| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Difficulty | Challenging +1.2 This is a standard Markov chain question requiring transition matrix construction, matrix powers for prediction, and equilibrium calculation via solving linear equations. While it involves multiple parts and some algebraic manipulation, these are routine techniques for Further Maths students with no novel conceptual challenges. |
| Spec | 4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = \begin{pmatrix}0.8 & 0.1 & 0\\0.2 & 0.6 & 0.15\\0 & 0.3 & 0.85\end{pmatrix}\) | B1 B1 B1 | Columns (or rows) correct; penalise if orientation wrong but consistent |
| Answer | Marks | Guidance |
|---|---|---|
| Initial vector (2007): \(\mathbf{p}_0 = \begin{pmatrix}0.6\\0.4\\0\end{pmatrix}\) | B1 | |
| Apply \(T\) repeatedly to find \(\mathbf{p}_7\) (for 2014, 7 years after 2007) | M1 | |
| \(\mathbf{p}_7 = T^7\mathbf{p}_0\); calculate step by step or use powers | M1 M1 | |
| Result: Division 2 most likely (showing relevant probabilities) | A1 A1 | Correct identification with supporting values |
| Answer | Marks |
|---|---|
| At equilibrium \(\mathbf{p} = T\mathbf{p}\): | M1 |
| \(0.8\pi_1+0.1\pi_2 = \pi_1 \Rightarrow -0.2\pi_1+0.1\pi_2=0 \Rightarrow \pi_2=2\pi_1\) | M1 |
| \(0.2\pi_1+0.6\pi_2+0.15\pi_3=\pi_2 \Rightarrow 0.2\pi_1-0.4\pi_2+0.15\pi_3=0\) | |
| Substituting \(\pi_2=2\pi_1\): \(0.2\pi_1-0.8\pi_1+0.15\pi_3=0 \Rightarrow \pi_3=4\pi_1\) | A1 |
| \(\pi_1+\pi_2+\pi_3=1 \Rightarrow \pi_1+2\pi_1+4\pi_1=1 \Rightarrow \pi_1=\frac{1}{7}\) | A1 |
| \(\pi_1=\frac{1}{7}, \pi_2=\frac{2}{7}, \pi_3=\frac{4}{7}\) | A1 |
| Answer | Marks |
|---|---|
| \(T_{new} = \begin{pmatrix}0.8&0.1&0&0\\0.2&0.6&0.15&0\\0&0.3&0.75&0\\0&0&0.1&1\end{pmatrix}\) | B1 B1 B1 |
| Answer | Marks |
|---|---|
| Start 2015 in Div 2: \(\mathbf{p}_0=(0,1,0,0)^T\) | B1 |
| Apply \(T_{new}^5\) to find probability still in league in 2020 | M1 M1 |
| Sum of first three components of \(\mathbf{p}_5\) | M1 |
| Answer (numerical value) | A1 |
| Answer | Marks |
|---|---|
| Find smallest \(n\) such that \((\mathbf{p}_n)_4 > 0.5\) where \((\mathbf{p}_n)_4\) is the "Out of league" probability | M1 M1 |
| Calculate successive years | M1 |
| State the first year | A1 |
# Question 5: (Option 5: Markov chains)
## Part (i)
States: Division 1, Division 2, Division 3
| $T = \begin{pmatrix}0.8 & 0.1 & 0\\0.2 & 0.6 & 0.15\\0 & 0.3 & 0.85\end{pmatrix}$ | B1 B1 B1 | Columns (or rows) correct; penalise if orientation wrong but consistent |
## Part (ii)
| Initial vector (2007): $\mathbf{p}_0 = \begin{pmatrix}0.6\\0.4\\0\end{pmatrix}$ | B1 | |
|---|---|---|
| Apply $T$ repeatedly to find $\mathbf{p}_7$ (for 2014, 7 years after 2007) | M1 | |
| $\mathbf{p}_7 = T^7\mathbf{p}_0$; calculate step by step or use powers | M1 M1 | |
| Result: Division 2 most likely (showing relevant probabilities) | A1 A1 | Correct identification with supporting values |
## Part (iii)
| At equilibrium $\mathbf{p} = T\mathbf{p}$: | M1 | |
|---|---|---|
| $0.8\pi_1+0.1\pi_2 = \pi_1 \Rightarrow -0.2\pi_1+0.1\pi_2=0 \Rightarrow \pi_2=2\pi_1$ | M1 | |
| $0.2\pi_1+0.6\pi_2+0.15\pi_3=\pi_2 \Rightarrow 0.2\pi_1-0.4\pi_2+0.15\pi_3=0$ | | |
| Substituting $\pi_2=2\pi_1$: $0.2\pi_1-0.8\pi_1+0.15\pi_3=0 \Rightarrow \pi_3=4\pi_1$ | A1 | |
| $\pi_1+\pi_2+\pi_3=1 \Rightarrow \pi_1+2\pi_1+4\pi_1=1 \Rightarrow \pi_1=\frac{1}{7}$ | A1 | |
| $\pi_1=\frac{1}{7}, \pi_2=\frac{2}{7}, \pi_3=\frac{4}{7}$ | A1 | |
## Part (iv)
States: Div 1, Div 2, Div 3, Out
| $T_{new} = \begin{pmatrix}0.8&0.1&0&0\\0.2&0.6&0.15&0\\0&0.3&0.75&0\\0&0&0.1&1\end{pmatrix}$ | B1 B1 B1 | |
## Part (v)
| Start 2015 in Div 2: $\mathbf{p}_0=(0,1,0,0)^T$ | B1 | |
|---|---|---|
| Apply $T_{new}^5$ to find probability still in league in 2020 | M1 M1 | |
| Sum of first three components of $\mathbf{p}_5$ | M1 | |
| Answer (numerical value) | A1 | |
## Part (vi)
| Find smallest $n$ such that $(\mathbf{p}_n)_4 > 0.5$ where $(\mathbf{p}_n)_4$ is the "Out of league" probability | M1 M1 | |
|---|---|---|
| Calculate successive years | M1 | |
| State the first year | A1 | |5 A local hockey league has three divisions. Each team in the league plays in a division for a year. In the following year a team might play in the same division again, or it might move up or down one division.
This question is about the progress of one particular team in the league. In 2007 this team will be playing in either Division 1 or Division 2. Because of its present position, the probability that it will be playing in Division 1 is 0.6 , and the probability that it will be playing in Division 2 is 0.4 .
The following transition probabilities apply to this team from 2007 onwards.
\begin{itemize}
\item When the team is playing in Division 1, the probability that it will play in Division 2 in the following year is 0.2 .
\item When the team is playing in Division 2, the probability that it will play in Division 1 in the following year is 0.1 , and the probability that it will play in Division 3 in the following year is 0.3 .
\item When the team is playing in Division 3, the probability that it will play in Division 2 in the following year is 0.15 .
\end{itemize}
This process is modelled as a Markov chain with three states corresponding to the three divisions.\\
(i) Write down the transition matrix.\\
(ii) Determine in which division the team is most likely to be playing in 2014.\\
(iii) Find the equilibrium probabilities for each division for this team.
In 2015 the rules of the league are changed. A team playing in Division 3 might now be dropped from the league in the following year. Once dropped, a team does not play in the league again.\\
-The transition probabilities from Divisions 1 and 2 remain the same as before.
\begin{itemize}
\item When the team is playing in Division 3, the probability that it will play in Division 2 in the following year is 0.15 , and the probability that it will be dropped from the league is 0.1 .
\end{itemize}
The team plays in Division 2 in 2015.\\
The new situation is modelled as a Markov chain with four states: 'Division1', 'Division 2', 'Division 3' and 'Out of league'.\\
(iv) Write down the transition matrix which applies from 2015.\\
(v) Find the probability that the team is still playing in the league in 2020.\\
(vi) Find the first year for which the probability that the team is out of the league is greater than 0.5 .
\hfill \mbox{\textit{OCR MEI FP3 2006 Q5 [24]}}