Year: 2019
Paper: 1
Question Number: 11
Course: LFM Stats And Pure
Section: Geometric Distribution
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; all of which are available from the STEP and Cambridge Examinations Board websites. 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 for several months beforehand. 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 it is that a candidate should never think that they are simply required to 'go through the motions' but must expect, sooner or later, to 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. When you read through the report and look at the solutions (either in the mark-scheme or the Hints & 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's paper produced the usual sorts of outcomes, with far too many candidates wasting valuable time by attempting more than six questions, and with many of these candidates picking up 0-4 marks on several 'false starts' which petered out the moment some understanding was required. Around one candidate in eight failed to hit the 30 mark overall, though this is an improvement on last year. Most candidates were able to produce good attempts at two or more questions. At the top end of the scale, around a hundred candidates scored 100 or more out of 120, with four hitting the maximum of 120 and many others not far behind. 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, although under two-thirds of the entry attempted it this year, and it also turned out to be the most successful question on the paper with a mean score of about 12 out of 20. In order of popularity, Q1 was followed by Qs.3, 4 and 2. Indeed, it was the pure maths questions in Section A that attracted the majority of attention from candidates, with the applied questions combined scoring fewer 'hits' than any one of the first four questions on its own. Though slightly more popular than the applied questions, the least successful question of all was Q5, on vectors. This question was attempted by almost 750 candidates, but 70% of these scored no more than 2 marks, leaving it with a mean score of just over 3 out of 20. Q9 (a statics question) was found only marginally more appetising, with a mean score of almost 3½ out of 20. In general, it was found that explanations were poorly supplied, with many candidates happy to overlook completely any requests for such details.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Two people adopt the following procedure for deciding where to go for a cup of tea: either to a hotel or to a tea shop. Each person has a coin which has a probability $p$ of showing heads and $q$ of showing tails (where $p+q = 1$). In each round of the procedure, both people toss their coins once. If both coins show heads, then both people go to the hotel; if both coins show tails, then both people go to the tea shop; otherwise, they continue to the next round. This process is repeated until a decision is made.
Show that the probability that they make a decision on the $n$th round is
$$(q^2 + p^2)(2qp)^{n-1}.$$
Show also that the probability that they make a decision on or before the $n$th round is at least
$$1 - \frac{1}{2^n}$$
whatever the value of $p$.
\item Three people adopt the following procedure for deciding where to go for a cup of tea: either to a hotel or to a tea shop. Each person has a coin which has a probability $p$ of showing heads and $q$ of showing tails (where $p + q = 1$). In the first round of the procedure, all three people toss their coins once. If all three coins show heads, then all three people go to the hotel; if all three coins show tails, then all three people go to the tea shop; otherwise, they continue to the next round.
In the next round the two people whose coins showed the same face toss again, but the third person just turns over his or her coin. If all three coins show heads, then all three people go to the hotel; if all three coins show tails, then all three people go to the tea shop; otherwise, they go to the third round.
Show that the probability that they make a decision on or before the second round is at least $\frac{7}{16}$, whatever the value of $p$.
\end{questionparts}\begin{questionparts}
\item The probability they don't make a decision in a round is $qp + pq = 2qp$ (TH and HT). The probability they make a decision in a round is $q^2+p^2$ (TT and HH). Therefore the probability they make a decision in the $n$th round is:
\[ (q^2+p^2)(2qp)^{n-1} \]
by having $n-1$ failures and one success.
The probability they make a decision on or before the $n$th round is the $1-$ the probability they don't, ie $1 - (2qp)^n$. Notice that $\sqrt{qp} \leq \frac{p+1}{2} = \frac12 \Rightarrow qp \leq \frac14$ so $1-(2pq)^n \leq 1 - \frac1{2^n}$
\item The probability it's decided in the first round is $p^3 + q^3$ (HHH, TTT). The probability it's decided in the second round is $3p^2q \cdot p^2 + 3qq^2 \cdot q^2 = 3pq(p^3+q^3)$ (HHT -> HHH) and (TTH -> TTT) with reorderings).
Therefore the probability of making a decision in the first or second round is $(p^3+q^3)(1 + 3pq)$ which is minimised when $p = q$ by Muirhead (or whatever your favourite inequality is). So $\frac{2}{8} \cdot \left ( 1 + \frac{3}{4} \right) = \frac{7}{16}$
\end{questionparts}
A relatively small number of candidates attempted this question, which was generally not well done. In part (i), candidates often failed to explain clearly how to get the given answer, with some simply calculating the probability for the first few values of n explicitly and spotting a pattern. Many candidates identified the given value as corresponding to p = 0.5, but lost marks for failing to explain why the probability was minimised at this point. Successful approaches to this part included differentiating, completing the square, or the AM-GM inequality. Most candidates struggled with the setup for part (ii). Most got the correct probability of finishing in the first round, but generally didn't properly account for the different cases in the second round depending on the first round tosses. One occasional misconception was to think that the process is similar to part (i), i.e. that all three coins are tossed again if the first round is inconclusive. This incorrect approach fortuitously produces the same probability as a correct approach, so candidates who successfully analysed this situation could still get most of the marks. Again, relatively few candidates explained clearly why p = 0.5 gives the minimal probability.