Year: 2018
Paper: 1
Question Number: 12
Course: LFM Stats And Pure
Section: Conditional Probability
In order to get the fullest picture, this document should be read in conjunction with the question paper, the marking scheme and (for comments on the underlying purpose and motivation for finding the right solution-approaches to questions) the Hints and Solutions document. The purpose of the STEPs is to learn what students are able to achieve mathematically when applying the knowledge, skills and techniques that they have learned within their standard A-level (or equivalent) courses … but seldom within the usual range of familiar settings. STEP questions require candidates to work at an extended piece of mathematics, often with the minimum of specific guidance, and to make the necessary connections. This requires a very different mind-set to that which is sufficient for success at A-level, and the requisite skills tend only to develop with prolonged and determined practice at such longer questions. One of the most crucial features of the STEPs is that the routine technical and manipulative skills are almost taken for granted; it is necessary for candidates to produce them with both speed and accuracy so that the maximum amount of time can be spent in thinking their way through the problem and the various hurdles and obstacles that have been set before them. Most STEP questions begin by asking the solver to do something relatively routine or familiar before letting them loose on the real problem. Almost always, such an opening has not been put there to allow one to pick up a few easy marks, but rather to point the solver in the right direction for what follows. Very often, the opening result or technique will need to be used, adapted or extended in the later parts of the question, with the demands increasing the further on that one goes. So a candidate should never think that they are simply required to 'go through the motions'; rather they will, sooner or later, be required to show either genuine skill or real insight in order to make a reasonably complete effort. The more successful candidates are the ones who manage to figure out how to move on from the given starting-point. Finally, reading through a finished solution is often misleading – even unhelpful – unless you have attempted the problem for yourself. This is because the thinking has been done for you. So, when you read through the report and look at the solutions (either in the mark scheme or the Hints and Solutions booklet), try to figure out how you could have arrived at the solution, learn from your mistakes and pick up as many tips as you can whilst working through past paper questions. This year far too many candidates wasted time by attempting more than six questions, with many of these candidates picking up 0-4 marks on several 'false starts' which petered out the moment some understanding was required. There were almost 2000 candidates for this SI paper. Almost one-sixth of candidates failed to reach a total of 30 and around two-thirds fell below half-marks overall. This highlights the fact that many candidates don't find this test an easy one. At the other end of the spectrum, almost one-in-ten managed a total of 84 out of 120 – these candidates usually marked out by their ability to complete whole questions – with almost 4% of the entry achieving the highly praiseworthy feat of getting into three-figures with their overall score. The paper is constructed so that question 1 is very approachable indeed, the intention being to get everyone started with some measure of success; unsurprisingly, Q1 was the most popular question of all, with almost all candidates attempting it, and it also turned out to be the most successful question on the paper with a mean score of more than 15 out of 20. Around 7% of candidates didn't make any kind of attempt at it at all. In order of popularity, Q1 was followed by Qs. 2, 7, 4 and 3. Indeed, it was the pure maths questions in Section A that attracted the majority of attention from candidates, with the most popular applied question (Q9, mechanics) still getting fewer 'hits' than the least popular pure question (Q5). Questions 10, 11 and 13 proved to attract very little attention from candidates and many of the attempts were minimal.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 4
A multiple-choice test consists of five questions. For each question, $n$ answers are given ($n\ge2$) only one of which is correct and candidates either attempt the question by choosing one of the $n$ given answers or do not attempt it.
For each question attempted, candidates receive two marks for the correct answer and lose one mark for an incorrect answer. No marks are gained or lost for questions that are not attempted. The pass mark is five.
Candidates A, B and C don't understand any of the questions
so, for any question which they attempt, they each choose one of the $n$ given answers at random, independently of their choices for any other question.
\begin{questionparts}
\item Candidate A chooses in advance to attempt exactly $k$ of the five questions, where $k=0, 1, 2, 3, 4$ or $5$. Show that, in order to have the greatest probability of passing the test, she should choose $k=4\,$.
\item Candidate B chooses at random the number of questions he will attempt, the six possibilities being equally likely. Given that Candidate B passed the test find, in terms of $n$, the probability that he attempted exactly four questions.
[Not on original test: Show that this probability is an increasing function of $n$.]
\item For each of the five questions Candidate C decides whether to attempt the question by tossing a biased coin. The coin has a probability of $\frac n{n+1}$ of showing a head, and she attempts the question if it shows a head. Find the probability, in terms of $n$, that Candidate C passes the test.
\end{questionparts}\begin{questionparts}
\item Her probability of passing if she answers $k \leq 2$ is $0$, since she can attain at most $4$ marks.
If she attempts $3$ questions, she needs to get all of them right, hence $\mathbb{P}(\text{gets all }3\text{ correct}) = \frac{1}{n^3}$.
If she attempts $4$ questions, we can afford to get one wrong
\begin{align*}
&& \mathbb{P}(\text{passes}|\text{attempts }4) &=\mathbb{P}(4/4) +\mathbb{P}(3/4) \\
&&&= \frac{1}{n^4} + 4\cdot\frac{1}{n^3} \cdot \frac{n-1}{n} \\
&&&= \frac{4n-3}{n^4}
\end{align*}
If she attempts $5$ questions she can get $5$ right (10), $4$ right, $1$ wrong (7), but $3$ right will not work ($6 - 2 = 4< 5$), hence:
\begin{align*}
&& \mathbb{P}(\text{passes}|\text{attempts }5) &=\mathbb{P}(5/5) +\mathbb{P}(4/5) \\
&&&= \frac{1}{n^5} + 5\cdot\frac{1}{n^4} \cdot \frac{n-1}{n} \\
&&&= \frac{5n-4}{n^5}
\end{align*}
If $4n-3 > n \Leftrightarrow n \geq 2$ then $4$ attempts is better than $3$. If $4n^2-3n > 5n-4 \Leftrightarrow 4n^2-8n+4 = 4(n-1)^2 > 0 \Leftrightarrow n \geq$ then $4$ is better than $5$, but $n$ is $\geq 2$ so, $4$ is the best option.
\item $\,$
\begin{align*}
&& \mathbb{P}(\text{passes}) &= \frac16 \cdot 0 + \frac16 \cdot 0 + \frac16 \cdot 0 + \frac16 \cdot \frac1{n^3} + \frac16 \frac{4n-3}{n^4} + \frac16 \frac{5n-4}{n^5} \\
&&&= \frac{n^2+4n^2-3n+5n-4}{6n^5} \\
&&&= \frac{5n^2+2n-4}{6n^5} \\
&& \mathbb{P}(\text{answered }4|\text{passes}) &= \frac{\mathbb{P}(\text{answered }4\text{ and passes})}{ \mathbb{P}(\text{passes})} \\
&&&= \frac{\frac16 \cdot \frac{4n-3}{n^4}}{\frac{5n^2-2n-4}{6n^5} } \\
&&&= \frac{4n^2-3n}{5n^2+2n-4}
\end{align*}
Notice that the function takes all values for $n$ between the roots of the denominator (which are either side of $0$ and below $3/4$. Therefore after $3/4$ the function must be increase since otherwise we would have a quadratic equation with more than $2$ roots.
\item $\,$ \begin{align*}
&&\mathbb{P}(C \text{ passes}) &= \binom{5}{3} \left ( \frac{n}{n+1} \right)^3 \left ( \frac{1}{n+1}\right)^2 \frac{1}{n^3} + \binom{5}{4} \left ( \frac{n}{n+1} \right)^4 \left ( \frac{1}{n+1}\right) \frac{4n-3}{n^4} +\\
&&&\quad \quad + \binom{5}{5} \left ( \frac{n}{n+1} \right)^5 \frac{5n-4}{n^5} \\
&&&= \frac{10}{(n+1)^5} + \frac{5(4n-3)}{(n+1)^5} + \frac{(5n-4)}{(n+1)^5} \\
&&&= \frac{10+20n-15+5n-4}{(n+1)^5}\\
&&&= \frac{25n-9}{(n+1)^5}\\
\end{align*}
\end{questionparts}
This was a reasonably popular question, with almost a third of the candidature making an attempt at it. Most of these candidates did (i) and (ii) well. It was quite common to state ∼ Bin 2, and write down the mean and variance of that. This was acceptable but, because the variance is given in the question, we required candidates to explicitly state the distribution rather than just giving the answers. A lot of candidates struggled with (iii), commonly thinking it was still binomial or failing to condition on the chosen coins. While the most common successful approach was to calculate the combined probabilities of different numbers of heads, some candidates found the mean correctly by averaging the means of pairs of coins. It was very rare to obtain the variance correctly with this latter approach. There were relatively few attempts at part (iv) which got the correct expression for the difference of the variances, and very few came up with the idea of completing the square. There was a very small number – no more than 5 – of correct solutions by the AM-GM inequality or more exotic methods. The condition for equality was often not justified.