Year: 2007
Paper: 2
Question Number: 13
Course: LFM Stats And Pure
Section: Independent Events
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: 1669.0
Banger Comparisons: 17
Given that $0 < r < n$ and $r$ is much smaller than $n$, show that $\dfrac {n-r}n \approx \e^{-r/n}$. There are $k$ guests at a party. Assuming that there are exactly 365 days in the year, and that the birthday of any guest is equally likely to fall on any of these days, show that the probability that there are at least two guests with the same birthday is approximately $1-\e^{-k(k-1)/730}$. Using the approximation $ \frac{253}{365} \approx \ln 2$, find the smallest value of $k$ such that the probability that at least two guests share the same birthday is at least $\frac12$. How many guests must there be at the party for the probability that at least one guest has the same birthday as the host to be at least $\frac12$?Given $0 < r \ll n$, then $\frac{r}{n}$ is small and so, $e^x \approx 1+x$, therefore: $\displaystyle e^{-r/n} \approx 1 - \frac{r}{n} = \frac{n-r}{n}$.
Line everyone in the room up in some order. The first person is always going to have a birthday we haven't seen before. The probability the second person has a new birthday is $\displaystyle 1 - \frac{1}{365}$ since they can't be born on the same day as the first person. The third person has a $\displaystyle 1 - \frac{2}{365}$ probability of having a birthday we've not seen before, since they can't share a birthday with either of the first two people. Similarly the $k$th person has a $\displaystyle 1 - \frac{k-1}{365}$ chance of having a unique birthday.
\begin{align*}
\prod_{i=1}^k \mathbb{P}(\text{the } i \text{th person has a new birthday}) &= \prod_{i=1}^k \l 1 - \frac{i-1}{365}\r \\
&\approx \prod_{i=1}^k \exp \l -\frac{i-1}{365}\r \\
&= \exp\l - \sum_{i=1}^k\frac{i-1}{365}\r \\
&= \exp\l - \frac{k(k-1)}{2\cdot365}\r \\
&= e^{-k(k-1)/730}
\end{align*}
But this the probability no-one shares a birthday, so the answer we are looking for is $1-$ this, ie $1 - e^{-k(k-1)/730}$
Suppose $1 - e^{-k(k-1)/730} = \frac12$, then
\begin{align*}
&& 1 - e^{-k(k-1)/730} &= \frac12 \\
\Rightarrow && e^{-k(k-1)/730} &= \frac12 \\
\Rightarrow && -k(k-1)/730 &= -\ln 2 \\
\Rightarrow && k(k-1)/730 &\approx \frac{253}{365} \\
\Rightarrow && k(k-1) &\approx 506
\end{align*}
Therefore since $22 \cdot 23 = 506$, we should expect the number to be approximately $23$. Since $e^{-r/n} > \frac{n-r}{n}$ we should expect this to be an overestimate, therefore $23$ should suffice.
This is a lovely approach to a well-known problem, and employs a very handy rational approximation to ln 2. Although it drew a small number of attempts, many of these were partial attempts at best, and few were seen of a good standard throughout. Disappointingly, several candidates arriving at the correct quadratic equation in the third part didn't seem to know how to go about solving it. As was mentioned earlier, regarding the end of Q11, working with approximations proved to be a particular obstacle for most candidates who made it to the last part here.