned. Having said that, it was clear that, in many of these cases, these partial attempts represented an abandonment of a question after a brief start, with the candidates presumably having decided that they were unlikely to make much successful further progress on it, and this is a much better employment of resources. As in recent years, most candidates' contributing question-scores came exclusively from attempts at the pure maths questions in Section A. Attempts at the mechanics and statistics questions were very much more of a rarity, although more (and better) attempts were seen at these than in other recent papers.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Bag $P$ and bag $Q$ each contain $n$ counters, where $n\ge2$. The counters are identical in shape and size, but coloured either black or white. First, $k$ counters ($0\le k\le n$) are drawn at random from bag $P$ and placed in bag $Q$. Then, $k$ counters are drawn at random from bag $Q$ and placed in bag $P$.
\begin{questionparts}
\item If initially $n-1$ counters in bag $P$ are white and one is black, and all $n$ counters in bag $Q$ are white, find the probability in terms of $n$ and $k$ that the black counter ends up in bag $P$.
Find the value or values of $k$ for which this probability is maximised.
\item If initially $n-1$ counters in bag $P$ are white and one is black, and $n-1$ counters in bag $Q$ are white and one is black, find the probability in terms of $n$ and $k$ that the black counters end up in the same bag.
Find the value or values of $k$ for which this probability is maximised.
\end{questionparts}\begin{questionparts}
\item $\,$
\begin{align*}
\mathbb{P}(\text{black counter in }P) &=
\mathbb{P}(\text{black counter moves twice})+\mathbb{P}(\text{black counter doesn't move}) \\
&= \mathbb{P}(\text{black counter moves out})\mathbb{P}(\text{black counter moves back}) + (1-\mathbb{P}(\text{black counter moves out})) \\
&= \frac{k}n\cdot \frac{k}{n+k}+\frac{n-k}{n} \\
&= \frac{k^2+n^2-k^2}{n(n+k)} \\
&= \frac{n^2}{n(n+k)} = \frac{n}{n+k}
\end{align*}
This is maximised if $k$ is as small as possibe, ie $k = 0$ (ie it doesn't leave it's bag)
\item $\,$
\begin{align*}
&& \mathbb{P}(\text{both counters in same bag}) &= \mathbb{P}(\text{both in }P)+ \mathbb{P}(\text{both in }Q) \\
&&&= \mathbb{P}(B_P \to Q \to P, B_Q \to P)+\mathbb{P}(B_P \text{ stays}, B_Q \to P)+\mathbb{P}(B_P \to Q, \text{both stay}) \\
&&&= \frac{k}{n} \cdot \frac{k(k-1)}{(n+k)(n+k-1)} + \frac{n-k}{n} \frac{k}{n+k} + \frac{k}{n} \frac{n(n-1)}{(n+k)(n+k-1)} \\
&&&= \frac{(k^3-k^2)+(n-k)k(n+k-1)+kn(n-1)}{n(n+k)(n+k-1)}\\
&&&= \frac{2kn(n-1)}{n(n+k)(n+k-1)}\\
&&&= \frac{2k(n-1)}{(n+k)(n+k-1)}
\end{align*}
\begin{align*}
&& \frac{P_{k+1}}{P_k} &= \frac{2(k+1)(n-1)}{(n+k+1)(n+k)} \frac{(n+k)(n+k-1)}{2k(n-1)} \\
&&&= \frac{(k+1)(n+k-1)}{k(n+k+1)} \\
&&& \geq 1 \\
\Leftrightarrow && (k+1)(n+k-1) &\geq k(n+k+1) \\
\Leftrightarrow && n-1 &\geq k \\
\end{align*}
Therefore this probability is increasing while $k \leq n-1$, ie it's maximised $k = n-1$ or $k=n$
\end{questionparts}
Perhaps encouraged by the ease with which they had managed Q12, many of these candidates went on to attempt this question also. Although the listing of relevant cases was a fairly straightforward exercise, the handling of the binomial coefficients – which certainly looked clumsy and unappealing – was coped with much less well, and many mistakes were made in the ensuing algebra. In the very final part of the question, the idea that the calculus could lead to a nice, neat answer (k = √(n(n−1))) that then needed to be interpreted in terms of integer values, was just one step too far for most takers. The eventual mean score of 8 on this question testifies to the difficulties found in the algebra by most of the candidates who attempted it.