Year: 2014
Paper: 1
Question Number: 12
Course: LFM Stats And Pure
Section: Discrete Probability Distributions
More than 1800 candidates sat this paper, which represents another increase in uptake for this STEP paper. The impression given, however, is that many of these extra candidates are just not sufficiently well prepared for questions which are not structured in the same way as are the A-level questions that they are, perhaps, more accustomed to seeing. Although STEP questions try to give all able candidates "a bit of an intro." into each question, they are not intended to be easy, and (at some point) imagination and real flair (as well as determination) are required if one is to score well on them. In general, it is simply not possible to get very far into a question without making some attempt to think about what is actually going on in the situation presented therein; and those students who expect to be told exactly what to do at each stage of a process are in for a shock. Too many candidates only attempt the first parts of many questions, restricting themselves to 3-6 marks on each, rather than trying to get to grips with substantial portions of work – the readiness to give up and try to find something else that is "easy pickings" seldom allows such candidates to acquire more than 40 marks (as was the case with almost half of this year's candidature, in fact). Poor preparation was strongly in evidence – curve-sketching skills were weak, inequalities very poorly handled, algebraic capabilities (especially in non-standard settings) were often pretty poor, and the ability to get to grips with extended bits of working lacking in the extreme; also, an unwillingness to be imaginative and creative, allied with a lack of thoroughness and attention to detail, made this a disappointing (and, possibly, very uncomfortable) experience for many of those students who took the paper. On the other side of the coin, there was a very pleasing number of candidates who produced exceptional pieces of work on 5 or 6 questions, and thus scored very highly indeed on the paper overall. Around 100 of them scored 90+ marks of the 120 available, and they should be very proud of their performance – it is a significant and noteworthy achievement.
Difficulty Rating: 1484.0
Difficulty Comparisons: 1
Banger Rating: 1441.7
Banger Comparisons: 4
A game in a casino is played with a fair coin and an unbiased cubical die whose faces are labelled $1, 1, 1, 2, 2$ and $3.$ In each round of the game, the die is rolled once and the coin is tossed once. The outcome of the round is a random variable $X$. The value, $x$, of $X$ is determined as follows. If the result of the toss is heads then $x= \vert ks -1\vert$, and if the result of the toss is tails then $x=\vert k-s\vert$, where $s$ is the number on the die and $k$ is a given number. Show that $\mathbb{E}(X^2) = k +13(k-1)^2 /6$. Given that both $\mathbb{E}(X^2)$ and $\mathbb{E}(X)$ are positive integers, and
that $k$ is a single-digit positive integer, determine the value of $k$, and write down the probability distribution of $X$.
A gambler pays $\pounds 1$ to play the game, which consists of two rounds. The gambler is paid:
\begin{itemize}
\item $\pounds w$, where $w$ is an integer, if the sum of the outcomes of the two rounds exceeds $25$;
\item
$\pounds 1$ if the sum of the outcomes equals $25$;
\item
nothing if the sum of the outcomes is less that $25$.
\end{itemize}
Find, in terms of $w$, an expression for the amount the gambler expects to be paid in a game, and deduce the maximum possible value of $w$, given that the casino's owners choose $w$ so that the game is in their favour.\begin{align*}
&& \mathbb{E}(X^2) &= \frac12 \left (\frac16 \left ( 3(k -1)^2+2(2k-1)^2+(3k-1)^2 \right) +\frac16 \left ( 3(k -1)^2+2(k-2)^2+(k-3)^2 \right) \right) \\
&&&= \frac12 \left (\frac16 \left (20k^2-20k+6 \right) + \frac16 \left ( 6k^2-20k+20\right) \right) \\
&&&= \frac1{12} \left (26k^2-40k+ 26\right) \\
&&&= \frac{13}{6} (k^2+1) - \frac{10}{3}k \\
&&&= \frac{13}{6}(k-1)^2+k
\end{align*}
Since $k$ a single digit positive number and $\mathbb{E}(X^2)$ is an integer, $6 \mid k-1 \Rightarrow k = 1, 7$.
\begin{align*}
\mathbb{E}(X | k=1) &= \frac12 \left (\frac16 \left ( 2+2 \right) +\frac16 \left ( 2+2 \right) \right) = \frac23 \not \in \mathbb{Z}\\
\mathbb{E}(X | k=7) &= \frac12 \left (\frac16 \left ( 3\cdot6+2\cdot13+20 \right) +\frac16 \left ( 3\cdot6+2\cdot5+4 \right) \right) = 8
\end{align*}
Therefore $k = 7$
The probability distribution is
\begin{align*}
&& \mathbb{P}(X=4) = \frac1{12} \\
&& \mathbb{P}(X=5) = \frac1{6} \\
&& \mathbb{P}(X=6) = \frac12 \\
&& \mathbb{P}(X=13) = \frac1{6} \\
&& \mathbb{P}(X=20)= \frac1{12} \\
\end{align*}
The only ways to score more than $25$ are:
$20+6, 20+13, 20+20, 13+13$
The only ways to score exactly $25$ are $20+5$
\begin{align*}
\mathbb{P}(>25) &= \frac1{12} \cdot\left(2\cdot \frac12+2\cdot\frac16+\frac1{12}\right) + \frac{1}{6^2} \\
&= \frac{7}{48} \\
\mathbb{P}(=25) &= \frac{2}{12 \cdot 6} = \frac{1}{36} \\
\\
\mathbb{E}(\text{payout}) &= \frac{7}{48}w + \frac{1}{36} = \frac{21w+4}{144}
\end{align*}
The casino needs $\frac{21w+4}{144} < 1 \Rightarrow 21w< 140 \Rightarrow w < \frac{20}{3}$
Q12 elicited 250 attempts, and marks were generally in line with those for Q9 at around 6½/20. The big hurdle in Q12 was the modulus function, which many candidates simply ignored. Those who were happy to use it properly generally gained the required result and used it to find that k = 7 (though some failed to discount k = 1 along the way). A further obstacle arose as most candidates who continued into the second part of the question failed to account for all six outcomes when calculating P(X > 25).