Year: 2015
Paper: 1
Question Number: 13
Course: LFM Stats And Pure
Section: Negative Binomial Distribution
The aim of this report is to account for how well, or how poorly, the candidates performed this year while at the same time attempting to indicate their corresponding areas of strength, or weakness. If I should concentrate marginally more in the direction of the candidates' weaknesses, the reader should understand that it is with the hope that both future STEP candidates, and the teachers preparing them for this examination in years to come, will have the opportunity to focus on those areas of common weakness in an effort to ensure a better preparedness for what is unquestionably the most demanding of examinations in the UK for students of pre-university years. For the record, the scripts are marked by a team of postgraduate mathematical students – many working towards a doctorate in this, or another, closely-related subject – who spend days poring over the scripts, working in small teams under the supervision of the Principal Examiner and carefully appointed "question captains". Their powers of concentration are truly phenomenal and they not only appreciate the need for mathematical rigour but (due to having once been in exactly this position themselves) are also deeply sympathetic towards the candidates; making every effort first to understand what has been presented to them by the candidates and then to reward genuinely good mathematics when it appears, no matter how hastily and/or messily it has been set down onto paper. Thus it is that the comments produced by the Principal Examiner within this Report are merely summaries of what these markers have passed on to him (or her) at the end of the marking period. Moreover, since the candidates' backgrounds are entirely unknown to the markers, any comments – critical or otherwise – cannot possibly be taken to have been directed towards specifically chosen targets. More than 2000 candidates sat SI this year, which represents another increase of around 10% over last year's entry numbers. Once again, however, it is sadly the case that many of these candidates have simply not prepared sufficiently well to be in a position to emerge from the experience with any amount of positive feelings of success at the results of their efforts. In the first instance, many candidates (this year, more than half of the entry) attempt more than the recommended six questions. This automatically penalises them for the time that they have spent on extra questions whose marks will not count towards their final total – remember that only the highest-scoring six questions count towards a candidate's final total; note also that a grade 1 can usually be obtained from four questions which have been completed reasonably successfully, or from two questions done completely correctly plus four "halves", or from any in-between combination of question-scores. Thus, it is strongly advised that candidates spend a few minutes at some stage of the examination reading the questions carefully with a view to deciding which of them they would best attempt. Overall, this year's paper worked out in very much the same way as had the 2014 paper, with a mean score of around 43-44% and with approximately 45% of candidates failing to exceed a total of 40 marks; though totals in excess of 100 marks were slightly down on 2014. Part of the reason for this is that the five applied maths questions were very much non-standard this year, and this prevented a lot of very easy marks (for routine beginnings) being picked up by candidates attempting these questions. The mean score for Qs.9-13 thus fell from 6½ in 2014 to 2½ in 2015. Another trend of recent years is the widespread dislike for the vectors questions, which have become both unpopular and very low-scoring for candidates. Points of general application regarding candidates' attempts this year are little different to usual – far too many candidates produce only fragmentary attempts (often, as mentioned above, to almost every question they attempt) at solutions, with little apparent intent to persevere beyond the first obstacle. Presentation was also particularly poor this year, with most candidates making life hard both for themselves and for the markers who genuinely wish to find credit-worthy mathematics in order to award the marks available. In long questions such as these, with the barest minimum of structure provided, candidates need a lot of prior practice at past STEP questions in learning how to supply their own. Curve-sketching skills continue to be a weakness, as candidates tend to veer away from justifying what they have drawn; algebra and calculus skills are very mixed, and it was especially clear this year how little candidates like being required to formulate their own solution-strategies – no doubt being the result of an over-reliance on being told exactly what to do, as is customarily the case in AS- and A-level papers. On the other side of the coin, there was a very pleasing number of candidates who produced exceptional pieces of work on 5 or 6 questions (or more), and thus scored very highly indeed on the paper overall. Around 80 of them scored 90+ marks of the 120 available, and they should be very proud of their performance – it is a significant and noteworthy achievement.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1501.1
Banger Comparisons: 4
A fair die with faces numbered $1, \ldots, 6$ is thrown repeatedly. The events $A$, $B$, $C$, $D$ and $E$ are defined as follows.
\begin{align*}
A: && \text{the first 6 arises on the $n$th throw.}\\
B: && \text{at least one 5 arises before the first 6.} \\
C: && \text{at least one 4 arises before the first 6.}\\
D: && \text{exactly one 5 arises before the first 6.}\\
E: && \text{exactly one 4 arises before the first 6.}
\end{align*}
Evaluate the following probabilities:
\begin{questionparts}
\item $\P(A)$
\item $\P(B)$
\item $\P(B\cap C)$
\item $\P(D)$
\item $\P(D\cup E)$
\end{questionparts}
For some parts of this question, you may want to make use of the binomial expansion in the form:
\[ (1-x)^{-n} = 1 +nx +\frac {n(n+1)}2 x^2 + \cdots + \frac {(n+r-1)!}{r! (n-1)!}x^r +\cdots\ .\]\begin{questionparts}
\item $\,$ \begin{align*}
\mathbb{P}(A) &= \mathbb{P}(\text{the first 6 arises on the $n$th throw.}) \\
&= \mathbb{P}(\text{$n-1$ not 6s, followed by a 6.})\\
&= \left ( \frac56\right)^{n-1} \cdot \frac16 = \frac{5^{n-1}}{6^n}
\end{align*}
\item There is nothing special about $5$ or $6$, so which comes first is $50:50$, therefore this probability is $\frac12$
\item There is nothing special about $4$, $5$ or $6$ so this is the probability that $6$ appears last out of these three numbers, hence $\frac13$
\item $\,$ \begin{align*}
\mathbb{P}(D) &= \mathbb{P}(\text{exactly one 5 arises before the first 6.}) \\
&=\sum_{n=2}^{\infty} \mathbb{P}(\text{exactly one 5 arises before the first 6 which appears on the $n$th roll.}) \\
&= \sum_{n=2}^{\infty} \binom{n-1}{1} \left ( \frac46 \right)^{n-2} \frac16 \cdot \frac16 \\
&= \frac1{36} \sum_{n=2}^{\infty} (n-1) \left ( \frac23 \right)^{n-2} \\
&= \frac1{36} \sum_{n=1}^{\infty} n \left ( \frac23 \right)^{n-1} \\
&= \frac1{36} \frac{1}{\left ( 1- \frac23 \right)^2} = \frac14
\end{align*}
\item $\,$ \begin{align*}
\mathbb{P}(D \cup E) &= \mathbb{P}(D) + \mathbb{P}(E) - \mathbb{P}(D \cap E) \\
&= \frac12 - \mathbb{P}(D \cap E) \\
&=\frac12 - \sum_{n=3}^{\infty} \mathbb{P}(\text{exactly one 5 and one 4 arises before the first 6 which appears on the $n$th roll.}) \\
&=\frac12 - \sum_{n=3}^{\infty} 2\binom{n-1}{2} \left ( \frac36 \right)^{n-3}\cdot \frac16 \cdot \frac16 \cdot \frac16 \\
&=\frac12 - \frac2{6^3}\sum_{n=3}^{\infty} \frac{(n-1)(n-2)}{2} \left ( \frac12 \right)^{n-3} \\
&=\frac12 - \frac2{6^3}\frac{1}{(1-\tfrac12)^3}\\
&= \frac12 - \frac{2}{27} \\
&= \frac{23}{54}
\end{align*}
\end{questionparts}
Most attempts at Q13 got little further on in the question than writing the simple, general term for P(A); namely (5/6)^(n−1)(1/6). Even though the Geometric Distribution is not expected here, the probability of the run of independent events "n – 1 failures followed by a success" should be within the scope of any STEP candidate (who has studied any small amount of probability and/or statistics). Many candidates decided that (ii) and (iii) also required a general expression for each of P(B) and P(B∩C), whereas these were clearly intended to be numerical, and a brief 'symmetry' argument quickly reveals their respective probabilities to be ½ and ⅓. Parts (iv) and (v) were tougher, but required the candidates to see that each was the sum of an infinite number of terms, and the helpfully given series expansion at the end of the question helped wrap these up. One third of all candidates made an attempt at this question, but almost none of them got around to either of these final two parts.