Year: 2019
Paper: 2
Question Number: 11
Course: LFM Stats And Pure
Section: Uniform Distribution
The Pure questions were again the most popular of the paper, with only one of the questions attempted by fewer than half of the candidates (of the remaining four questions, only question 9 was attempted by more than a quarter of the candidates). In many of the questions candidates were often unable to make good use of the results shown in the earlier parts of the question in order to solve the more complex later parts. Nevertheless, some good solutions were seen to all of the questions. For many of the questions, solutions were seen in which the results were reached, but without sufficient justification of some of the steps.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item The three integers $n_1$, $n_2$ and $n_3$ satisfy $0 < n_1 < n_2 < n_3$ and $n_1 + n_2 > n_3$. Find the number of ways of choosing the pair of numbers $n_1$ and $n_2$ in the cases $n_3 = 9$ and $n_3 = 10$.
Given that $n_3 = 2n + 1$, where $n$ is a positive integer, write down an expression (which you need not prove is correct) for the number of ways of choosing the pair of numbers $n_1$ and $n_2$. Simplify your expression.
Write down and simplify the corresponding expression when $n_3 = 2n$, where $n$ is a positive integer.
\item You have $N$ rods, of lengths $1, 2, 3, \ldots, N$ (one rod of each length). You take the rod of length $N$, and choose two more rods at random from the remainder, each choice of two being equally likely. Show that, in the case $N = 2n + 1$ where $n$ is a positive integer, the probability that these three rods can form a triangle (of non-zero area) is
$$\frac{n - 1}{2n - 1}.$$
Find the corresponding probability in the case $N = 2n$, where $n$ is a positive integer.
\item You have $2M + 1$ rods, of lengths $1, 2, 3, \ldots, 2M + 1$ (one rod of each length), where $M$ is a positive integer. You choose three at random, each choice of three being equally likely. Show that the probability that the rods can form a triangle (of non-zero area) is
$$\frac{(4M + 1)(M - 1)}{2(2M + 1)(2M - 1)}.$$
Note: $\sum_{k=1}^{K} k^2 = \frac{1}{6}K(K + 1)(2K + 1)$.
\end{questionparts}\begin{questionparts}
\item If $n_3 = 9$ and we are looking for $0 < n_1 < n_2 < n_3$ we can consider values for each $n_2$.
\begin{array}{clc|c}
n_2 & \text{range} & \text{count} \\ \hline
6 & 4-5 & 2 \\
7 & 3-6 & 4 \\
8 & 2-7 & 6 \\ \hline
& & 12
\end{array}
When $n_3 = 10$
\begin{array}{clc|c}
n_2 & \text{range} & \text{count} \\ \hline
6 & 5 & 1 \\
7 & 4-6 & 3 \\
8 & 3-7 & 5 \\
9 & 2-8 & 7 \\ \hline
& & 16
\end{array}
When $n_3 = 2n+1$ we can have $2 + 4 + \cdots + 2n-2 = n(n-1)$
When $n_3 = 2n$ we can have $1 + 3 + \cdots + 2n-3 = (n-1)^2$
\item For the 3 rods to form a triangle, it suffices for the sum of the lengths of the shorter rods to be larger than $N$. When $N = 2n+1$ there are $n(n-1)$ ways this can happen, out of $\binom{2n}{2}$ ways to choos the numbers, ie
\begin{align*}
&& P &= \frac{n(n-1)}{\frac{2n(2n-1)}{2}} \\
&&&= \frac{n-1}{2n-1}
\end{align*}
When $N = 2n$ there are $(n-1)^2$ ways this can happen, out of $\binom{2n-1}{2}$ ways, ie
\begin{align*}
&& P &= \frac{(n-1)^2}{\frac{(2n-1)(2n-2)}{2}} \\
&&&= \frac{n-1}{2n-1}
\end{align*}
\item The number of ways this can happen is:
\begin{align*}
C &= \sum_{k=3}^{2M+1} \# \{ \text{triangles where }k\text{ is largest} \} \\
&= \sum_{k=1}^{M} \# \{ \text{triangles where }2k+1\text{ is largest} \} +\sum_{k=1}^{M} \# \{ \text{triangles where }2k\text{ is largest} \}\\
&= \sum_{k=1}^{M} n(n-1)+\sum_{k=1}^{M} (n-1)^2\\
&= \sum_{k=1}^{M} (2n^2-3n+1)\\
&= \frac26M(M+1)(2M+1) - \frac32M(M+1) + M \\
&= \frac16 M(4M+1)(M-1)
\end{align*}
Therefore the probability is
\begin{align*}
&& P &= \frac{M(4M+1)(M-1)}{6 \binom{2M+1}{3}} \\
&&&= \frac{M(4M+1)(M-1)}{(2M+1)2M(2M-1)} \\
&&&= \frac{(4M+1)(M-1)}{2(2M+1)(2M-1)}
\end{align*}
\end{questionparts}
Candidates got the correct number of pairs in the special cases n₃ = 9 and n₃ = 10 but sometimes the working was very unclear. A large majority found the expressions for general n, the most common error being a shift n → n+1 in the answer. Those who could obtain the result given for odd n in part (ii) were generally able to find the corresponding result for even n too. A common error was to double count the number of pairs of rods and not to double the number of pairs which made a triangle. Many candidates failed to explain why the conditions of part (i) were relevant for forming triangles. The most successful candidates in part (iii) counted the number of triples which make a triangle using a sum, and divided by the appropriate combinatorial factor, while those who conditioned on the largest rod and used conditional probability did less well. A common conceptual error was to assume that each integer was equally likely to appear as the largest rod, and candidates making this assumption lost many marks. Otherwise, algebraic errors were the most common. Candidates should remember that when an answer is given in the question, they need to take care to fully justify their answers.