2011 Paper 1 Q12

Year: 2011
Paper: 1
Question Number: 12

Course: LFM Stats And Pure
Section: Probability Definitions

Difficulty: 1500.0 Banger: 1470.2

Problem

I am selling raffle tickets for \(\pounds1\) per ticket. In the queue for tickets, there are \(m\) people each with a single \(\pounds1\) coin and \(n\) people each with a single \(\pounds2\) coin. Each person in the queue wants to buy a single raffle ticket and each arrangement of people in the queue is equally likely to occur. Initially, I have no coins and a large supply of tickets. I stop selling tickets if I cannot give the required change.
  1. In the case \(n=1\) and \(m\ge1\), find the probability that I am able to sell one ticket to each person in the queue.
  2. By considering the first three people in the queue, show that the probability that I am able to sell one ticket to each person in the queue in the case \(n=2\) and \(m\ge2\) is \(\dfrac{m-1}{m+1}\,\).
  3. Show that the probability that I am able to sell one ticket to each person in the queue in the case \(n=3\) and \(m\ge3\) is \(\dfrac{m-2}{m+1}\,\).

Solution

  1. The only way you wont be able to sell to them is if they are first, ie \(\frac1{m+1}\)
  2. If \(n=2\), the the only way you fail to sell to them is if one comes first or they both appear before two people with pound coins, ie \(2\) or \(122\). These have probabilities \(\frac{2}{m+2}\) and \(\frac{m}{m+2} \cdot \frac{2}{m+1} \frac{1}{m} = \frac{2}{(m+1)(m+2)}\). Therefore the total probability you don't sell all the tickets is \(\frac{2}{m+2}\left ( 1 + \frac{1}{m+1} \right) = \frac{2}{m+2} \frac{m+2}{m+1} = \frac{2}{m+1}\). Therefore the probability you do sell all the tickets is \(1 - \frac{2}{m+1} = \frac{m-1}{m+1}\)
  3. The only ways to fail when \(n=3\) are: \(2\), \(122\), or if all three \(2\)s appear before three \(1\)s. this can happen in \(11222\), \(12122\) These happen with probability: \begin{align*} 2: && \frac{3}{m+3} \\ 122: && \frac{m}{m+3} \cdot \frac{3}{m+2} \cdot \frac{2}{m+1} \\ 11222: && \frac{m(m-1) 6}{(m+3)(m+2)(m+1)m(m-1)} \\ 12122: && \frac{m(m-1) 6}{(m+3)(m+2)(m+1)m(m-1)} \\ \end{align*} Therefore the total probability is: \begin{align*} P &= \frac{1}{(m+3)(m+2)(m+1)} \left (3(m+2)(m+1)+6m + 12 \right) \\ &= \frac{1}{(m+3)(m+2)(m+1)} \left (3(m+1)(m+2) \right) \\ &= \frac{3}{m+1} \end{align*} and the result follows
Examiner's report
— 2011 STEP 1, Question 12
Mean: 11 / 20 ~23% attempted (inferred) Inferred ~23% from 'fewer than one-quarter'. Median 11, upper quartile 16; one of most successfully answered.

Generally, Statistics questions are generally the least popular on STEP papers, and this year was no exception; this question was answered by fewer than one-quarter of candidates. In spite of this (or perhaps because of it), it was one of the most successfully answered on the paper, with a median mark of 11 and an upper quartile of 16. (i) A few candidates failed to get started and scored no marks at all, or misread the question and gave the probability of failure instead, but the majority gained full marks on this part. (ii) Most candidates were able to enumerate the possible cases of success or the cases of failure. However, a significant proportion did not provide any justification that they had considered all possible cases, and so lost at least 2 marks. In part (iii), this lack led a significant number to overlook one or more cases, and so get the wrong answer. Several candidates failed to realise that the probability of the second person having a £2 coin depends upon the first person's coin. (iii) Those who succeeded on part (ii) generally made good progress on this part as well; see the comments above.

There were again significantly more candidates attempting this paper than last year (just over 1100), but the scores were significantly lower than last year: fewer than 2% of candidates scored above 100 marks, and the median mark was only 44, compared to 61 last year. It is not clear why this was the case. One possibility is that Questions 2 and 3, which superficially looked straightforward, turned out to be both popular and far harder than candidates anticipated. The only popular and well-answered questions were 1 and 4. The pure questions were the most popular as usual, though there was noticeable variation: questions 1–4 were the most popular, while question 7 (on differential equations) was fairly unpopular. Just over half of all candidates attempted at least one mechanics question, which one-third attempted at least one probability question, an increase on last year. The marks were surprising, though: the two best-answered questions were the pure questions 1 and 4, but the next best were statistics question 12 and mechanics question 9. The remainder of the questions were fairly similar in their marks. A number of candidates ignored the advice on the front cover and attempted more than six questions, with a fifth of candidates trying eight or more questions. A good number of those extra attempts were little more than failed starts, but still suggest that some candidates are not very effective at question-picking. This is an important skill to develop during STEP preparation. Nevertheless, the good marks and the paucity of candidates who attempted the questions in numerical order does suggest that the majority are being wise in their choices. Because of the abortive starts, I have generally restricted my attention to those attempts which counted as one of the six highest-scoring answers in the detailed comments. On occasions, candidates spent far longer on some questions than was wise. Often, this was due to an algebraic slip early on, and they then used time which could have been far better spent tackling another question. It is important to balance the desire to finish a question with an appreciation of when it is better to stop and move on. Many candidates realised that for some questions, it was possible to attempt a later part without a complete (or any) solution to an earlier part. An awareness of this could have helped some of the weaker students to gain vital marks when they were stuck; it is generally better to do more of one question than to start another question, in particular if one has already attempted six questions. It is also fine to write "continued later" at the end of a partial attempt and then to continue the answer later in the answer booklet. As usual, though, some candidates ignored explicit instructions to use the previous work, such as "Hence", or "Deduce". They will get no credit if they do not do what they are asked to! (Of course, a question which has the phrase "or otherwise" gives them the freedom to use any method of their choosing; often the "hence" will be the easiest, though.) It is wise to remember that 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. There were a number of common errors and issues which appeared across the whole paper. The first was a lack of fluency in algebraic manipulations. STEP questions often use more variables than A-level questions (which tend to be more numerical), and therefore require candidates to be comfortable engaging in extended sequences of algebraic manipulations with determination and, crucially, accuracy. This is a skill which requires plenty of practice to master. Another area of weakness is logic. A lack of confidence in this area showed up several times. In particular, a candidate cannot possibly gain full marks on a question which reads "Show that X if and only if Y" unless they provide an argument which shows that Y follows from X and vice versa. Along with this comes the need for explanations in English: a sequence of formulæ or equations with no explicit connections between them can leave the reader (and writer) confused as to the meaning. Brief connectives or explanations would help, and sometimes longer sentences are necessary. Another related issue continues to be legibility. Many candidates at some point in the paper lost marks through misreading their own writing. One frequent error was dividing by zero. On several occasions, an equation of the form xy = xz appeared, and candidates blithely divided by x to reach the conclusion y = z. Again, I give a strong reminder that it is vital to draw appropriate, clear, accurate diagrams when attempting some questions, mechanics questions in particular.

Source: Cambridge STEP 2011 Examiner's Report · 2011-full.pdf
Rating Information

Difficulty Rating: 1500.0

Difficulty Comparisons: 0

Banger Rating: 1470.2

Banger Comparisons: 2

Show LaTeX source
Problem source
I am selling raffle tickets for $\pounds1$ per ticket.
In the  queue for tickets,
there are $m$ people each with a single $\pounds1$ coin and $n$
people each with a single $\pounds2$ coin. Each person in the 
queue wants to buy a single raffle ticket and each arrangement
of people in the queue is equally likely to occur.
Initially, I have no coins and
a large supply of tickets. I stop selling tickets if 
I cannot give the required change.
\begin{questionparts}
\item In the case $n=1$ and $m\ge1$, find the probability
that I am able to sell one ticket to each person in the queue.
\item By considering the first three people in the queue, 
show that the probability that I am able to sell one ticket
to each person in the queue in the case $n=2$ and $m\ge2$ is
$\dfrac{m-1}{m+1}\,$.
\item 
Show that the probability that I am able to sell one ticket
to each person in the queue in the case $n=3$ and $m\ge3$ is
$\dfrac{m-2}{m+1}\,$.
\end{questionparts}
Solution source
\begin{questionparts}
\item The only way you wont be able to sell to them is if they are first, ie $\frac1{m+1}$

\item If $n=2$, the the only way you fail to sell to them is if one comes first or they both appear before two people with pound coins, ie $2$ or $122$. These have probabilities $\frac{2}{m+2}$ and $\frac{m}{m+2} \cdot \frac{2}{m+1} \frac{1}{m} = \frac{2}{(m+1)(m+2)}$. Therefore the total probability you don't sell all the tickets is $\frac{2}{m+2}\left ( 1 + \frac{1}{m+1} \right) = \frac{2}{m+2} \frac{m+2}{m+1} = \frac{2}{m+1}$. Therefore the probability you do sell all the tickets is $1 - \frac{2}{m+1} = \frac{m-1}{m+1}$

\item The only ways to fail when $n=3$ are:

$2$, $122$, or if all three $2$s appear before three $1$s. this can happen in $11222$, $12122$

These happen with probability:

\begin{align*}
2: && \frac{3}{m+3} \\
122: && \frac{m}{m+3} \cdot \frac{3}{m+2} \cdot \frac{2}{m+1} \\
11222: && \frac{m(m-1) 6}{(m+3)(m+2)(m+1)m(m-1)} \\
12122: && \frac{m(m-1) 6}{(m+3)(m+2)(m+1)m(m-1)} \\
\end{align*}

Therefore the total probability is:

\begin{align*}
P &= \frac{1}{(m+3)(m+2)(m+1)} \left (3(m+2)(m+1)+6m + 12 \right) \\
&=  \frac{1}{(m+3)(m+2)(m+1)} \left (3(m+1)(m+2) \right) \\
&=  \frac{3}{m+1} 
\end{align*}
and the result follows
\end{questionparts}