Year: 2009
Paper: 1
Question Number: 13
Course: LFM Stats And Pure
Section: Discrete Probability Distributions
There were significantly more candidates attempting this paper again this year (over 900 in total), and the scores were pleasing: fewer than 5% of candidates failed to get at least 20 marks, and the median mark was 48. The pure questions were the most popular as usual; about two-thirds of candidates attempted each of the pure questions, with the exceptions of question 2 (attempted by about 90%) and question 5 (attempted by about one third). The mechanics questions were only marginally more popular than the probability and statistics questions this year; about one quarter of the candidates attempted each of the mechanics questions, while the statistics questions were attempted by about one fifth of the candidates. A significant number of candidates ignored the advice on the front cover and attempted more than six questions. In general, those candidates who submitted answers to eight or more questions did fairly poorly; very few people who tackled nine or more questions gained more than 60 marks overall (as only the best six questions are taken for the final mark). This suggests that a skill lacking in many students attempting STEP is the ability to pick questions effectively. This is not required for A-levels, so must become an important part of STEP preparation. Another "rubric"-type error was failing to follow the instructions in the question. In particular, when a question says "Hence", the candidate must make (significant) use of the preceding result(s) in their answer if they wish to gain any credit. In some questions (such as question 2), many candidates gained no marks for the final part (which was worth 10 marks) as they simply quoted an answer without using any of their earlier work. There were a number of common errors which appeared across the whole paper. These included a noticeable weakness in algebraic manipulations, sometimes indicating a serious lack of understanding of the mathematics involved. As examples, one candidate tried to use the misremembered identity cos β = sin √(1 − β²), while numerous candidates made deductions of the form "if a² + b² = c², then a + b = c" at some point in their work. Fraction manipulations are also notorious in the school classroom; the effects of this weakness were felt here, too. Another common problem was a lack of direction; writing a whole page of algebraic manipulations with no sense of purpose was unlikely to either reach the requested answer or gain the candidate any marks. It is a good idea when faced with a STEP question to ask oneself, "What is the point of this (part of the) question?" or "Why has this (part of the) question been asked?" Thinking about this can be a helpful guide. One aspect of this is evidenced by pages of formulæ and equations with no explanation. It is very good practice to explain why you are doing the calculation you are, and to write sentences in English to achieve this. It also forces one to focus on the purpose of the calculations, and may help avoid some dead ends. Finally, there is a tendency among some candidates when short of time to write what they would do at this point, rather than using the limited time to actually try doing it. Such comments gain no credit; marks are only awarded for making progress in a question. STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1504.1
Banger Comparisons: 4
I seat $n$ boys and $3$ girls in a line at random, so
that each order of the $n+3$ children
is as likely to occur as any other. Let $K$ be the maximum number of
consecutive girls in the line so, for example, $K=1$ if
there is at least one boy between
each pair of girls.
\begin{questionparts}
\item Find $\P(K=3)$.
\item Show that
\[\P(K=1)=
\frac{n(n-1)}{(n+2)(n+3)}\,.
\]
\item Find $\E(K)$.
\end{questionparts}\begin{questionparts}
\item If all the girls are say together there are $n+1$ ways to place the block of 3 girls. There are $\binom{n+3}{3}$ ways to choose where to place the girls in total, therefore:
\begin{align*}
&& \mathbb{P}(K =3) &= \frac{n+1}{\binom{n+3}3} \\
&&&= \frac{6(n+1)}{(n+3)(n+2)(n+1)} \\
&&&= \frac{6}{(n+3)(n+2)}
\end{align*}
\item If $K= 1$ then all of the girls are separated. We can place three girls and two boys separating them, then we are allocating $N-2$ boys to $4$ gaps, ie $\binom{N-2+3}{3} = \binom{N+1}{3}$.
\begin{align*}
&& \mathbb{P}(K=3) &= \frac{\binom{n+1}{3}}{\binom{n+3}{3}} \\
&&&= \frac{(n+1)n(n-1)}{(n+3)(n+2)(n+1)} \\
&&&= \frac{n(n-1)}{(n+3)(n+2)}
\end{align*}
\item $\,$
\begin{align*}
\mathbb{E}(K) &= \sum_{k=1}^3 k \mathbb{P}(K=k) \\
&= \frac{6}{(n+3)(n+2)} + 2 \left (1 - \frac{6}{(n+3)(n+2)} - \frac{n(n-1)}{(n+3)(n+2)} \right) + 3\frac{n(n-1)}{(n+3)(n+2)} \\
&= 2+\frac{6-12+n(n-1)}{(n+3)(n+2)} \\
&= 2 + \frac{n^2-n-6}{(n+2)(n+3)}\\
&= 2 + \frac{(n-3)(n+2)}{(n+2)(n+3)} \\
&= 2 + \frac{n-3}{n+3} \\
&= \frac{2n}{n+3}
\end{align*}
\end{questionparts}
This was a very unpopular question and most attempts were fragmentary, while a small number were essentially perfect. Most attempts were very weak at conveying their combinatorial ideas clearly, leading to confused and incorrect arguments. A few candidates completely misunderstood what was being asked and tried to answer something different entirely. For part (i), very few candidates realised that the best way to begin part (i) was to regard the girls as a block dividing the boys into two sections. Even those who did spot this still struggled to deduce the required probability. Most of those who counted the total number of arrangements as (n + 3)! and the number of arrangements of the girls as 3! failed to see that the girls could be positioned as a block in only (n + 1) positions. For part (ii), most offerings were vain attempts to reach the given answer, but demonstrated no understanding of the problem.