Year: 2010
Paper: 3
Question Number: 12
Course: LFM Stats And Pure
Section: Geometric Distribution
About 80% of candidates attempted at least five questions, and well less than 20% made genuine attempts at more than six. Those attempting more than six questions fell into three camps which were those weak candidates who made very little progress on any question, those with four or five fair solutions casting about for a sixth, and those strong candidates that either attempted 7th or even 8th questions as an "insurance policy" against a solution that seemed strong but wasn't, or else for entertainment!
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The infinite series $S$ is given by
\[
S = 1 + (1 + d)r + (1 + 2d)r^2 + \cdots + (1+nd)r^n +\cdots\;
,\]
for $\vert r \vert <1\,$.
By considering $S - rS$, or otherwise, prove that
\[
S = \frac 1{1-r} + \frac {rd}{(1-r)^2}
\,.\]
Arthur and Boadicea shoot arrows at a target. The probability that an
arrow shot by Arthur hits the target is $a$; the probability that an arrow
shot by Boadicea hits the target is $b$. Each shot is independent of all
others. Prove that the expected number of shots it takes Arthur to hit
the target is $1/a$.
Arthur and Boadicea now have a contest. They take alternate shots,
with Arthur going first. The winner is the one who hits the target first.
The probability
that Arthur wins the contest is $\alpha$ and
the probability that Boadicea wins is
$\beta$. Show that
\[
\alpha = \frac a {1-a'b'}\,,
\]
where $a' = 1-a$ and $b'=1-b$, and find $\beta$.
Show that the expected number of shots in the contest is
$\displaystyle \frac \alpha a + \frac \beta b\,.$Notice that
\begin{align*}
&& S - rS &= 1 + dr + dr^2 + \cdots \\
&&&= 1 + dr(1 + r+r^2+ \cdots) \\
&&&= 1 + \frac{rd}{1-r} \\
\Rightarrow && S &= \frac{1}{1-r} + \frac{rd}{(1-r)^2}
\end{align*}
The number of shots Arthur takes is $\textrm{Geo}(a)$, so it's expectation is $1/a$.
The probability Arthur wins is:
\begin{align*}
\alpha &= a + a'b'a + (a'b')^2a + \cdots \\
&= a(1+a'b' + \cdots) \\
&= \frac{a}{1-a'b'} \\
\\
\beta &= a'b + a'b'a'b + \cdots \\
&= a'b(1+b'a' + (b'a')^2 + \cdots ) \\
&= \frac{a'b}{1-a'b'}
\end{align*}
The expected number of shots in the contest is:
\begin{align*}
E &= a + 2a'b + 3a'b'a + 4a'b'a'b + \cdots \\
&= a(1 + 3a'b' + 5(a'b')^2 + \cdots) + 2a'b(1 + 2(a'b') + 3(a'b')^2 + \cdots) \\
&= a \left ( \frac{1}{1-a'b'} + \frac{2a'b'}{(1-a'b')^2} \right) + 2a'b \left ( \frac{1}{1-a'b'} + \frac{a'b'}{(1-a'b')^2}\right) \\
&= \frac{a}{1-a'b'} \left (1 + \frac{2a'b'}{(1-a'b')} \right) + 2\frac{a'b}{1-a'b'} \left ( 1 + \frac{a'b'}{(1-a'b')}\right) \\
&= \alpha \frac{1+a'b'}{1-a'b'} + \beta \frac{2}{1-a'b'} \\
&= \alpha \frac{1+1-a-b+ab}{1-a'b'} + \beta \frac{2}{1-a'b'} \\
\end{align*}
Although this was marginally less popular than question 11, the success achieved was similar to that on the first two questions. A small number of candidates didn't get started but most found the first parts straightforward and dealt with the manipulation and summation of the geometric series correctly. Many found an incorrect "shortcut" on the last part, despite having a good idea how to attempt it correctly having completed the earlier parts.