Year: 2011
Paper: 3
Question Number: 13
Course: LFM Stats And Pure
Section: Negative Binomial Distribution
The percentages attempting larger numbers of questions were higher this year than formerly. More than 90% attempted at least five questions and there were 30% that didn't attempt at least six questions. About 25% made substantive attempts at more than six questions, of which a very small number indeed were high scoring candidates that had perhaps done extra questions (well) for fun, but mostly these were cases of candidates not being able to complete six good solutions.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
In this question, the notation $\lfloor x \rfloor$
denotes the greatest integer less than or equal to $x$,
so for example $\lfloor \pi\rfloor = 3$ and $\lfloor 3 \rfloor =3$.
\begin{questionparts}
\item
A bag contains $n$ balls, of which $b$ are black. A sample of $k$ balls is drawn, one after another, at random \textit{with} replacement. The random variable $X$ denotes the number of black balls in the sample.
By considering
\[
\frac{\P(X=r+1)}{\P(X=r)}\,,
\]
show that, in the case that it is unique,
the most probable number of black balls
in the sample is
\[
\left\lfloor \frac{(k+1)b}{n}\right\rfloor.
\]
Under what circumstances is the answer not unique?
\item A bag contains
$n$ balls, of which $b$ are black. A sample of $k$ balls
(where $k\le b$)
is drawn, one after another, at random \textit{without} replacement.
Find, in the case that it is unique, the most probable number of black
balls in the sample.
Under what circumstances is the answer not unique?
\end{questionparts}\begin{questionparts}
\item $\mathbb{P}(X = r) = \binom{k}{r}p^rq^{k-r}$ where $p = \frac{b}{n}, q = 1-p$.
Therefore
\begin{align*}
&& \frac{\mathbb{P}(X=r+1)}{\mathbb{P}(X=r)} &= \frac{\binom{k}{r+1}p^{r+1}q^{k-r-1}}{\binom{k}{r}p^rq^{k-r}} \\
&&&= \frac{(k-r)p}{(r+1)q} \\
&&&= \frac{(k-r)b}{(r+1)(n-b)}
\end{align*}
Comparing this to $1$ we find:
\begin{align*}
&& 1 & < \frac{\mathbb{P}(X=r+1)}{\mathbb{P}(X=r)} \\
\Leftrightarrow && 1 &< \frac{(k-r)b}{(r+1)(n-b)} \\
\Leftrightarrow && (r+1)(n-b) &<(k-r)b \\
\Leftrightarrow && rn& < (k+1)b - n \\
\Leftrightarrow && r &< \frac{(k+1)b}{n} - 1\\
\end{align*}
If this equation is true, then $\mathbb{P}(X=r+1)$ is larger, so $r_{max} = \left \lfloor \frac{(k+1)b}{n} \right \rfloor$
\item Let $Y$ be the number of black balls in our sample, ie $\mathbb{P}(Y = r) = \binom{b}{r}\binom{n-b}{k-r}/\binom{n}{k}$, so
\begin{align*}
&& \frac{\mathbb{P}(Y = r+1)}{\mathbb{P}(Y=r)} &= \frac{\binom{b}{r+1}\binom{n-b}{k-(r+1)}/\binom{n}{k}}{\binom{b}{r}\binom{n-b}{k-r}/\binom{n}{k}} \\
&&&= \frac{b-r}{r+1} \frac{k-r}{n-b-k+r+1} \\
&& 1 &< \frac{\mathbb{P}(Y = r+1)}{\mathbb{P}(Y=r)} \\
\Leftrightarrow && (r+1)(n-b-k+r+1) &< (b-r)(k-r) \\
\Leftrightarrow &&r(n-b-k+1)+(n-b-k+1) &< -r(b+k)+bk \\
\Leftrightarrow &&r(n+1) &< bk+b+k+1-n \\
\Leftrightarrow && r &< \frac{(b+1)(k+1)}{n+1} - \frac{n}{n+1}
\end{align*}
Therefore $r = \left \lfloor \frac{ (b+1)(k+1)}{n+1}\right \rfloor$, it is not unique if $n+1$ divides $(b+1)(k+1)$
\end{questionparts}
This too was fairly unpopular, being attempted by about 10% of the candidates. Of these no more than a dozen got it largely correct, but there was only one totally correct solution as the detail for the non unique case frequently tripped even the better candidates. The mean score was only about a third of the marks available as most candidates got part (i) largely correct, barring some simplifying errors and not obtaining the non unique solution. Fewer candidates had the correct probabilities for part (ii) and so were unable to proceed, though a few were wrong merely by a constant which cancelled to give the correct ratio.