Year: 2007
Paper: 2
Question Number: 12
Course: LFM Stats And Pure
Section: Geometric Distribution
Although the paper was by no means an easy one, it was generally found a more accessible paper than last year's, with most questions clearly offering candidates an attackable starting-point. The candidature represented the usual range of mathematical talents, with a pleasingly high number of truly outstanding students; many more who were able to demonstrate a thorough grasp of the material in at least three questions; and the few whose three-hour long experience was unlikely to have been a particularly pleasant one. However, even for these candidates, many were able to make some progress on at least two of the questions chosen. Really able candidates generally produced solid attempts at five or six questions, and quite a few produced outstanding efforts at up to eight questions. In general, it would be best if centres persuaded candidates not to spend valuable time needlessly in this way – it is a practice that is not to be encouraged, as it uses valuable examination time to little or no avail. Weaker brethren were often to be found scratching around at bits and pieces of several questions, with little of substance being produced on more than a couple. It is an important examination skill – now more so than ever, with most candidates now not having to employ such a skill on the modular papers which constitute the bulk of their examination experience – for candidates to spend a few minutes at some stage of the examination deciding upon their optimal selection of questions to attempt. As a rule, question 1 is intended to be accessible to all takers, with question 2 usually similarly constructed. In the event, at least one – and usually both – of these two questions were among candidates' chosen questions. These, along with questions 3 and 6, were by far the most popularly chosen questions to attempt. The majority of candidates only attempted questions in Section A (Pure Maths), and there were relatively few attempts at the Applied Maths questions in Sections B & C, with Mechanics proving the more popular of the two options. It struck me that, generally, the working produced on the scripts this year was rather better set-out, with a greater logical coherence to it, and this certainly helps the markers identify what each candidate thinks they are doing. Sadly, this general remark doesn't apply to the working produced on the Mechanics questions, such as they were. As last year, the presentation was usually appalling, with poorly labelled diagrams, often with forces missing from them altogether, and little or no attempt to state the principles that the candidates were attempting to apply.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
I have two identical dice. When I throw either one of them, the
probability of it showing a 6 is $p$ and the probability of it not showing a 6 is $q$, where $p+q=1$. As an experiment to determine $p$, I throw the dice simultaneously until at least one die shows a 6. If both dice show a six on this throw, I stop. If just one die shows a six, I throw the other die until it shows a 6 and then stop.
\begin{questionparts}
\item Show that the probability that I stop after $r$ throws
is $pq^{r-1}(2-q^{r-1}-q^r)$, and find an expression for the
expected number of throws.
[{\bf Note:} You may use the result $\ds \sum_{r=0}^\infty rx^r =
x(1-x)^{-2}$.]
\item In a large number of such experiments, the mean number of throws was $m$. Find an estimate for $p$ in terms of $m$.
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
\mathbb{P}(\text{stop after r}) &= \mathbb{P}(\text{both stop at r}) + 2\mathbb{P}(\text{first stops before r second stops at r})\\
&= (q^2)^{r-1} p^2 + 2\cdot q^{r-1} p\cdot(1-q^{r-1}) \\
&= q^{r-1}p\left (2-2q^{r-1}+pq^{r-1} \right) \\
&= q^{r-1}p\left (2-q^{r-1}(1+p+q-p) \right) \\
&= q^{r-1}p\left (2-q^{r-1}-q^r\right) \\
\end{align*}
\begin{align*}
\E[\text{throws}] &= \sum_{r=1}^{\infty} r \mathbb{P}(\text{stop after r}) \\
&= \sum_{r=1}^{\infty} r q^{r-1}p\left (2-q^{r-1}-q^r\right) \\
&= \sum_{r=1}^{\infty} 2r q^{r-1}p-\sum_{r=1}^{\infty}r pq^{2r-2}-\sum_{r=1}^{\infty}r q^{2r-1}p \\
&=2p \sum_{r=1}^{\infty} r q^{r-1}-pq^{-2}\sum_{r=1}^{\infty}r q^{2r}-pq^{-1}\sum_{r=1}^{\infty}r q^{2r} \\
&= 2p(1-q)^{-2} - pq^{-2}q^2(1-q^2)^{-2}-pq^{-1}q^2(1-q^2)^{-2} \\
&= 2pp^{-2} -p(1+q)(1-q^2)^{-2} \\
&= 2p^{-1}-p(1+q)(1+q)^{-2}p^{-2} \\
&= 2p^{-1}-p^{-1}(1+q)^{-1} \\
&= \frac{2(1+q)-1}{p(1+q)} \\
&= \frac{1+2q}{p(1+q)} \\
&= \frac{3-2p}{p(2-p)}
\end{align*}
\item $\,$ \begin{align*}
&& m &= \frac{3-2p}{p(2-p)} \\
\Rightarrow && 2mp-mp^2 &= 3-2p \\
\Rightarrow && 0 &= mp^2-(2m+2)p + 3 \\
\Rightarrow && p &= \frac{2m+2 \pm \sqrt{(2m+2)^2-12m}}{2m} \\
&&&= \frac{m+1- \sqrt{m^2-m + 1}}{m} \\
\end{align*}
If we are looking for an approximation, we could say $p^2 \approx 0$ and $p \approx \frac{3}{2(m+1)}$
\end{questionparts}
This was the least popular of the Statistics questions, with very few attempts seen on it. Most of these tended to consist of muddled or unexplained reasoning which led to the fiddling of (i)'s given result. Progress into (ii) was either non-existent or sketchy as a consequence.