| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Difficulty | Easy -2.5 This question is mislabeled - it's actually a standard Markov chains problem from statistics, not group theory from Further Pure Mathematics. The tasks involve routine matrix operations (writing transition matrix, computing powers), basic probability calculations, and finding steady states - all mechanical procedures requiring no novel insight. This would be trivial for the stated topic (Groups/FP3) but is still below-average difficulty even as a statistics question. |
| Spec | 4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{P} = \begin{pmatrix} 0.9 & 0.3 & 0.1 \\ 0.07 & 0.6 & 0.7 \\ 0.03 & 0.1 & 0.2 \end{pmatrix}\) | B2 | Give B1 for two columns correct |
| [2] | Allow tolerance of \(\pm 0.0001\) in probabilities throughout this question; Do not penalise answers given to more than 4 dp |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{P}^9\begin{pmatrix}1\\0\\0\end{pmatrix}\) | M1 | For \(\mathbf{P}^9\) (allow \(\mathbf{P}^{10}\)) |
| M1 | For initial column matrix (or first column of \(\mathbf{P}^9\)) — Dependent on previous M1 | |
| \(P(A)=0.7268\quad P(B)=0.2189\quad P(C)=0.0544\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{P}^{11} = \begin{pmatrix}0.7242 & \cdots & \cdots \\ 0.2211 & \cdots & \cdots \\ 0.0547 & \cdots & \cdots\end{pmatrix}\) | M1 | Appropriate elements from \(\mathbf{P}^{11}\) (allow \(\mathbf{P}^{12}\)) |
| M1 | Diagonal elements from \(\mathbf{P}\) — Dependent on previous M1M1 | |
| \(0.7242\times0.9 + 0.2211\times0.6 + 0.0547\times0.2 = 0.7954\) | M1 A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((0.9\quad 0.6\quad 0.2)\mathbf{P}^{n-1}\begin{pmatrix}1\\0\\0\end{pmatrix} = (0.8009)\) when \(n=7\) | M1 | Repeating (iii) for another value of \(n\) |
| \(= (0.7986)\) when \(n=8\) | M1 | Obtaining values both sides of 0.8 — Valid method required here |
| Smallest value is \(n=8\) | B1 | |
| Probability is \(0.7986\) | A1 | \(0.7986\) implies M1M1 |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Expected run length is \(\frac{1}{1-0.9} = 10\) | M1 | Using \(\frac{1}{1-p}\) or \(\frac{p}{1-p}\) with \(p=0.9\) |
| A1 | ||
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{P}^n \to \begin{pmatrix}0.7225 & 0.7225 & 0.7225\\ 0.2225 & 0.2225 & 0.2225\\ 0.0549 & 0.0549 & 0.0549\end{pmatrix}\) | B2 | Give B1 for 6 elements correct to 3 dp |
| \(P(A)=0.7225\quad P(B)=0.2225\quad P(C)=0.0549\) | B1 | FT if columns agree to 4 dp |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.7225 \times 0.9 \times 0.9 = 0.5853\) | M1, A1 [2] | FT \(P(A) \times 0.81\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix}1-2x & y & 0.1\\x & 1-2y & 0.7\\x & y & 0.2\end{pmatrix}\begin{pmatrix}0.5\\0.4\\0.1\end{pmatrix} = \begin{pmatrix}0.5\\0.4\\0.1\end{pmatrix}\) | M1 | First or second column correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(x=0.06, \quad y=0.125\) | A1, M1 | Obtaining values for \(x\) and \(y\) |
| Transition matrix is \(\begin{pmatrix}0.88 & 0.125 & 0.1\\0.06 & 0.75 & 0.7\\0.06 & 0.125 & 0.2\end{pmatrix}\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \((0.5 \quad 0.4 \quad 0.1)\begin{pmatrix}1-2x & x & x\\y & 1-2y & y\\0.1 & 0.7 & 0.2\end{pmatrix} = (0.5 \quad 0.4 \quad 0.1)\) | M1 | First or second row correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(x=0.06, \quad y=0.125\) | A1, M1 | Obtaining values for \(x\) and \(y\) |
| Transition matrix is \(\begin{pmatrix}0.88 & 0.06 & 0.06\\0.125 & 0.75 & 0.125\\0.1 & 0.7 & 0.2\end{pmatrix}\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{P} = \begin{pmatrix}0.9 & 0.07 & 0.03\\0.3 & 0.6 & 0.1\\0.1 & 0.7 & 0.2\end{pmatrix}\) | B2 [2] | Give B1 for two rows correct; Allow tolerance of \(\pm 0.0001\) in probabilities throughout this question; Do not penalise answers given to more than 4 dp |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(A)=0.7268 \quad P(B)=0.2189 \quad P(C)=0.0544\) | M1, M1, A1 [3] | For \(\mathbf{P}^9\) (allow \(\mathbf{P}^{10}\)); For initial row matrix (or first row of \(\mathbf{P}^9\)); Dependent on previous M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.7242\times0.9 + 0.2211\times0.6 + 0.0547\times0.2 = 0.7954\) | M1, M1, M1, A1 [4] | Appropriate elements from \(\mathbf{P}^{11}\) (allow \(\mathbf{P}^{12}\)); Diagonal elements from \(\mathbf{P}\); Dependent on previous M1M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Smallest value is \(n=8\), Probability is \(0.7986\) | M1, M1, B1, A1 [4] | Repeating (iii) for another value of \(n\); Obtaining values both sides of 0.8; Valid method required here; \(0.7986\) implies M1M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Expected run length is \(\dfrac{1}{1-0.9} = 10\) | M1, A1 [2] | Using \(\dfrac{1}{1-p}\) or \(\dfrac{p}{1-p}\) with \(p=0.9\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(A)=0.7225 \quad P(B)=0.2225 \quad P(C)=0.0549\) | B2, B1 [3] | Give B1 for 6 elements correct to 3 dp; FT if rows agree to 4 dp |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{P} = \begin{pmatrix} 0.9 & 0.3 & 0.1 \\ 0.07 & 0.6 & 0.7 \\ 0.03 & 0.1 & 0.2 \end{pmatrix}$ | B2 | Give B1 for two columns correct |
| **[2]** | | Allow tolerance of $\pm 0.0001$ in probabilities throughout this question; Do not penalise answers given to more than 4 dp |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{P}^9\begin{pmatrix}1\\0\\0\end{pmatrix}$ | M1 | For $\mathbf{P}^9$ (allow $\mathbf{P}^{10}$) |
| | M1 | For initial column matrix (or first column of $\mathbf{P}^9$) — Dependent on previous M1 |
| $P(A)=0.7268\quad P(B)=0.2189\quad P(C)=0.0544$ | A1 | |
| **[3]** | | |
## Question 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{P}^{11} = \begin{pmatrix}0.7242 & \cdots & \cdots \\ 0.2211 & \cdots & \cdots \\ 0.0547 & \cdots & \cdots\end{pmatrix}$ | M1 | Appropriate elements from $\mathbf{P}^{11}$ (allow $\mathbf{P}^{12}$) |
| | M1 | Diagonal elements from $\mathbf{P}$ — Dependent on previous M1M1 |
| $0.7242\times0.9 + 0.2211\times0.6 + 0.0547\times0.2 = 0.7954$ | M1 A1 | |
| **[4]** | | |
## Question 5(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(0.9\quad 0.6\quad 0.2)\mathbf{P}^{n-1}\begin{pmatrix}1\\0\\0\end{pmatrix} = (0.8009)$ when $n=7$ | M1 | Repeating (iii) for another value of $n$ |
| $= (0.7986)$ when $n=8$ | M1 | Obtaining values both sides of 0.8 — Valid method required here |
| Smallest value is $n=8$ | B1 | |
| Probability is $0.7986$ | A1 | $0.7986$ implies M1M1 |
| **[4]** | | |
## Question 5(v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Expected run length is $\frac{1}{1-0.9} = 10$ | M1 | Using $\frac{1}{1-p}$ or $\frac{p}{1-p}$ with $p=0.9$ |
| | A1 | |
| **[2]** | | |
## Question 5(vi):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{P}^n \to \begin{pmatrix}0.7225 & 0.7225 & 0.7225\\ 0.2225 & 0.2225 & 0.2225\\ 0.0549 & 0.0549 & 0.0549\end{pmatrix}$ | B2 | Give B1 for 6 elements correct to 3 dp |
| $P(A)=0.7225\quad P(B)=0.2225\quad P(C)=0.0549$ | B1 | FT if columns agree to 4 dp |
| **[3]** | | |
## Question 5:
### Part (vii):
$0.7225 \times 0.9 \times 0.9 = 0.5853$ | M1, A1 [2] | FT $P(A) \times 0.81$
---
### Part (viii) [Column multiplication version]:
$\begin{pmatrix}1-2x & y & 0.1\\x & 1-2y & 0.7\\x & y & 0.2\end{pmatrix}\begin{pmatrix}0.5\\0.4\\0.1\end{pmatrix} = \begin{pmatrix}0.5\\0.4\\0.1\end{pmatrix}$ | M1 | First or second column correct
$0.5(1-2x)+0.4y+0.01=0.5$
$0.5x+0.4(1-2y)+0.07=0.4$
$0.5x+0.4y+0.02=0.1$
$x=0.06, \quad y=0.125$ | A1, M1 | Obtaining values for $x$ and $y$
Transition matrix is $\begin{pmatrix}0.88 & 0.125 & 0.1\\0.06 & 0.75 & 0.7\\0.06 & 0.125 & 0.2\end{pmatrix}$ | A1 [4] |
---
### Part (viii) [Row multiplication version]:
$(0.5 \quad 0.4 \quad 0.1)\begin{pmatrix}1-2x & x & x\\y & 1-2y & y\\0.1 & 0.7 & 0.2\end{pmatrix} = (0.5 \quad 0.4 \quad 0.1)$ | M1 | First or second row correct
$0.5(1-2x)+0.4y+0.01=0.5$
$0.5x+0.4(1-2y)+0.07=0.4$
$0.5x+0.4y+0.02=0.1$
$x=0.06, \quad y=0.125$ | A1, M1 | Obtaining values for $x$ and $y$
Transition matrix is $\begin{pmatrix}0.88 & 0.06 & 0.06\\0.125 & 0.75 & 0.125\\0.1 & 0.7 & 0.2\end{pmatrix}$ | A1 [4] |
---
### Part (i):
$\mathbf{P} = \begin{pmatrix}0.9 & 0.07 & 0.03\\0.3 & 0.6 & 0.1\\0.1 & 0.7 & 0.2\end{pmatrix}$ | B2 [2] | Give B1 for two rows correct; Allow tolerance of $\pm 0.0001$ in probabilities throughout this question; Do not penalise answers given to more than 4 dp
---
### Part (ii):
$(1 \quad 0 \quad 0)\mathbf{P}^9$
$P(A)=0.7268 \quad P(B)=0.2189 \quad P(C)=0.0544$ | M1, M1, A1 [3] | For $\mathbf{P}^9$ (allow $\mathbf{P}^{10}$); For initial row matrix (or first row of $\mathbf{P}^9$); Dependent on previous M1
---
### Part (iii):
$\mathbf{P}^{11} = \begin{pmatrix}0.7242 & 0.2211 & 0.0547\\ \cdots & \cdots & \cdots\\ \cdots & \cdots & \cdots\end{pmatrix}$
$0.7242\times0.9 + 0.2211\times0.6 + 0.0547\times0.2 = 0.7954$ | M1, M1, M1, A1 [4] | Appropriate elements from $\mathbf{P}^{11}$ (allow $\mathbf{P}^{12}$); Diagonal elements from $\mathbf{P}$; Dependent on previous M1M1
---
### Part (iv):
$(1 \quad 0 \quad 0)\mathbf{P}^{n-1}\begin{pmatrix}0.9\\0.6\\0.2\end{pmatrix} = (0.8009)$ when $n=7$
$= (0.7986)$ when $n=8$
Smallest value is $n=8$, Probability is $0.7986$ | M1, M1, B1, A1 [4] | Repeating (iii) for another value of $n$; Obtaining values both sides of 0.8; Valid method required here; $0.7986$ implies M1M1
---
### Part (v):
Expected run length is $\dfrac{1}{1-0.9} = 10$ | M1, A1 [2] | Using $\dfrac{1}{1-p}$ or $\dfrac{p}{1-p}$ with $p=0.9$
---
### Part (vi):
$\mathbf{P}^n \to \begin{pmatrix}0.7225 & 0.2225 & 0.0549\\0.7225 & 0.2225 & 0.0549\\0.7225 & 0.2225 & 0.0549\end{pmatrix}$
$P(A)=0.7225 \quad P(B)=0.2225 \quad P(C)=0.0549$ | B2, B1 [3] | Give B1 for 6 elements correct to 3 dp; FT if rows agree to 4 dp5 In this question, give probabilities correct to 4 decimal places.\\
The speeds of vehicles are measured on a busy stretch of road and are categorised as A (not more than 30 mph ), B (more than 30 mph but not more than 40 mph ) or C (more than 40 mph ).
\begin{itemize}
\item Following a vehicle in category A , the probabilities that the next vehicle is in categories $\mathrm { A } , \mathrm { B } , \mathrm { C }$ are $0.9,0.07,0.03$ respectively.
\item Following a vehicle in category B , the probabilities that the next vehicle is in categories $\mathrm { A } , \mathrm { B } , \mathrm { C }$ are $0.3,0.6,0.1$ respectively.
\item Following a vehicle in category C , the probabilities that the next vehicle is in categories $\mathrm { A } , \mathrm { B } , \mathrm { C }$ are $0.1,0.7,0.2$ respectively.
\end{itemize}
This is modelled as a Markov chain with three states corresponding to the categories A, B, C. The speed of the first vehicle is measured as 28 mph .\\
(i) Write down the transition matrix $\mathbf { P }$.\\
(ii) Find the probabilities that the 10th vehicle is in each of the three categories.\\
(iii) Find the probability that the 12th and 13th vehicles are in the same category.\\
(iv) Find the smallest value of $n$ for which the probability that the $n$th and $( n + 1 )$ th vehicles are in the same category is less than 0.8, and give the value of this probability.\\
(v) Find the expected number of vehicles (including the first vehicle) in category A before a vehicle in a different category.\\
(vi) Find the limit of $\mathbf { P } ^ { n }$ as $n$ tends to infinity, and hence write down the equilibrium probabilities for the three categories.\\
(vii) Find the probability that, after many vehicles have passed by, the next three vehicles are all in category A.
On a new stretch of road, the same categories are used but some of the transition probabilities are different.
\begin{itemize}
\item Following a vehicle in category A , the probability that the next vehicle is in category B is equal to the probability that it is in category C .
\item Following a vehicle in category B , the probability that the next vehicle is in category A is equal to the probability that it is in category C .
\item Following a vehicle in category C , the probabilities that the next vehicle is in categories $\mathrm { A } , \mathrm { B } , \mathrm { C }$ are $0.1,0.7,0.2$ respectively.
\end{itemize}
In the long run, the proportions of vehicles in categories A, B, C are 50\%, 40\%, 10\% respectively.\\
(viii) Find the transition matrix for the new stretch of road.
\hfill \mbox{\textit{OCR MEI FP3 2014 Q5 [24]}}