Year: 2010
Paper: 2
Question Number: 13
Course: LFM Stats And Pure
Section: Tree Diagrams
There were just under 1000 entries for paper II this year, almost exactly the same number as last year. Of this number, more than 60 scored over 90% while, at the other end of the scale, almost 200 failed to score more than 40 marks. In hindsight, many of the pure maths questions were a little too accessible and lacked a sufficiently tough 'difficulty gradient', so that scores were slightly higher than anticipated. This was reflected in the grade boundaries for the "1" and the "2" (around ten marks higher than is generally planned) in particular. Next year's questions may be expected to be a little bit more demanding, but only in the sense that the final 5 or 6 marks on each question should have rather more bite to them: it should certainly not be the case that all questions are tougher to get into at the outset. Most candidates attempted the requisite number of questions (six), although many of the weaker brethren made seven or eight attempts, most of which were feeble at best and they generally only picked up a maximum of 5 or 6 marks per question. It is a truth universally acknowledged that practice maketh if not perfect then at least a whole lot better prepared, and choosing to waste time on a couple of extra questions is not a good strategy on the STEPs. The major down-side of the present modular examination system is that students are not naturally prepared to approach the subject holistically; ally this to the current practice of setting highly-structured, fully-guided questions requiring no imagination, insight, depth or planning from A-level candidates in a system that fails almost nobody and rewards even the most modestly able with high grades in a manner reminiscent of a dentist giving lollipops to kids who have done little more than been brave and seen the course through, it is even more important to ensure a full and thorough preparation for these papers. The 20% of the entry who seem to be either unprepared for the rigours of a STEP, or unwittingly possessed of only a smattering of basic advanced-level skills, seems to be remarkably steady year-on-year, even in a year when their more suitably prepared compatriots found the paper appreciably easier than usual. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1502.2
Banger Comparisons: 2
Rosalind wants to join the Stepney Chess Club. In order to be accepted, she must play a challenge match consisting of several games against Pardeep (the Club champion) and Quentin (the Club secretary), in which she must win at least one game against each of Pardeep and Quentin. From past experience, she knows that the probability of her winning a single game against Pardeep is $p$ and the probability of her winning a single game against Quentin is $q$, where $0 < p < q < 1$.
\begin{questionparts}
\item The challenge match consists of three games. Before the match begins, Rosalind must choose either to play Pardeep twice and Quentin once or to play Quentin twice and Pardeep once. Show that she should choose to play Pardeep twice.
\item In order to ease the entry requirements, it is decided instead that the challenge match will consist of four games.
Now, before the match begins, Rosalind must choose whether to play Pardeep three times and Quentin once (strategy 1), or to play Pardeep twice and Quentin twice (strategy 2) or to play Pardeep once and Quentin three times (strategy 3). Show that, if $q-p > \frac 12$, Rosalind should choose strategy 1. If $q-p<\frac12$, give examples of values of $p$ and $q$ to show that strategy 2 can be better or worse than strategy 1.
\end{questionparts}\begin{questionparts}
\item If she plays $P$ twice her probability is $q(p^2+2p(1-p)) = qp(2-p)$. If she plays $Q$ twice her probability is $pq(2-q)$. Since $p < q$ she should play $P$ twice.
\item
Under strategy 1, her probability is $q(p^3+3p^2(1-p)+3p(1-p)^2) = qp(p^2+3p-3p^2+3-6p+3p^2) = qp(3-3p+p^2)$
Under strategy 2 her probability is $(p^2+2p(1-p))(q^2+2q(1-q)) = pq(2-p)(2-q)$.
Under strategy 3 her probability is $qp(3-3q+q^2)$
\begin{align*}
&& q - p &> \frac12 \\
\Rightarrow && (2-p)(2-q) & < (2-p)(\frac32 - p) \\
&&&= 3 - \frac72p + p^2 \\
&&&< 3- 3p + p^2
\end{align*}
Therefore Strategy 1 dominates if $q-p > \frac12$.
If $p = \frac14, q = \frac12$ then $(2-p)(2-q) =\frac74 \cdot \frac32 = \frac{21}{8}$ and $3-3p + p^2 = 3 - \frac34 + \frac1{16} = \frac{48-12+1}{16} = \frac{37}{16} < \frac{42}{16}$ so strategy 2 dominates.
Notice that strategy 1 always dominates strategy 3 since $f(x) = 3-3x+x^2$ is decreasing for $x < 1.5$.
If $p = \frac14, q = \frac12$ then $(2-p)(2-q) =\frac74 \cdot \frac32 = \frac{21}{8}$ and $3-3p + p^2 = 3 - \frac34 + \frac1{16} = \frac{48-12+1}{16} = \frac{37}{16} < \frac{42}{16}$ so strategy 2 dominates.
For strategy 1 to dominate, we need $3-3p+p^2 > (2-q)(2-p)$ or $\frac{3-3p+p^2}{2-p} > 2-q$. When $p = \frac12$ this is $\frac{3-\frac32 + \frac14}{2 - \frac12} = \frac{\frac{7}{4}}{\frac{3}{2}} = \frac76 = 2-\frac{5}{6}$ so take any value of $q$ larger than $\frac56$.
\end{questionparts}
This question worked almost exactly like Q12, in the sense that the start required some straightforward (probability) work, followed by an extension that needed only careful handling, before finishing with some poorly-handled inequalities work. The biggest problem for candidates lay in their lack of care to show that their chosen values of p and q actually satisfied any claimed conditions.