Year: 2025
Paper: 2
Question Number: 11
Course: LFM Stats And Pure
Section: Geometric Distribution
As is commonly the case, the vast majority of candidates focused on the Pure questions in Section A of the paper, with a good number of attempts made on all of those questions. Candidates that attempted the Mechanics questions in Section B generally answered both questions. More candidates attempted Question 11 in Section C than either Mechanics question, but very few attempted Question 12 in that section. There were a large number of good responses seen for all the questions, but a significant number of responses lacked sufficient detail in the presentation, particularly when asked to prove a given result or provide an explanation. Candidates who did well on this paper generally: gave careful explanations of each step within their solutions; indicated all points of interest on graphs and other diagrams clearly; made clear comments about the approach that needed to be taken, particularly when having to explore a number of cases as part of the solution to a question; used mathematical terminology accurately within their solutions. Candidates who did less well on this paper generally: made errors with basic algebraic manipulation, such as incorrect processing of indices; produced sketches of graphs in which significant points were difficult to see clearly because of the chosen scale; skipped important lines within lengthy sections of algebraic reasoning.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item By considering the sum of a geometric series, or otherwise, show that
\[\sum_{r=1}^{\infty} rx^{r-1} = \frac{1}{(1-x)^2} \quad \text{for } |x| < 1.\]
\item Ali plays a game with a fair $2k$-sided die. He rolls the die until the first $2k$ appears. Ali wins if all the numbers he rolls are even.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that Ali wins the game.
If Ali wins the game, he earns £1 for each roll, including the final one. If he loses, he earns nothing.
\item Find Ali's expected earnings from playing the game.
\end{enumerate}
\item Find a simplified expression for
\[1 + 2\binom{n}{1}x + 3\binom{n}{2}x^2 + \ldots + (n+1)x^n,\]
where $n$ is a positive integer.
\item Zen plays a different game with a fair $2k$-sided die. She rolls the die until the first $2k$ appears, and wins if the numbers rolled are strictly increasing in size. For example, if $k = 3$, she wins if she rolls 2, 6 or 1, 4, 5, 6, but not if she rolls 1, 4, 2, 6 or 1, 3, 3, 6.
If Zen wins the game, she earns £1 for each roll, including the final one. If she loses, she earns nothing.
Find Zen's expected earnings from playing the game.
\item Using the approximation
\[\left(1 + \frac{1}{n}\right)^n \approx e \quad \text{for large } n,\]
show that, when $k$ is large, Zen's expected earnings are a little over 35\% more than Ali's expected earnings.
\end{questionparts}
\begin{questionparts}
\item Note that,
\begin{align*}
&& \sum_{r = 0}^\infty x^r &= \frac{1}{1-x} && |x| < 1\\
\underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \sum_{r = 0}^\infty rx^{r-1} &= \frac{1}{(1-x)^2} && |x| < 1\\
&& \sum_{r = 1}^\infty rx^{r-1} &= \frac{1}{(1-x)^2} && |x| < 1\\
\end{align*}
\item \begin{enumerate}
\item \begin{align*}
&& \mathbb{P}(\text{Ali wins in }s\text{ rounds}) &= \left ( \frac{k-1}{2k} \right)^{s-1} \frac{1}{2k} \\
\Rightarrow && \mathbb{P}(\text{Ali wins}) &= \sum_{s=1}^\infty \mathbb{P}(\text{Ali wins in }s\text{ rounds}) \\
&&&=\sum_{s=1}^\infty \left ( \frac{k-1}{2k} \right)^{s-1} \frac{1}{2k} \\
&&&= \frac{1}{2k} \sum_{s=0}^\infty \left ( \frac{k-1}{2k} \right)^{s} \\
&&&= \frac{1}{2k} \frac{1}{1 - \frac{k-1}{2k}} \\
&&&= \frac{1}{2k - (k-1)} \\
&&&= \frac{1}{k+1}
\end{align*}
\item \begin{align*}
\mathbb{E}(\text{Ali score}) &= \sum_{s=1}^{\infty} s \mathbb{P}(\text{Ali wins in }s\text{ rounds}) \\
&= \sum_{s=1}^{\infty} s \left ( \frac{k-1}{2k} \right)^{s-1} \frac{1}{2k} \\
&= \frac{1}{2k} \frac{1}{\left (1 - \frac{k-1}{2k} \right)^2} \\
&= \frac{2k}{(k+1)^2}
\end{align*}
\end{enumerate}
\item \begin{align*}
&& (1+x)^{n} &= \sum_{k=0}^n \binom{n}{k} x^k \\
\Rightarrow && x(1+x)^n &= \sum_{k=0}^n \binom{n}{k} x^{k+1} \\
\Rightarrow && (1+x)^n + nx(1+x)^{n-1} &= \sum_{k=0}^n (k+1)\binom{n}{k} x^k \\
\Rightarrow && (1+x)^{n-1}(1+(n+1)x) &= 1 + 2\binom{n}{1}x + 3\binom{n}{2}x^2 + \ldots + (n+1)x^n
\end{align*}
\item \begin{align*}
\mathbb{E}(\text{Zen score}) &= \sum_{s=1}^{2k} s \mathbb{P} \left ( \text{Zen gets }s\text{ numbers in increasing order ending with }2k \right) \\
&= \sum_{s=1}^{2k} s \binom{2k-1}{s-1} \frac{1}{(2k)^s} \\
&= \frac{1}{2k}\sum_{s=0}^{2k-1} (s+1) \binom{2k-1}{s} \frac{1}{(2k)^s} \\
&= \frac{1}{2k} \left ( 1 + \frac{1}{2k} \right)^{2k-2} \left ( 1 + (2k-1+1) \frac{1}{2k} \right) \\
&= \frac{1}{k}\left ( 1 + \frac{1}{2k} \right)^{2k-2}
\end{align*}
\item Therefore as $k \to \infty$
\begin{align*}
\frac{\mathbb{E}(\text{Zen score})}{\mathbb{E}(\text{Ali score}) } &= \frac{1}{k}\left ( 1 + \frac{1}{2k} \right)^{2k-2} \big / \frac{2k}{(k+1)^2} \\
&= \frac{(k+1)^2}{2k^2} \cdot \left ( 1 + \frac{1}{2k} \right)^{2k} \cdot \left ( 1 + \frac{1}{2k} \right)^{-2} \\
&\to \frac12 e \approx 2.7/2 = 1.35
\end{align*} ie Zen's expected earnings are $\approx 35\%$ more.
\end{questionparts}
A large number of very good solutions to this question were seen. Part (i) was completed well by the majority of candidates. However, many candidates did not identify the correct probabilities to use in the calculations for part (ii). Attempts at part (ii) (b) often applied a correct method for calculating the expected value, but used the incorrect value for the probability that had been used in part (ii) (a) and so gained the method mark for part (ii) (b). Most candidates who attempted part (iii) were able to complete it successfully, usually by applying a similar approach to the one used in part (i). Combinatorial errors were common in part (iv), with candidates often confusing 2k and 2k − 1 in their calculations or incorrectly accounting for the requirements of the order. Part (v) was usually completed well by those candidates that had previously obtained the correct expressions for the expected values.