| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Difficulty | Challenging +1.8 This is a comprehensive Markov chain question requiring matrix operations, probability calculations, and limit analysis. While it involves multiple parts and extended calculations (matrix powers, absorbing state probabilities), the techniques are standard for Further Maths and each step follows methodically from the previous one without requiring novel insights. The computational intensity and length push it above average, but it remains a structured application of known methods. |
| Spec | 4.03a Matrix language: terminology and notation4.03b Matrix operations: addition, multiplication, scalar |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{P} = \begin{pmatrix}1&0.5&0&0&0\\0&0.05&0.5&0&0\\0&0.45&0.05&0.5&0\\0&0&0.45&0.05&0\\0&0&0&0.45&1\end{pmatrix}\) | B3 | Give B2 for four columns correct; Give B1 for two columns correct |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{P}^8\begin{pmatrix}0\\\frac{1}{3}\\\frac{1}{3}\\\frac{1}{3}\\0\end{pmatrix} = \begin{pmatrix}0.5042\\0.0230\\0.0278\\\mathbf{0.02071}\\0.4242\end{pmatrix}\), \(P(\text{3 lives}) = 0.0207\) (4 dp) | M1, E1 | For \(\mathbf{P}^8\) (allow \(\mathbf{P}^7\) or \(\mathbf{P}^9\)) and initial column matrix; Correctly shown |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(q(n) = P(\text{not yet ended after }n\text{ tasks})\) | ||
| \(= (0\ 1\ 1\ 1\ 1\ 0)\mathbf{P}^n\begin{pmatrix}0\\\frac{1}{3}\\\frac{1}{3}\\\frac{1}{3}\\0\end{pmatrix}\) | M1, M1 | Obtaining probabilities after 10 tasks; Adding probabilities of 1, 2, 3 lives |
| \(q(10) = 0.0371\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(q(9) - q(10)\) | M1 | Using \(q(9)\) and \(q(10)\) |
| \(= 0.05072 - 0.03709\) | M1 | Evaluating \(q(9)\) |
| \(= 0.0136\) | A1 | |
| [3] | ||
| OR: \(\mathbf{P}^9\begin{pmatrix}0\\\frac{1}{3}\\\frac{1}{3}\\\frac{1}{3}\\0\end{pmatrix} = \begin{pmatrix}\cdot\\0.01506\\\cdot\\0.01355\\\cdot\end{pmatrix}\) | M1 | Probs of 1 and 3 lives after 9 tasks |
| \(0.01506\times0.5 + 0.01355\times0.45 = 0.0136\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(q(13) = 0.01374\) | M1 | Evaluating \(q(n)\) for some \(n>10\) |
| \(q(14) = 0.00998\) | M1 | Consecutive values each side of 0.01 |
| Smallest \(N\) is 14 | A1 | Must be clear that answer is 14 |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{P}^n \to \begin{pmatrix}1&0.7880&0.5525&0.2908&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0.2120&0.4475&0.7092&1\end{pmatrix} = \mathbf{L}\) | B2 | Give B1 for any element correct to 3 dp (other than 0 or 1) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{L}\begin{pmatrix}0\\\frac{1}{3}\\\frac{1}{3}\\\frac{1}{3}\\0\end{pmatrix} = \begin{pmatrix}0.5438\\0\\0\\0\\0.4562\end{pmatrix}\) | M1M1 | Using \(\mathbf{L}\) and the initial column matrix |
| \(P(\text{wins a prize}) = 0.4562\) | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Maximum probability is \(0.7092\) | B1 ft | |
| Always start with 3 lives | B1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} 0 \\ 0.1 \\ p \\ q \\ 0 \end{pmatrix} \mathbf{L} = \begin{pmatrix} 0.4 \\ 0 \\ 0 \\ 0 \\ 0.6 \end{pmatrix}\) | M1 | |
| \(0.7880 \times 0.1 + 0.5525p + 0.2908(0.9 - p) = 0.4\) | M1 | Or \(0.0212 + 0.4475p + 0.7092(0.9-p) = 0.6\); Allow use of \(p + q = 1\) |
| \(P(\text{2 lives}) = 0.2273\), \(P(\text{3 lives}) = 0.6727\) | A1 | Obtaining a value for \(p\) or \(q\); Accept values rounding to \(0.227\), \(0.673\) [3] |
| Answer | Marks | Guidance |
|---|---|---|
| *Post-multiplication by transition matrix* | Allow tolerance of \(\pm 0.0001\) in probabilities throughout this question |
| Answer | Marks | Guidance |
|---|---|---|
| \[\mathbf{P} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0.5 & 0.05 & 0.45 & 0 & 0 \\ 0 & 0.5 & 0.05 & 0.45 & 0 \\ 0 & 0 & 0.5 & 0.05 & 0.45 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}\] | B3 | Give B2 for four rows correct; Give B1 for two rows correct [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 \end{pmatrix} \mathbf{P}^8\) | M1 | For \(\mathbf{P}^8\) (allow \(\mathbf{P}^7\) or \(\mathbf{P}^9\)) and initial row matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{3 lives}) = 0.0207\) (4 dp) | E1 | Correctly shown [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \begin{pmatrix} 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 \end{pmatrix} \mathbf{P}^n \begin{pmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 0 \end{pmatrix}\) | M1 | Obtaining probabilities after 10 tasks; Allow M1 for using \(\mathbf{P}^9\) or \(\mathbf{P}^{11}\) |
| M1 | Adding probabilities of 1, 2, 3 lives | |
| \(q(10) = 0.0371\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(q(9) - q(10)\) | M1 | Using \(q(9)\) and \(q(10)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.0136\) | M1, A1 | Evaluating \(q(9)\) [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \begin{pmatrix} \cdot & 0.01506 & \cdot & 0.01355 & \cdot \end{pmatrix}\) | M1 | Probs of 1 and 3 lives after 9 tasks |
| \(0.01506 \times 0.5 + 0.01355 \times 0.45 = 0.0136\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Smallest \(N\) is \(14\) | M1, M1, A1 | Evaluating \(q(n)\) for some \(n > 10\); Consecutive values each side of \(0.01\); Must be clear that their answer is 14; Just \(N = 14\) www earns B3 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \[\mathbf{P}^n \to \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0.7880 & 0 & 0 & 0 & 0.2120 \\ 0.5525 & 0 & 0 & 0 & 0.4475 \\ 0.2908 & 0 & 0 & 0 & 0.7092 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} = \mathbf{L}\] | B2 | Give B1 for any element correct to 3 dp (other than 0 or 1) [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 \end{pmatrix} \mathbf{L}\) | M1M1 | Using \(\mathbf{L}\) and the initial row matrix |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{wins a prize}) = 0.4562\) | A1 | [3] |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{P} = \begin{pmatrix}1&0.5&0&0&0\\0&0.05&0.5&0&0\\0&0.45&0.05&0.5&0\\0&0&0.45&0.05&0\\0&0&0&0.45&1\end{pmatrix}$ | B3 | Give B2 for four columns correct; Give B1 for two columns correct | Allow tolerance of $\pm0.0001$ throughout |
| **[3]** | | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{P}^8\begin{pmatrix}0\\\frac{1}{3}\\\frac{1}{3}\\\frac{1}{3}\\0\end{pmatrix} = \begin{pmatrix}0.5042\\0.0230\\0.0278\\\mathbf{0.02071}\\0.4242\end{pmatrix}$, $P(\text{3 lives}) = 0.0207$ (4 dp) | M1, E1 | For $\mathbf{P}^8$ (allow $\mathbf{P}^7$ or $\mathbf{P}^9$) and initial column matrix; Correctly shown |
| **[2]** | | |
## Question 5(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $q(n) = P(\text{not yet ended after }n\text{ tasks})$ | | |
| $= (0\ 1\ 1\ 1\ 1\ 0)\mathbf{P}^n\begin{pmatrix}0\\\frac{1}{3}\\\frac{1}{3}\\\frac{1}{3}\\0\end{pmatrix}$ | M1, M1 | Obtaining probabilities after 10 tasks; Adding probabilities of 1, 2, 3 lives | Allow M1 for using $\mathbf{P}^9$ or $\mathbf{P}^{11}$ |
| $q(10) = 0.0371$ | A1 | |
| **[3]** | | |
## Question 5(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $q(9) - q(10)$ | M1 | Using $q(9)$ and $q(10)$ |
| $= 0.05072 - 0.03709$ | M1 | Evaluating $q(9)$ |
| $= 0.0136$ | A1 | |
| **[3]** | | |
| **OR:** $\mathbf{P}^9\begin{pmatrix}0\\\frac{1}{3}\\\frac{1}{3}\\\frac{1}{3}\\0\end{pmatrix} = \begin{pmatrix}\cdot\\0.01506\\\cdot\\0.01355\\\cdot\end{pmatrix}$ | M1 | Probs of 1 and 3 lives after 9 tasks |
| $0.01506\times0.5 + 0.01355\times0.45 = 0.0136$ | M1, A1 | |
## Question 5(v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $q(13) = 0.01374$ | M1 | Evaluating $q(n)$ for some $n>10$ |
| $q(14) = 0.00998$ | M1 | Consecutive values each side of 0.01 |
| Smallest $N$ is 14 | A1 | Must be clear that answer is 14 | Just $N=14$ www earns B3 |
| **[3]** | | |
## Question 5(vi):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{P}^n \to \begin{pmatrix}1&0.7880&0.5525&0.2908&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0.2120&0.4475&0.7092&1\end{pmatrix} = \mathbf{L}$ | B2 | Give B1 for any element correct to 3 dp (other than 0 or 1) |
| **[2]** | | |
## Question 5(vii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{L}\begin{pmatrix}0\\\frac{1}{3}\\\frac{1}{3}\\\frac{1}{3}\\0\end{pmatrix} = \begin{pmatrix}0.5438\\0\\0\\0\\0.4562\end{pmatrix}$ | M1M1 | Using $\mathbf{L}$ and the initial column matrix |
| $P(\text{wins a prize}) = 0.4562$ | A1 | |
| **[3]** | | |
# Question 5:
## Part (viii)
Maximum probability is $0.7092$ | B1 ft |
Always start with 3 lives | B1 | **[2]**
## Part (ix)
$\begin{pmatrix} 0 \\ 0.1 \\ p \\ q \\ 0 \end{pmatrix} \mathbf{L} = \begin{pmatrix} 0.4 \\ 0 \\ 0 \\ 0 \\ 0.6 \end{pmatrix}$ | M1 |
$0.7880 \times 0.1 + 0.5525p + 0.2908(0.9 - p) = 0.4$ | M1 | Or $0.0212 + 0.4475p + 0.7092(0.9-p) = 0.6$; Allow use of $p + q = 1$
$P(\text{2 lives}) = 0.2273$, $P(\text{3 lives}) = 0.6727$ | A1 | Obtaining a value for $p$ or $q$; Accept values rounding to $0.227$, $0.673$ **[3]**
---
*Post-multiplication by transition matrix* | | Allow tolerance of $\pm 0.0001$ in probabilities throughout this question
---
## Part (i)
$$\mathbf{P} = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0.5 & 0.05 & 0.45 & 0 & 0 \\ 0 & 0.5 & 0.05 & 0.45 & 0 \\ 0 & 0 & 0.5 & 0.05 & 0.45 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}$$ | B3 | Give B2 for four rows correct; Give B1 for two rows correct **[3]**
## Part (ii)
$\begin{pmatrix} 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 \end{pmatrix} \mathbf{P}^8$ | M1 | For $\mathbf{P}^8$ (allow $\mathbf{P}^7$ or $\mathbf{P}^9$) and initial row matrix
$= \begin{pmatrix} 0.5042 & 0.0230 & 0.0278 & \mathbf{0.02071} & 0.4242 \end{pmatrix}$
$P(\text{3 lives}) = 0.0207$ (4 dp) | E1 | Correctly shown **[2]**
## Part (iii)
Let $q(n) = P(\text{not yet ended after } n \text{ tasks})$
$= \begin{pmatrix} 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 \end{pmatrix} \mathbf{P}^n \begin{pmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 0 \end{pmatrix}$ | M1 | Obtaining probabilities after 10 tasks; Allow M1 for using $\mathbf{P}^9$ or $\mathbf{P}^{11}$
| M1 | Adding probabilities of 1, 2, 3 lives
$q(10) = 0.0371$ | A1 | **[3]**
## Part (iv)
$q(9) - q(10)$ | M1 | Using $q(9)$ and $q(10)$
$= 0.05072 - 0.03709$
$= 0.0136$ | M1, A1 | Evaluating $q(9)$ **[3]**
**OR** $\begin{pmatrix} 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 \end{pmatrix} \mathbf{P}^9$
$= \begin{pmatrix} \cdot & 0.01506 & \cdot & 0.01355 & \cdot \end{pmatrix}$ | M1 | Probs of 1 and 3 lives after 9 tasks
$0.01506 \times 0.5 + 0.01355 \times 0.45 = 0.0136$ | M1, A1 |
## Part (v)
$q(13) = 0.01374$
$q(14) = 0.00998$
Smallest $N$ is $14$ | M1, M1, A1 | Evaluating $q(n)$ for some $n > 10$; Consecutive values each side of $0.01$; Must be clear that their answer is 14; Just $N = 14$ www earns B3 **[3]**
## Part (vi)
$$\mathbf{P}^n \to \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0.7880 & 0 & 0 & 0 & 0.2120 \\ 0.5525 & 0 & 0 & 0 & 0.4475 \\ 0.2908 & 0 & 0 & 0 & 0.7092 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix} = \mathbf{L}$$ | B2 | Give B1 for any element correct to 3 dp (other than 0 or 1) **[2]**
## Part (vii)
$\begin{pmatrix} 0 & \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 \end{pmatrix} \mathbf{L}$ | M1M1 | Using $\mathbf{L}$ and the initial row matrix
$= \begin{pmatrix} 0.5438 & 0 & 0 & 0 & 0.4562 \end{pmatrix}$
$P(\text{wins a prize}) = 0.4562$ | A1 | **[3]**5 In this question, give probabilities correct to 4 decimal places.\\
A contestant in a game-show starts with one, two or three 'lives', and then performs a series of tasks. After each task, the number of lives either decreases by one, or remains the same, or increases by one. The game ends when the number of lives becomes either four or zero. If the number of lives is four, the contestant wins a prize; if the number of lives is zero, the contestant loses and leaves with nothing.
At the start, the number of lives is decided at random, so that the contestant is equally likely to start with one, two or three lives. The tasks do not involve any skill, and after every task:
\begin{itemize}
\item the probability that the number of lives decreases by one is 0.5 ,
\item the probability that the number of lives remains the same is 0.05 ,
\item the probability that the number of lives increases by one is 0.45 .
\end{itemize}
This is modelled as a Markov chain with five states corresponding to the possible numbers of lives. The states corresponding to zero lives and four lives are absorbing states.\\
(i) Write down the transition matrix $\mathbf { P }$.\\
(ii) Show that, after 8 tasks, the probability that the contestant has three lives is 0.0207 , correct to 4 decimal places.\\
(iii) Find the probability that, after 10 tasks, the game has not yet ended.\\
(iv) Find the probability that the game ends after exactly 10 tasks.\\
(v) Find the smallest value of $N$ for which the probability that the game has not yet ended after $N$ tasks is less than 0.01 .\\
(vi) Find the limit of $\mathbf { P } ^ { n }$ as $n$ tends to infinity.\\
(vii) Find the probability that the contestant wins a prize.
The beginning of the game is now changed, so that the probabilities of starting with one, two or three lives can be adjusted.\\
(viii) State the maximum possible probability that the contestant wins a prize, and how this can be achieved.\\
(ix) Given that the probability of starting with one life is 0.1 , and the probability of winning a prize is 0.6 , find the probabilities of starting with two lives and starting with three lives.
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\hfill \mbox{\textit{OCR MEI FP3 2013 Q5 [24]}}