Year: 2017
Paper: 2
Question Number: 13
Course: LFM Stats And Pure
Section: Negative Binomial Distribution
This year's paper was, perhaps, slightly more straightforward than usual, with more helpful guidance offered in some of the questions. Thus the mark required for a "1", a Distinction, was 80 (out of 120), around ten marks higher than that which would customarily be required to be awarded this grade. Nonetheless, a three‐figure mark is still a considerable achievement and, of the 1330 candidates sitting the paper, there were 89 who achieved this. At the other end of the scale, there were over 350 who scored 40 or below, including almost 150 who failed to exceed a total score of 25. As a general strategy for success in a STEP examination, candidates should be looking to find four "good" questions to work at (which may be chosen freely by the candidates from a total of 13 questions overall). It is unfortunately the case that so many low‐scoring candidates flit from one question to another, barely starting each one before moving on. There needs to be a willingness to persevere with a question until a measure of understanding as to the nature of the question's purpose and direction begins to emerge. Many low‐scoring candidates fail to deal with those parts of questions which cover routine mathematical processes ‐ processes that should be standard for an A‐level candidate. The significance of the "rule of four" is that four high‐scoring questions (15‐20 marks apiece) obtains you up to around the total of 70 that is usually required for a "1"; and with a couple of supporting starts to questions, such a total should not be beyond a good candidate who has prepared adequately. This year, significantly more than 10% of candidates failed to score at least half marks on any one question; and, given that Q1 (and often Q2 also) is (are) specifically set to give all candidates the opportunity to secure some marks, this indicates that these candidates are giving up too easily. Mathematics is about more than just getting to correct answers. It is about communicating clearly and precisely. Particularly with "show that" questions, candidates need to distinguish themselves from those who are just tracking back from given results. They should also be aware that convincing themselves is not sufficient, and if they are using a result from 3 pages earlier, they should make this clear in their working. A few specifics: In answers to mechanics questions, clarity of diagrams would have helped many students. If new variables or functions are introduced, it is important that students clearly define them. One area which is very important in STEP but which was very poorly done is dealing with inequalities. Although a wide range of approaches such as perturbation theory were attempted, at STEP level having a good understanding of the basics – such as changing the inequality if multiplying by a negative number – is more than enough. In fact, candidates who used more advanced methods rarely succeeded.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
In a television game show, a contestant has to open a door using a key. The contestant is given a bag containing $n$ keys, where $n\ge2$. Only one key in the bag will open the door. There are three versions of the game. In each version, the contestant starts by choosing a key at random from the bag.
\begin{questionparts}
\item In version 1, after each failed attempt at opening the door the key that has been tried is put back into the bag and the contestant again selects a key at random from the bag. By considering the binomial expansion of $( 1 - q)^{-2}$, or otherwise, find the expected number of attempts required to open the door.
\item In version 2, after each failed attempt at opening the door the key that has been tried is put aside and the contestant selects another key at random from the bag. Find the expected number of attempts required to open the door.
\item In version 3, after each failed attempt at opening the door the key that has been tried is put back into the bag and another incorrect key is added to the bag.
The contestant then selects a key at random from the bag. Show that the probability that the contestant draws the correct key at the $k$th attempt is
\[ \frac{n-1}{(n+k-1)(n+k-2)} \,.\]
Show also, using partial fractions, that the expected number of attempts required to open the door is infinite.
You may use without proof the result that $\displaystyle\sum_{m=1}^N \dfrac 1 m \to \infty \,$ as $N\to \infty\,$.
\end{questionparts}\begin{questionparts}
\item The probability they pull the key out on the $k$th attempt will be $\left ( \frac{n-1}{n} \right)^{k-1} \frac1n$, so we want:
\begin{align*}
\E[G_1] &= \sum_{k=1}^{\infty} k \cdot \left ( \frac{n-1}{n} \right)^{k-1} \frac1n \\
&= \frac{1}n \sum_{k=1}^{\infty} k \cdot \left ( \frac{n-1}{n} \right)^{k-1} \\
&= \frac1n \frac{1}{\left (1 - \frac{n-1}{n} \right)^2} \\
&= \frac{1}{n} \frac{n^2}{1^2} = n
\end{align*}
\item In version 2, the probability the correct key comes out at the $k$th attempt is $\frac1n$ (assume we take out all the keys, then the correct key is equally likely to appear in all of the space). Therefore $\E[G_2] = \frac1n (1 + 2 + \cdots + n) = \frac{n+1}{2}$
\item The probability the key comes out on the correct attempt is:
\begin{align*}
&& \mathbb{P}(G_3 = k) &= \frac{n-1}{n} \cdot \frac{n}{n+1} \cdot \frac{n+1}{n+2} \cdots \frac{n+k-3}{n+k-2} \cdot \frac{1}{n+k-1} \\
&&&= \frac{n-1}{(n+k-2)(n+k-1)} \\
\\
&&k \cdot \mathbb{P}(G_3 = k) &= \frac{k(n-1)}{(n+k-2)(n+k-1)} \\
&&&= \frac{(n-1)(2-n)}{n+k-2} + \frac{(n-1)^2}{n+k-1} \\
&&&= \frac{(n-1)^2}{n+k-1} - \frac{(n-1)^2}{n+k-2} + \frac{n-1}{n-k+2} \\
\Rightarrow && \E[G_3] &= \sum_{k=1}^{\infty} k \cdot \mathbb{P}(G_3 = k) \\
&&&= \sum_{k=1}^{\infty} \left ( \frac{(n-1)^2}{n+k-1} - \frac{(n-1)^2}{n+k-2} + \frac{n-1}{n+k-2} \right) \\
&&&= \sum_{k=1}^{\infty} \left ( \frac{(n-1)^2}{n+k-1} - \frac{(n-1)^2}{n+k-2} \right) +\underbrace{\sum_{k=1}^{\infty} \frac{n-1}{n-k+2}}_{\to \infty} \\
\end{align*}
\end{questionparts}
The first two parts of this question were reasonably straightforward, but this was only marginally more popular than question 12. A surprising number of candidates did not seem to be confident dealing with telescoping fractions – a fairly common tool in probability. The algebraic demands of the final part proved challenging for many candidates.