Year: 2020
Paper: 3
Question Number: 12
Course: LFM Stats And Pure
Section: Geometric Distribution
No solution available for this problem.
In spite of the change to criteria for entering the paper, there was still a very healthy number of candidates, and the vast majority handled the protocols for the online testing very well. Just over half the candidates attempted exactly six questions, and whilst about 10% attempted a seventh, hardly any did more than seven. With 20% attempting five questions, and 10% attempting only four, overall, there were very few candidates not attempting the target number. There was a spread of popularity across the questions, with no question attracting more than 90% of candidates and only one less than 10%, but every question received a good number of attempts. Likewise, there was a spread of success on the questions, though every question attracted at least one perfect solution.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
$A$ and $B$ both toss the same biased coin. The probability that the coin shows heads is $p$, where $0 < p < 1$, and the probability that it shows tails is $q = 1 - p$.
Let $X$ be the number of times $A$ tosses the coin until it shows heads. Let $Y$ be the number of times $B$ tosses the coin until it shows heads.
\begin{questionparts}
\item The random variable $S$ is defined by $S = X + Y$ and the random variable $T$ is the maximum of $X$ and $Y$. Find an expression for $\mathrm{P}(S = s)$ and show that
\[ \mathrm{P}(T = t) = pq^{t-1}(2 - q^{t-1} - q^t). \]
\item The random variable $U$ is defined by $U = |X - Y|$, and the random variable $W$ is the minimum of $X$ and $Y$. Find expressions for $\mathrm{P}(U = u)$ and $\mathrm{P}(W = w)$.
\item Show that $\mathrm{P}(S = 2 \text{ and } T = 3) \neq \mathrm{P}(S = 2) \times \mathrm{P}(T = 3)$.
\item Show that $U$ and $W$ are independent, and show that no other pair of the four variables $S$, $T$, $U$ and $W$ are independent.
\end{questionparts}
As well as the popularity of this question being similar to that of question 11, the success was very similar too. It was just below question 11 with its mean score. Very few scored full marks, partly because very few recognised the need to consider the case U=0 separately in parts (ii) and (iv), and of those who did, many made mistakes in other places or forgot to also consider it in (iv) after correctly considering it in (ii). However, the question was also rather forgiving, in the sense that it was possible to make substantial progress on the question even with errors in the earlier parts. A common error in parts (i) and (ii) was to "double count" the case X=Y, when finding the distribution of T and U. It was also rather common for candidates to think that Y was the number of tosses until B got a tail (rather than a head). Many candidates identified correct counter-examples for the last part of (iv), but a significant proportion failed to justify that their joint probabilities were equal to zero. There were also a number of candidates who made their lives significantly harder by injudicious choice of counterexamples; e.g. candidates who chose S=2, and then U=0 who then had to do much more work to prove the probabilities were not equal, than if they had made any other choice of U would give a contradiction simply and immediately.