Year: 2008
Paper: 2
Question Number: 12
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: 1500.0
Banger Comparisons: 0
In the High Court of Farnia, the outcome of each case is determined by three judges: the ass, the beaver and the centaur.
Each judge decides its verdict independently. Being simple creatures, they make their decisions entirely at random. Past verdicts show that the ass gives a guilty verdict with probability $p$, the beaver gives a guilty verdict with probability $p/3$ and the centaur gives a guilty verdict with probability $p^2$.
Let $X$ be the number of guilty verdicts given by the three judges in a case. Given that $\E(X)= 4/3$, find the value of $p$. The probability that a defendant brought to trial is guilty is $t$. The King pronounces that the defendant is guilty if at least two of the judges give a guilty verdict; otherwise, he pronounces the defendant not guilty.
Find the value of $t$ such that the probability that the King pronounces correctly is $1/2$.\begin{align*}
&& \mathbb{E}(X) &= p + \frac{p}{3} + p^2 = \frac43p+p^2 \\
\Rightarrow && \frac43 &= \frac43p+p^2 \\
\Rightarrow && 0 &= 3p^2+4p-4 \\
&&&= (3p-2)(p+2) \\
\Rightarrow && p &= \frac23, -2
\end{align*}
Since $p \in [0,1]$ we must have $p = \frac23$.
\begin{align*}
&& \mathbb{P}(\text{correct verdict}) &= t p+ (1-t) (1-p) \\
&&&= t(2p-1)+(1-p)\\
\Rightarrow && \frac12 &= t(2p-1)+(1-p) \\
\Rightarrow && t &= \frac{\frac12-(1-p)}{2p-1} \\
&&&= \frac{2p-1}{2(2p-1)} = \frac12
\end{align*}
(so it doesn't depend at all on what the judges are doing, the only way to be fair is if the trials happen at random!)
There were almost 200 attempts to Q12, and the mean score was the highest – at 14 – of all the applied questions. This was partly due to the fact that the result of the first part could be largely circumvented by anyone who knew a little bit about expectation algebra, enabling them to write down E(X) straightaway. The simple combinations of events, and their associated probabilities, in the final part of the question were very confidently and competently handled by most candidates and many polished the question off in its entirety relatively quickly.