| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Recurrence relation solution |
| Difficulty | Standard +0.8 This is a multi-part Markov chain question requiring transition matrices, matrix powers, equilibrium calculations, and expected value from geometric series. While systematic, it demands careful probability tracking across multiple scenarios and understanding of absorbing states, placing it moderately above average difficulty for Further Maths. |
| Answer | Marks | Guidance |
|---|---|---|
| \[\mathbf{P} = \begin{pmatrix} 0 & \frac{1}{3} & 0 \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{2} \end{pmatrix}\] | B2 | B1 for two out of three columns correct |
| Answer | Marks | Guidance |
|---|---|---|
| \[\begin{pmatrix} 0.1388\dot{8} & - & - \\ - & - & - \\ - & - & - \end{pmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix} \text{ gives } 0.1388\dot{8}\left(=\frac{5}{36}\right)\] | M1 | Cube \(P\) |
| A1 | Sight of matrix soi | |
| A1 | Allow M1 for \(P^4\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(p = 0.5 \times 0.4285 + 0.5 \times 0.4286 = 0.4286\) | M1 | Using diagonal elements from \(P^5\) |
| M1 | Using probabilities from 2nd day | |
| A1 | Ft | |
| A1 | Cao |
| Answer | Marks | Guidance |
|---|---|---|
| Over a long period these are the probabilities that on any day at random the inspector is at these factories | M1 | Obtaining equations |
| B1 | Sight of \(x + y + z = 1\) soi | A2 for probabilities, A1 one error |
| A1 | ||
| B1 |
| Answer | Marks |
|---|---|
| \[\left(=\frac{5}{11}, \frac{3}{11}, \frac{3}{11}\right)\] | B1 |
| M1 | Solve or consider \(Q^n\) for large \(n\) |
| A1 |
| Answer | Marks |
|---|---|
| \[\Rightarrow \frac{\alpha}{1-\alpha} = \frac{0.8}{0.2} = 4\] | B1 |
| M1 | For using \(\dfrac{\alpha}{1-\alpha}\) or \(\dfrac{1}{1-\alpha}\) |
| A1 |
| Answer | Marks | Guidance |
|---|---|---|
| So 10 days later (which is day 25). | M1 | Method by "trial" with new matrix |
| A1 | For sight of one value | |
| A1 |
| Answer | Marks |
|---|---|
| If it goes to A then it stays there. | B1 |
| B1 | oe |
# Question 5:
## Part (i):
$$\mathbf{P} = \begin{pmatrix} 0 & \frac{1}{3} & 0 \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{2} \end{pmatrix}$$ | **B2** | B1 for two out of three columns correct
---
## Part (ii)(A):
$\mathbf{P}^3\mathbf{p}$
$$\begin{pmatrix} 0.1388\dot{8} & - & - \\ - & - & - \\ - & - & - \end{pmatrix}\begin{pmatrix}1\\0\\0\end{pmatrix} \text{ gives } 0.1388\dot{8}\left(=\frac{5}{36}\right)$$ | **M1** | Cube $P$ |
| **A1** | Sight of matrix soi |
| **A1** | | Allow M1 for $P^4$
---
## Part (ii)(B):
$P = M^5p$
$$\mathbf{P}^5 = \begin{pmatrix} \cdots & - & - \\ - & 0.4285 & - \\ - & - & 0.4286 \end{pmatrix}$$
$p = 0.5 \times 0.4285 + 0.5 \times 0.4286 = 0.4286$ | **M1** | Using diagonal elements from $P^5$ |
| **M1** | Using probabilities from 2nd day |
| **A1** | Ft |
| **A1** | Cao |
---
## Part (iii):
$0.143, \ 0.429, \ 0.429$
$$\left(=\frac{1}{7}, \frac{3}{7}, \frac{3}{7}\right)$$
Over a long period these are the probabilities that on any day at random the inspector is at these factories | **M1** | Obtaining equations | Or **M1** considering $P^n$ where $n$ is large (10 or more) |
| **B1** | Sight of $x + y + z = 1$ soi | **A2** for probabilities, A1 one error |
| **A1** | |
| **B1** | |
---
## Part (iv):
$$\mathbf{Q} = \begin{pmatrix} 0.8 & \frac{1}{3} & 0 \\ 0.1 & \frac{1}{3} & \frac{1}{2} \\ 0.1 & \frac{1}{3} & \frac{1}{2} \end{pmatrix}$$
$0.455, \ 0.273, \ 0.273$
$$\left(=\frac{5}{11}, \frac{3}{11}, \frac{3}{11}\right)$$ | **B1** | |
| **M1** | Solve or consider $Q^n$ for large $n$ |
| **A1** | |
---
## Part (v):
$P(\text{from A to A}) = 0.8$. So $\alpha = 0.8$
$$\Rightarrow \frac{\alpha}{1-\alpha} = \frac{0.8}{0.2} = 4$$ | **B1** | |
| **M1** | For using $\dfrac{\alpha}{1-\alpha}$ or $\dfrac{1}{1-\alpha}$ |
| **A1** | |
---
## Part (vi):
New transition matrix:
$$\mathbf{R} = \begin{pmatrix} 1 & \frac{1}{3} & 0 \\ 0 & \frac{1}{3} & \frac{1}{2} \\ 0 & \frac{1}{3} & \frac{1}{2} \end{pmatrix}$$
We need $3^\text{rd}$ entry of $2^\text{nd}$ row to be $< 0.1$
$$\mathbf{R}^9 = \begin{pmatrix}\cdots & \cdots & \cdots \\ \cdots & \cdots & 0.1162 \\ \cdots & \cdots & \cdots\end{pmatrix}, \quad \mathbf{R}^{10} = \begin{pmatrix}\cdots & \cdots & \cdots \\ \cdots & \cdots & 0.0969 \\ \cdots & \cdots & \cdots\end{pmatrix}$$
So 10 days later (which is day 25). | **M1** | Method by "trial" with new matrix |
| **A1** | For sight of one value |
| **A1** | |
---
## Part (vii):
A is the absorbing state.
If it goes to A then it stays there. | **B1** | |
| **B1** | oe |5 An inspector has three factories, A, B, C, to check. He spends each day in one of the factories. He chooses the factory to visit on a particular day according to the following rules.
\begin{itemize}
\item If he is in A one day, then the next day he will never choose A but he is equally likely to choose B or C .
\item If he is in B one day, then the next day he is equally likely to choose $\mathrm { A } , \mathrm { B }$ or C .
\item If he is in C one day, then the next day he will never choose A but he is equally likely to choose B or C .
\begin{enumerate}[label=(\roman*)]
\item Write down the transition matrix, $\mathbf { P }$.
\item On Day 1 the inspector chooses A.\\
(A) Find the probability that he will choose A on Day 4.\\
(B) Find the probability that the factory he chooses on Day 7 is the same factory that he chose on Day 2.
\item Find the equilibrium probabilities and explain what they mean.
\end{itemize}
The inspector is not satisfied with the number of times he visits A so he changes the rules as follows.
\begin{itemize}
\item If he is in A one day, then the next day he will choose $\mathrm { A } , \mathrm { B } , \mathrm { C }$, with probabilities $0.8,0.1,0.1$, respectively.
\item If he is in B or C one day, then the probabilities for choosing the factory the next day remain as before.
\item Write down the new transition matrix, $\mathbf { Q }$, and find the new equilibrium probabilities.
\item On a particular day, the inspector visits factory A. Find the expected number of consecutive further days on which he will visit factory A.
\end{itemize}
Still not satisfied, the inspector changes the rules as follows.
\begin{itemize}
\item If he is in A one day, then the next day he will choose $\mathrm { A } , \mathrm { B } , \mathrm { C }$, with probabilities $1,0,0$, respectively.
\item If he is in B or C one day, then the probabilities for choosing the factory the next day remain as before.
\end{itemize}
The new transition matrix is $\mathbf { R }$.
\item On Day 15 he visits C . Find the first subsequent day for which the probability that he visits B is less than 0.1.
\item Show that in this situation there is an absorbing state, explaining what this means.
\section*{END OF QUESTION PAPER}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP3 2015 Q5 [24]}}