| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Applied recurrence modeling |
| Difficulty | Challenging +1.2 This is a standard Further Maths Markov chain question with absorbing states. While it requires multiple calculations (transition matrix, powers of matrices, absorbing state probabilities), all techniques are routine for FP3 students. The multi-part structure and computational length add some difficulty, but no novel insight is required—just systematic application of learned methods. |
| Spec | 4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{P} = \begin{pmatrix} 1 & 0.07 & 0 & 0 \\ 0 & 0.8 & 0.15 & 0 \\ 0 & 0.13 & 0.75 & 0 \\ 0 & 0 & 0.1 & 1 \end{pmatrix}\) | B2 | Allow tolerance of \(\pm 0.0001\) in probabilities throughout this question; Give B1 for two columns correct (2 marks) |
| Answer | Marks |
|---|---|
| If system enters an absorbing state, it remains in that state | B1 |
| A and D are absorbing states | B1 (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{P}^9 \begin{pmatrix} 0 \\ 0.4 \\ 0.6 \\ 0 \end{pmatrix} = \begin{pmatrix} 0.2236 \\ 0.2505 \\ 0.1998 \\ 0.3261 \end{pmatrix}\) | M1M1 | For \(\mathbf{P}^9\) (or \(\mathbf{P}^{10}\)) and |
| Prob(owned by B) = 0.2505 | A1 (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{P}^4 = \begin{pmatrix} 1 & \ldots & \ldots & \ldots \\ \ldots & 0.4818 & \ldots & \ldots \\ \ldots & \ldots & 0.3856 & \ldots \\ \ldots & \ldots & \ldots & 1 \end{pmatrix}\) | M1 | Using diagonal elements from \(\mathbf{P}^4\) |
| \(0.2236 + 0.2505 \times 0.4818 + 0.1998 \times 0.3856 + 0.3261 = 0.7474\) | M1 A1 | Using probabilities for 10th day (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \((1 \quad 0 \quad 0 \quad 0) \mathbf{P}^n \begin{pmatrix} 0 \\ 0.4 \\ 0.6 \\ 0 \end{pmatrix}\) | M1 | Considering \(\mathbf{P}^n\) for some \(n > 20\) |
| Evaluating Prob(A or D) for some values of \(n\) | M1 | |
| \(= (0.8971)\) when \(n = 26\) or \(= (0.9057)\) when \(n = 27\) i.e. on the 28th day | A1 ft | Identifying \(n = 26\) or \(n = 27\) (Implies M1M1) |
| \(= (0.9057)\) when \(n = 27\) i.e. on the 28th day | A1 (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{P}^n \rightarrow \begin{pmatrix} 1 & 0.5738 & 0.3443 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0.4262 & 0.6557 & 1 \end{pmatrix} = \mathbf{Q}\) | B2 | Give B1 for two bold elements correct (to 3 dp) (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{Q} \begin{pmatrix} 0 \\ 0.4 \\ 0.6 \\ 0 \end{pmatrix} = \begin{pmatrix} 0.4361 \\ 0 \\ 0 \\ 0.5639 \end{pmatrix}\) | M1M1 | Using Q and \(\begin{pmatrix} 0 \\ 0.4 \\ 0.6 \\ 0 \end{pmatrix}\) |
| Prob(eventually owned by A) = 0.4361 | A1 (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{Q} \begin{pmatrix} 0 \\ p \\ q \\ 0 \end{pmatrix} = \begin{pmatrix} 0.5 \\ 0 \\ 0 \\ 0.5 \end{pmatrix}\) (where \(q = 1 - p\)) | M1M1 | For LHS and RHS |
| \(0.5738p + 0.3443q = 0.5\) or \(0.4262p + 0.6557q = 0.5\) | A1 ft | Or \(0.4262p + 0.6557q = 0.5\) |
| \(p = 0.6786\), \(q = 0.3214\) | M1 A1 | Solving to obtain a value of \(p\); Allow 0.678 – 0.679, 0.321 – 0.322 (5 marks) |
### 5(i)
**Matrix P:**
$\mathbf{P} = \begin{pmatrix} 1 & 0.07 & 0 & 0 \\ 0 & 0.8 & 0.15 & 0 \\ 0 & 0.13 & 0.75 & 0 \\ 0 & 0 & 0.1 & 1 \end{pmatrix}$ | B2 | Allow tolerance of $\pm 0.0001$ in probabilities throughout this question; Give B1 for two columns correct (2 marks)
### 5(ii)
If system enters an absorbing state, it remains in that state | B1 |
A and D are absorbing states | B1 (2 marks)
### 5(iii)
$\mathbf{P}^9 \begin{pmatrix} 0 \\ 0.4 \\ 0.6 \\ 0 \end{pmatrix} = \begin{pmatrix} 0.2236 \\ 0.2505 \\ 0.1998 \\ 0.3261 \end{pmatrix}$ | M1M1 | For $\mathbf{P}^9$ (or $\mathbf{P}^{10}$) and
Prob(owned by B) = 0.2505 | A1 (3 marks)
### 5(iv)
$\mathbf{P}^4 = \begin{pmatrix} 1 & \ldots & \ldots & \ldots \\ \ldots & 0.4818 & \ldots & \ldots \\ \ldots & \ldots & 0.3856 & \ldots \\ \ldots & \ldots & \ldots & 1 \end{pmatrix}$ | M1 | Using diagonal elements from $\mathbf{P}^4$
$0.2236 + 0.2505 \times 0.4818 + 0.1998 \times 0.3856 + 0.3261 = 0.7474$ | M1 A1 | Using probabilities for 10th day (3 marks)
### 5(v)
$(1 \quad 0 \quad 0 \quad 0) \mathbf{P}^n \begin{pmatrix} 0 \\ 0.4 \\ 0.6 \\ 0 \end{pmatrix}$ | M1 | Considering $\mathbf{P}^n$ for some $n > 20$
Evaluating Prob(A or D) for some values of $n$ | M1 |
$= (0.8971)$ when $n = 26$ or $= (0.9057)$ when $n = 27$ i.e. on the 28th day | A1 ft | Identifying $n = 26$ or $n = 27$ (Implies M1M1)
$= (0.9057)$ when $n = 27$ i.e. on the 28th day | A1 (4 marks)
### 5(vi)
$\mathbf{P}^n \rightarrow \begin{pmatrix} 1 & 0.5738 & 0.3443 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0.4262 & 0.6557 & 1 \end{pmatrix} = \mathbf{Q}$ | B2 | Give B1 for two bold elements correct (to 3 dp) (2 marks)
### 5(vii)
$\mathbf{Q} \begin{pmatrix} 0 \\ 0.4 \\ 0.6 \\ 0 \end{pmatrix} = \begin{pmatrix} 0.4361 \\ 0 \\ 0 \\ 0.5639 \end{pmatrix}$ | M1M1 | Using Q and $\begin{pmatrix} 0 \\ 0.4 \\ 0.6 \\ 0 \end{pmatrix}$
Prob(eventually owned by A) = 0.4361 | A1 (3 marks)
### 5(viii)
$\mathbf{Q} \begin{pmatrix} 0 \\ p \\ q \\ 0 \end{pmatrix} = \begin{pmatrix} 0.5 \\ 0 \\ 0 \\ 0.5 \end{pmatrix}$ (where $q = 1 - p$) | M1M1 | For LHS and RHS
$0.5738p + 0.3443q = 0.5$ or $0.4262p + 0.6557q = 0.5$ | A1 ft | Or $0.4262p + 0.6557q = 0.5$
$p = 0.6786$, $q = 0.3214$ | M1 A1 | Solving to obtain a value of $p$; Allow 0.678 – 0.679, 0.321 – 0.322 (5 marks)5 In this question, give probabilities correct to 4 decimal places.\\
Alpha and Delta are two companies which compete for the ownership of insurance bonds. Boyles and Cayleys are companies which trade in these bonds. When a new bond becomes available, it is first acquired by either Boyles or Cayleys. After a certain amount of trading it is eventually owned by either Alpha or Delta. Change of ownership always takes place overnight, so that on any particular day the bond is owned by one of the four companies. The trading process is modelled as a Markov chain with four states, as follows.
On the first day, the bond is owned by Boyles or Cayleys, with probabilities $0.4,0.6$ respectively.\\
If the bond is owned by Boyles, then on the next day it could be owned by Alpha, Boyles or Cayleys, with probabilities $0.07,0.8,0.13$ respectively.
If the bond is owned by Cayleys, then on the next day it could be owned by Boyles, Cayleys or Delta, with probabilities $0.15,0.75,0.1$ respectively.
If the bond is owned by Alpha or Delta, then no further trading takes place, so on the next day it is owned by the same company.\\
(i) Write down the transition matrix $\mathbf { P }$.\\
(ii) Explain what is meant by an absorbing state of a Markov chain. Identify any absorbing states in this situation.\\
(iii) Find the probability that the bond is owned by Boyles on the 10th day.\\
(iv) Find the probability that on the 14th day the bond is owned by the same company as on the 10th day.\\
(v) Find the day on which the probability that the bond is owned by Alpha or Delta exceeds 0.9 for the first time.\\
(vi) Find the limit of $\mathbf { P } ^ { n }$ as $n$ tends to infinity.\\
(vii) Find the probability that the bond is eventually owned by Alpha.
The probabilities that Boyles and Cayleys own the bond on the first day are changed (but all the transition probabilities remain the same as before). The bond is now equally likely to be owned by Alpha or Delta at the end of the trading process.\\
(viii) Find the new probabilities for the ownership of the bond on the first day.
\hfill \mbox{\textit{OCR MEI FP3 2011 Q5 [24]}}