| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| 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 matrices, steady-state calculations, and multi-step probability computations. While it has multiple parts and requires careful bookkeeping, all techniques are routine for Further Maths students: writing transition matrices from verbal descriptions, solving for steady states (part ii), matrix powers for multi-step transitions (parts iii-iv), and geometric series for expected values (part v). The concepts are well-practiced in FP3, making this slightly above average difficulty but not requiring novel insight. |
| Answer | Marks | Guidance |
|---|---|---|
| \[\begin{pmatrix} 0.75 & 0 & p \\ 0.125 & 0.5 & \frac{1-p}{2} \\ 0.125 & 0.5 & \frac{1-p}{2} \end{pmatrix}\] | B1, M1, A1 | B1 for first two columns; M1 for making 3rd column sum to 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \[\begin{pmatrix} 0.75 & 0 & p \\ 0.125 & 0.5 & \frac{1-p}{2} \\ 0.125 & 0.5 & \frac{1-p}{2} \end{pmatrix} \begin{pmatrix} \frac{1}{3} \\ \frac{1}{3} \\ \frac{1}{3} \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \\ \frac{1}{3} \\ \frac{1}{3} \end{pmatrix}\] | M1, A1 | M1 for equilibrium probs; A1 for equation |
| \(0.75 + p = 1\) | A1 | Correct equation implies M1A1A1 |
| \(p = 0.25\) | A1 | Just answer: B4 |
| Answer | Marks | Guidance |
|---|---|---|
| \[\text{P(A on day 5)} = \begin{pmatrix} 0.75 & 0 & 0.25 \\ 0.125 & 0.5 & 0.375 \\ 0.125 & 0.5 & 0.375 \end{pmatrix}^4 \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\] | M1 | For power 4 |
| \[= \begin{pmatrix} 0.435059 & \cdots & \cdots \\ \cdots & \cdots & \cdots \\ \cdots & \cdots & \cdots \end{pmatrix}\] | A1 | At least one right, soi |
| \(= 0.435\) | A1 | Just answer: B3 |
| Answer | Marks | Guidance |
|---|---|---|
| \[\mathbf{P} = \begin{pmatrix} 0.75 & 0 & 0.4 \\ 0.125 & 0.5 & 0.3 \\ 0.125 & 0.5 & 0.3 \end{pmatrix}, \quad \mathbf{P}^3 = \begin{pmatrix} 0.536875 & 0.31 & 0.431 \\ 0.231563 & 0.345 & 0.2845 \\ 0.231563 & 0.345 & 0.2845 \end{pmatrix}\] | M1 | For using \(P^3\) |
| \[\mathbf{P}^3\begin{pmatrix}1\\0\\0\end{pmatrix} = \begin{pmatrix}0.536875\\0.231563\\0.231563\end{pmatrix}\] | M1, A1 | First column of \(P^3\) |
| \[p = 0.536875 \times 0.536875 + 0.345 \times 0.2315625 + 0.2845 \times 0.2315625\] | M1, A1 | |
| \(= 0.434(003)\ldots\) | [5 marks total] |
| Answer | Marks | Guidance |
|---|---|---|
| P(from A to A) \(= 0.75\) so \(\alpha = 0.75\) | B1 | |
| Expected number \(= \dfrac{\alpha}{1-\alpha} = \dfrac{0.75}{0.25} = 3\) | M1, A1 | Using \(\dfrac{\alpha}{1-\alpha}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \[\begin{pmatrix} 0 & 0.5 & 0.5 \\ 0 & 0 & 0.5 \\ 1 & 0.5 & 0 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}x\\y\\z\end{pmatrix}\] | B1 | New transition matrix |
| \(0.5y + 0.5z = x, \quad 0.5z = y, \quad x + 0.5y = z\) | M1, A1 | |
| \(x + y + z = 1\) | M1 | |
| \(x = \dfrac{1}{3},\quad y = \dfrac{2}{9},\quad z = \dfrac{4}{9}\) | A2 | \(-1\) each error |
| OR: New transition matrix; considering a high power (at least 20); \(P(A)=0.333,\ P(B)=0.222,\ P(C)=0.444\) | B1, M2, A1A1A1 | Give M1 for at least 10 |
## Question 5:
### Part (i):
$$\begin{pmatrix} 0.75 & 0 & p \\ 0.125 & 0.5 & \frac{1-p}{2} \\ 0.125 & 0.5 & \frac{1-p}{2} \end{pmatrix}$$ | B1, M1, A1 | B1 for first two columns; M1 for making 3rd column sum to 1
---
### Part (ii):
$$\begin{pmatrix} 0.75 & 0 & p \\ 0.125 & 0.5 & \frac{1-p}{2} \\ 0.125 & 0.5 & \frac{1-p}{2} \end{pmatrix} \begin{pmatrix} \frac{1}{3} \\ \frac{1}{3} \\ \frac{1}{3} \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \\ \frac{1}{3} \\ \frac{1}{3} \end{pmatrix}$$ | M1, A1 | M1 for equilibrium probs; A1 for equation
$0.75 + p = 1$ | A1 | Correct equation implies M1A1A1
$p = 0.25$ | A1 | Just answer: B4
---
### Part (iii):
$$\text{P(A on day 5)} = \begin{pmatrix} 0.75 & 0 & 0.25 \\ 0.125 & 0.5 & 0.375 \\ 0.125 & 0.5 & 0.375 \end{pmatrix}^4 \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$$ | M1 | For power 4
$$= \begin{pmatrix} 0.435059 & \cdots & \cdots \\ \cdots & \cdots & \cdots \\ \cdots & \cdots & \cdots \end{pmatrix}$$ | A1 | At least one right, soi
$= 0.435$ | A1 | Just answer: B3
---
### Part (iv):
$$\mathbf{P} = \begin{pmatrix} 0.75 & 0 & 0.4 \\ 0.125 & 0.5 & 0.3 \\ 0.125 & 0.5 & 0.3 \end{pmatrix}, \quad \mathbf{P}^3 = \begin{pmatrix} 0.536875 & 0.31 & 0.431 \\ 0.231563 & 0.345 & 0.2845 \\ 0.231563 & 0.345 & 0.2845 \end{pmatrix}$$ | M1 | For using $P^3$
$$\mathbf{P}^3\begin{pmatrix}1\\0\\0\end{pmatrix} = \begin{pmatrix}0.536875\\0.231563\\0.231563\end{pmatrix}$$ | M1, A1 | First column of $P^3$
$$p = 0.536875 \times 0.536875 + 0.345 \times 0.2315625 + 0.2845 \times 0.2315625$$ | M1, A1 |
$= 0.434(003)\ldots$ | | [5 marks total]
---
### Part (v):
P(from A to A) $= 0.75$ so $\alpha = 0.75$ | B1 |
Expected number $= \dfrac{\alpha}{1-\alpha} = \dfrac{0.75}{0.25} = 3$ | M1, A1 | Using $\dfrac{\alpha}{1-\alpha}$
---
### Part (vi):
$$\begin{pmatrix} 0 & 0.5 & 0.5 \\ 0 & 0 & 0.5 \\ 1 & 0.5 & 0 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}x\\y\\z\end{pmatrix}$$ | B1 | New transition matrix
$0.5y + 0.5z = x, \quad 0.5z = y, \quad x + 0.5y = z$ | M1, A1 |
$x + y + z = 1$ | M1 |
$x = \dfrac{1}{3},\quad y = \dfrac{2}{9},\quad z = \dfrac{4}{9}$ | A2 | $-1$ each error
**OR:** New transition matrix; considering a high power (at least 20); $P(A)=0.333,\ P(B)=0.222,\ P(C)=0.444$ | B1, M2, A1A1A1 | Give M1 for at least 105 Each day that Adam is at work he carries out one of three tasks A, B or C. Each task takes a whole day. Adam chooses the task to carry out on each day according to the following set of three rules.
\begin{enumerate}
\item If, on any given day, he has worked on task A then the next day he will choose task A with probability 0.75 , and tasks B and C with equal probability.
\item If, on any given day, he has worked on task B then the next day he will choose task B or task C with equal probability but will never choose task A .
\item If, on any given day, he has worked on task C then the next day he will choose task A with probability $p$ and tasks B and C with equal probability.\\
(i) Write down the transition matrix.\\
(ii) Over a long period Adam carries out the tasks $\mathrm { A } , \mathrm { B }$ and C with equal frequency. Find the value of $p$.\\
(iii) On day 1 Adam chooses task A . Find the probability that he also chooses task A on day 5 .
\end{enumerate}
Adam decides to change rule 3 as follows.\\
If, on any given day, he has worked on task C then the next day he will choose tasks $\mathrm { A } , \mathrm { B } , \mathrm { C }$ with probabilities $0.4,0.3,0.3$ respectively.\\
(iv) On day 1 Adam chooses task A. Find the probability that he chooses the same task on day 7 as he did on day 4 .\\
(v) On a particular day, Adam chooses task A. Find the expected number of consecutive further days on which he will choose A.
Adam changes all three rules again as follows.
\begin{itemize}
\item If he works on A one day then on the next day he chooses C .
\item If he works on B one day then on the next day he chooses A or C each with probability 0.5.
\item If he works on C one day then on the next day he chooses A or B each with probability 0.5 .\\
(vi) Find the long term probabilities for each task.
\end{itemize}
\hfill \mbox{\textit{OCR MEI FP3 2016 Q5 [24]}}