Year: 2011
Paper: 2
Question Number: 12
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. After the relatively easy time candidates experienced on last year's paper, this year's questions had been toughened up significantly, with particular attention made to ensure that candidates had to be prepared to invest more thought at the start of each question – last year saw far too many attempts from the weaker brethren at little more than the first part of up to ten questions, when the idea is that they should devote 25-40 minutes on four to six complete questions in order to present work of a substantial nature. It was also the intention to toughen up the final "quarter" of questions, so that a complete, or nearly-complete, conclusion to any question represented a significant (and, hopefully, satisfying) mathematical achievement. Although such matters are always best assessed with the benefit of hindsight, our efforts in these areas seem to have proved entirely successful, with the vast majority of candidates concentrating their efforts on four to six questions, as planned. Moreover, marks really did have to be earned: only around 20 candidates managed to gain or exceed a score of 100, and only a third of the entry managed to hit the half-way mark of 60. 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. Questions 1 and 2 were attempted by almost all candidates; 3 and 4 by around three-quarters of them; 6, 7 and 9 by around half; the remaining questions were less popular, and some received almost no "hits". Overall, the highest scoring questions (averaging over half-marks) were 1, 2 and 9, along with 13 (very few attempts, but those who braved it scored very well). This at least is indicative that candidates are being careful in exercising some degree of thought when choosing (at least the first four) 'good' questions for themselves, although finding six successful questions then turned out to be a key discriminating factor of candidates' abilities from the examining team's perspective. Each of questions 4-8, 11 & 12 were rather poorly scored on, with average scores of only 5.5 to 6.6.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
Xavier and Younis are playing a match.
The match consists of a series of games and each game consists of three points.
Xavier has probability $p$ and Younis has probability $1-p$ of winning the first point of any game. In the second and third points of each game, the player who won the previous point has probability $p$ and the player who lost the previous point has probability $1-p$ of winning the point.
If a player wins two consecutive points in a single game, the match ends and that player has won; otherwise the match continues with another game.
\begin{questionparts}
\item Let $w$ be the probability that Younis wins the match.
Show that, for $p\ne0$,
\[
w = \frac{1-p^2}{2-p}.
\]
Show that $w>\frac12$ if $p<\frac12$, and $w<\frac12$ if $p>\frac12$. Does $w$ increase whenever
$p$ decreases?
\item If Xavier wins the match, Younis gives him $\pounds1$; if
Younis wins the match, Xavier gives him $\pounds k$.
Find the value of $k$ for which the game is `fair'
in the case when $p =\frac23$.
\item What happens when $p = 0$?
\end{questionparts}\begin{questionparts}
\item We can be in several states.
\begin{enumerate}
\item No points played
\item Y just won the last point
\item X just won the last point
\item Y won the game
\item X won the game
\end{enumerate}
The probability $Y$ wins from any of these states are:
\begin{align*}
&&P_{-} &= p P_X + (1-p) P_Y &= w \\
&&P_Y &= p + (1-p)P_X \\
&&P_X &= (1-p)P_Y \\
\\
\Rightarrow &&& \begin{cases} P_Y - (1-p)P_X &= p \\ (1-p)P_Y -P_X &= 0\end{cases} \\
\Rightarrow && P_Y &= \frac{1}{1-(1-p)^2} \cdot p \\
&&&= \frac{1}{2-p} \\
&& P_X &= \frac{1-p}{2-p} \\
&& w &= \frac{p(1-p) + (1-p)}{2-p} \\
&&&= \frac{1-p^2}{2-p}
\end{align*}
\item If $p = \frac23$ then $w = \frac{1-\frac49}{2-\frac23} = \frac{5}{12}$.
The game is fair if $\mathbb{E}(result) = 0$, ie $\frac{5}{12} \cdot k - \frac{7}{12} 1 \Rightarrow k = \frac{7}{5} = 1.4$
\item If $p = 0$ then they will keep playing forever, since no-one can win two points in a row.
\end{questionparts}
Around a quarter of all candidates made an attempt at this question, though the average score was very low. Parts (ii) and (iii) were managed quite comfortably, on the whole, but it was (i) that proved to be difficult for most of those who attempted the question. The real difficulty lay in establishing the given result for w, as the event to which it corresponded was defined recurrently. As it happens, most wayward solutions left the straight-and-narrow by misreading the rules of the match to begin with.