Year: 2022
Paper: 2
Question Number: 11
Course: LFM Stats And Pure
Section: Discrete Probability Distributions
No solution available for this problem.
Candidates appeared to be generally well prepared for most topics within the examination, but there were a few situations in questions where some did not appear to be as proficient in standard techniques as needed. In particular, the method for finding invariant lines required in question 8 and the manipulation of trigonometric functions that were needed in question 10 caused considerable difficulties for some candidates. An additional issue that occurred at numerous points in the paper relates to the direction in which a deduction is required. It is important that candidates make sure that they know which statement is the one that they should start from as they deduce the other and that it is clear in their solution that the logic has gone in the correct direction. Clarity of solution is also an issue that candidates should be aware of, especially in the situations where the result to be reached has been given. It is important to check that there are no special cases that need to be considered separately, and when dividing by a function it is necessary to confirm that the function cannot be equal to 0 (and in the case of inequalities that the function always has the same sign). When drawing diagrams and sketching graphs it is useful if significant points that need to be clear are not drawn over the lines on the page as these can be difficult to interpret during the marking process.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A batch of $N$ USB sticks is to be used on a network. Each stick has the same unknown probability $p$ of being infected with a virus. Each stick is infected, or not, independently of the others.
The network manager decides on an integer value of $T$ with $0 \leqslant T < N$. If $T = 0$ no testing takes place and the $N$ sticks are used on the network, but if $T > 0$, the batch is subject to the following procedure.
\begin{itemize}
\item Each of $T$ sticks, chosen at random from the batch, undergoes a test during which it is destroyed.
\item If any of these $T$ sticks is infected, all the remaining $N - T$ sticks are destroyed.
\item If none of the $T$ sticks is infected, the remaining $N - T$ sticks are used on the network.
\end{itemize}
If any stick used on the network is infected, the network has to be disinfected at a cost of $\pounds D$, where $D > 0$. If no stick used on the network is infected, there is a gain of $\pounds 1$ for each of the $N - T$ sticks. There is no cost to testing or destroying a stick.
\begin{questionparts}
\item Find an expression in terms of $N$, $T$, $D$ and $q$, where $q = 1 - p$, for the expected net loss.
\item Let $\alpha = \dfrac{DT}{N(N - T + D)}$. Show that $0 \leqslant \alpha < 1$.
Show that, for fixed values of $N$, $D$ and $T$, the greatest value of the expected net loss occurs when $q$ satisfies the equation $q^{N-T} = \alpha$.
Show further that this greatest value is $\pounds\dfrac{D(N-T)\,\alpha^k}{N}$, where $k = \dfrac{T}{N-T}$.
\item For fixed values of $N$ and $D$, show that there is some $\beta > 0$ so that for all $p < \beta$, the expression for the expected loss found in part (i) is an increasing function of $T$. Deduce that, for small enough values of $p$, testing no sticks minimises the expected net loss.
\end{questionparts}
Some candidates made good progress with this question. However, a large proportion of attempts did not make much progress and many were unable to score more than one or two marks. Many candidates were able to correctly identify the way in which the required expression in part (i) should be constructed and they were often successful in putting these together correctly. In part (ii) most candidates were able to consider the definition of α and demonstrate the inequality required, although some candidates only showed one side of the inequality. Candidates who recognised that differentiation of the expected net loss function was a useful approach were often able to make good progress towards finding the required results for the remainder of this question. A small number of candidates attempted part (iii) of the question and many of these were able to show the first result. The explanations of the final deduction were often not convincing however.