Year: 2016
Paper: 1
Question Number: 12
Course: LFM Stats And Pure
Section: Probability Definitions
This year, more than 2000 candidates signed up to sit this paper, though just under 2000 actually sat it. This figure is about the same as the entry figure for 2015, though the number of candidates opting to sit STEP I has risen significantly over recent years; for instance, it was around 1500 in 2013. There is no doubt that the purpose of the STEPs is to learn which students can genuinely use their mathematical knowledge, skills and techniques in an arena that demands of them a level of performance that exceeds anything they will have encountered within the standard A-level (or equivalent) assessments. The ability to work at an extended piece of mathematical work, often with the minimum of specific guidance, allied to the need for both determination and the ability to "make connections" at speed and under considerable time pressure, are characteristics that only follow from careful preparation, and there is a great benefit to be had from an early encounter with, and subsequent prolonged exposure to, these kinds of questions. It is not always easy to say what level of preparation has been undertaken by candidates, but the minimum expected requirement is the ability to undertake routine A-level-standard tasks and procedures with speed and accuracy. At the top end of the scale, almost 100 candidates produced a three-figure score to the paper, which is a phenomenal achievement; and around 250 others scored a mark of 70+, which is also exceptionally impressive. At the other end of the scale, over 400 candidates failed to reach a total of 40 marks out of the 120 available. For STEP I, the most approachable questions are always set as Qs.1 & 2 on the paper, with Q1 in particular intended to afford every candidate the opportunity to get something done successfully. So it is perfectly reasonable for a candidate, upon opening the paper, to make an immediate start at the first and/or second question(s) before looking around to decide which of the remaining 10 or 11 questions they feel they can tackle. It is very important that candidates spend a few minutes – possibly at the beginning – reading through the questions to decide which six they intend to work, since they will ultimately only be credited with their best six question marks. Many students spend time attempting seven, eight, or more questions and find themselves giving up too easily on a question the moment the going gets tough, and this is a great pity, since they are not allowing themselves thinking time, either on the paper as a whole or on individual questions. The other side to the notion of strategy is that most candidates clearly believe that they need to attempt (at least) six questions when, in fact, four questions (almost) completely done would guarantee a Grade 1 (Distinction), especially if their score on these first four questions were then to be supplemented by a couple of early attempts at the starting parts of a couple more questions (for the first five or six marks); attempts which need not take longer than, say, ten minutes of their time. It is thus perfectly reasonable to suggest to candidates, in their preparations, that they can spend more than 30 minutes on a question, but only IF they think they are going to finish it off satisfactorily, although it might be best if they were advised to spend absolutely no more than 40-45 minutes on any single question; if they haven't finished by then, it really is time to move on. Curve-sketching skills are usually a common weakness, but were only tested on this paper in Q3. The other common area of weakness – algebra – was tested relatively frequently, and proved to be as testing as usual. Calculus skills were generally "okay" although the integration of first-order differential equations by the separation of variables, as appearing repeatedly in Q4, was found challenging by many of the candidates who attempted this question. The most noticeable deficiency, however, was in the widespread inability to construct an argument, particularly in Qs. 5, 7 & 8. Vectors are often poorly handled, and this year proved no exception.
Difficulty Rating: 1516.0
Difficulty Comparisons: 1
Banger Rating: 1484.7
Banger Comparisons: 3
\begin{questionparts}
\item Alice tosses a fair coin twice and Bob tosses a fair coin three times.
Calculate the probability that Bob gets more heads than Alice.
\item Alice tosses a fair coin three times
and Bob tosses a fair coin four times.
Calculate the probability that Bob gets more heads than Alice.
\item
Let $p_1$ be the probability that Bob gets the same number of heads as
Alice, and let~$p_2$ be the probability that Bob gets more
heads than Alice, when
Alice and Bob each toss a fair coin $n$ times.
Alice tosses a fair coin $n$ times and Bob tosses a fair coin $n+1$ times.
Express the probability that Bob gets more heads than Alice
in terms of $p_1$ and $p_2$, and hence obtain a generalisation of the
results of parts (i) and (ii).
\end{questionparts}\begin{questionparts}
\item There are several possibilities
\begin{array}{c|c|c}
\text{Alice} & \text{Bob} & P \\ \hline
0 & 1 & \frac1{2^2} \cdot 3 \cdot \frac{1}{2^3} = \frac{3}{2^5} \\
0 & 2 & \frac1{2^2} \cdot 3 \cdot \frac{1}{2^3} = \frac{3}{2^5} \\
0 & 3 & \frac1{2^2} \cdot \frac{1}{2^3} = \frac{1}{2^5} \\
1 & 2 & 2 \cdot \frac1{2^2} \cdot 3 \cdot \frac{1}{2^3} = \frac{6}{2^5} \\
1 & 3 & 2\cdot \frac1{2^2} \cdot \frac{1}{2^3} = \frac{2}{2^5} \\
2 & 3 & \frac1{2^2} \cdot \frac{1}{2^3} = \frac{1}{2^5} \\ \hline
&& \frac{1}{2^5}(3+3+1+6+2+1) = \frac{16}{2^5} = \frac12
\end{array}
\item There are several possibilities
\begin{array}{c|c|c}
A & B & \text{count} \\ \hline
0 & 1 & 4 \\
0 & 2 & 6 \\
0 & 3 & 4 \\
0 & 4 & 1 \\
1 & 2 & 3\cdot6 \\
1 & 3 & 3\cdot4 \\
1 & 4 & 3 \\
2 & 3 & 3\cdot4 \\
2 & 4 & 3 \\
3 & 4 & 1 \\ \hline
&& 64
\end{array}
Therefore the total probability is $\frac12$
\item $\mathbb{P}(\text{Bob more than Alice}) = p_1 \cdot \underbrace{\frac12}_{\text{he wins by breaking the tie on his last flip}} + p_2$
If $p_3$ is the probability that Alice gets more heads than Bob, then by symmetry $p_3 = p_2$ and $p_1 + p_2 + p_3 = 1$. Therefore $p_1 + 2p_2 = 1$. ie $\frac12 p_1 + p_2 = \frac12$ therefore the answer is always $\frac12$ for all values of $n$.
\end{questionparts}
This question proved to be an especially popular choice (almost 45% uptake) and a high-scoring one at that (mean score of over 9 out of 20). However, most of these medium-successful marks came from attempts at the first two (specific) cases, in which Bob threw 2 coins, then 3. There were 6 marks for successful attempts at each of these preliminary parts, and candidates were happy to work them through reasonably successfully, although some of these attempts were very poorly explained; and occasionally it was the case that there was no explanation at all. Although they may have been getting the correct answers, it is imperative to ensure that answers are sufficiently coherent for the reader to be able to follow their reasoning and/or structure and hence understand how the answers have been arrived at. The final part of Q12 elicited little more than a few half-efforts at the general situation, but the general reasoning required proved beyond most candidates. A small number of candidates thought they were being asked for an inductive proof of some kind, using parts (i) and (ii) as base-line cases.