Year: 2018
Paper: 2
Question Number: 13
Course: LFM Stats And Pure
Section: Tree Diagrams
The pure questions were again the most popular of the paper, with only two of those questions being attempted by fewer than half of the candidates (none of the other questions was attempted by more than half of the candidates). Good responses were seen to all of the questions, but in many cases, explanations lacked sufficient detail to be awarded full marks. Candidates should ensure that they are demonstrating that the results that they are attempting to apply are valid in the cases being considered. In several of the questions, later parts involve finding solutions to situations that are similar to earlier parts of the question. In general candidates struggled to recognise these similarities and therefore spent a lot of time repeating work that had already been done, rather than simply observing what the result must be.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1502.8
Banger Comparisons: 2
Four children, $A$, $B$, $C$ and $D$, are playing a version of the game `pass the parcel'. They stand in a circle, so that $ABCDA$ is the clockwise order. Each time a whistle is blown, the child holding the parcel is supposed to pass the parcel immediately exactly one place clockwise. In fact each child, independently of any other past event, passes the parcel clockwise with probability $\frac{1}{4}$, passes it anticlockwise with probability $\frac{1}{4}$ and fails to pass it at all with probability $\frac{1}{2}$. At the start of the game, child $A$ is holding the parcel.
The probability that child $A$ is holding the parcel just after the whistle has been blown for the $n$th time is $A_n$, and $B_n$, $C_n$ and $D_n$ are defined similarly.
\begin{questionparts}
\item Find $A_1$, $B_1$, $C_1$ and $D_1$. Find also $A_2$, $B_2$, $C_2$ and $D_2$.
\item By first considering $B_{n+1}+D_{n+1}$, or otherwise,
find $B_n$ and $D_n$.
Find also expressions for $A_n$ and $C_n$ in terms of $n$.
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& A_1 &= \frac12 \\
&& B_1 &= \frac14 \\
&& C_1 &= 0 \\
&& D_1 &= \frac14
\end{align*}
\begin{align*}
&& A_2 &= \frac12 \cdot \frac12 + 2 \cdot \frac14 \cdot \frac14 = \frac38 \\
&& B_2 &= \frac14 \cdot \frac12 + \frac12 \cdot \frac14 = \frac14 \\
&& C_2 &=2 \cdot \frac14 \cdot \frac14 =\frac18 \\
&& D_2 &= B_2 = \frac14
\end{align*}
\item \begin{align*}
&& A_{n+1} &= \frac12 A_n+ \frac14(B_n + D_n) \\
&& B_{n+1} &= \frac12 B_n+ \frac14(A_n + C_n) \\
&& C_{n+1} &= \frac12 C_n+ \frac14(D_n +B_n) \\
&& D_{n+1} &= \frac12 D_n+ \frac14(C_n +A_n) \\
\\
\Rightarrow && B_{n+1}+D_{n+1} &= \frac12 (B_n+D_n) + \frac12(A_n+C_n) \\
&&&= \frac12 \\
\Rightarrow && B_{n+1}&=D_{n+1} = \frac14 \\
\\
&& C_{n+1} &= \frac12C_n + \frac14 \cdot \frac12 \\
&&&= \frac12 C_n + \frac18\\
&&&= \frac12 C_{n-1} + \frac1{8} + \frac1{16} \\
&&&= \frac1{8} + \frac{1}{16} + \cdots + \frac{1}{8 \cdot 2^{n-1}} \\
&&&= \frac18 \left (1 + \frac12 + \cdots + \frac1{2^{n-1}} \right) \\
&&&= \frac18\left ( \frac{1-\frac1{2^n}}{1-\frac12} \right) \\
&&&= \frac18 \left (2 - \frac{1}{2^{n-1}} \right) \\
&&&= \frac14 - \frac{1}{2^{n-1}} \\
\Rightarrow && A_n &= \frac14 + \frac1{2^{n-1}}
\end{align*}
\end{questionparts}
This was the more popular of the two probability and statistics questions on the paper and many good responses were seen. Candidates were generally able to work out the probabilities required in the first part of the question accurately. A considerable number of candidates were not able to make significant progress beyond that point, but those who did were often able to identify the relationships clearly and make use of the symmetry of the problem. Some attempts to tackle the second part of the question through counting arguments were seen, but these were not successful. The proportion of candidates achieving full marks for this question was higher than any other.