Question 3 - A-Level Maths

Edexcel FP2 2022 June — Question 3 8 marks

Exam Board	Edexcel
Module	FP2 (Further Pure Mathematics 2)
Year	2022
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Sequences and series, recurrence and convergence
Type	Applied recurrence modeling
Difficulty	Standard +0.8 This is a multi-part FP2 question involving probability modeling with recurrence relations and proof by induction. Part (a) requires basic understanding, part (b) requires a complete induction proof with algebraic manipulation of the recurrence relation, and part (c) requires limit analysis. While the induction itself is relatively standard, the context of deriving and proving a formula for a stochastic process elevates this above typical A-level questions. It's moderately challenging for Further Maths but not exceptionally difficult.
Spec	4.01a Mathematical induction: construct proofs

3. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{9516df6d-0e85-45d8-afb0-281c80450159-08_321_615_294_726} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} There are three lily pads on a pond. A frog hops repeatedly from one lily pad to another.
The frog starts on lily pad A, as shown in Figure 1.
In a model, the frog hops from its position on one lily pad to either of the other two lily pads with equal probability. Let $p _ { n }$ be the probability that the frog is on lily pad A after $n$ hops.

Explain, with reference to the model, why $p _ { 1 } = 0$ The probability $p _ { n }$ satisfies the recurrence relation $$p _ { n + 1 } = \frac { 1 } { 2 } \left( 1 - p _ { n } \right) \quad n \geqslant 1 \quad \text { where } p _ { 1 } = 0$$
Prove by induction that, for $n \geqslant 1$ $$p _ { n } = \frac { 2 } { 3 } \left( - \frac { 1 } { 2 } \right) ^ { n } + \frac { 1 } { 3 }$$
Use the result in part (b) to explain why, in the long term, the probability that the frog is on lily pad A is $\frac { 1 } { 3 }$

Show mark scheme Show mark scheme source

Question 3:

Part (a)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Implies moves from A: has to hop to different lily pad to A / will not hop on same pad / they can't stay on lily pad A / goes to B or C but not A / can't be where it started / can't stay on A	B1	Explains has to move from A
(1)

Part (b)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$p_1=\frac{2}{3}\left(-\frac{1}{2}\right)^1+\left(\frac{1}{3}\right)=0$; minimum needed $p_1=\frac{2}{3}\left(-\frac{1}{2}\right)+\left(\frac{1}{3}\right)=0$	B1	Shows statement true for $n=1$. Needs to show $p_1=0$ and conclusion true for $n=1$
Assume result is true for $n=k$: $p_k=\frac{2}{3}\left(-\frac{1}{2}\right)^k+\left(\frac{1}{3}\right)$	M1	Makes statement assuming result is true for $n=k$
Finds expression for $p_{k+1}=\frac{1}{2}\left(1-\left[\frac{2}{3}\left(-\frac{1}{2}\right)^k+\left(\frac{1}{3}\right)\right]\right)$	M1	Finds expression for $p_{k+1}$ using recurrence relation and substitutes in $p_k$
$p_{k+1}=\frac{1}{3}+\frac{2}{3}\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)^k$ or $\frac{1}{2}-\frac{1}{6}-\frac{2}{3}\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)^k$ $p_{k+1}=\frac{1}{3}-\frac{1}{3}\times-2\left(-\frac{1}{2}\right)^{k+1}$ or $\frac{1}{2}-\frac{1}{6}-\frac{1}{3}\times-2\left(-\frac{1}{2}\right)^{k+1}$	M1	Shows correct intermediate stage $p_{k+1}=\frac{1}{3}+\frac{2}{3}\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)^k$; condone $\frac{1}{3}-\frac{2}{3}\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)^k$
$p_{k+1}=\frac{2}{3}\left(-\frac{1}{2}\right)^{k+1}+\left(\frac{1}{3}\right)$	A1	$p_{k+1}=\frac{2}{3}\left(-\frac{1}{2}\right)^{k+1}+\left(\frac{1}{3}\right)$ cso
If true for $n=k$ then it is true for $n=k+1$ and as it is true for $n=1$, the statement is true for all $n$ (allow 'for all values')	A1	Correct complete conclusion, dependent on previous four marks. Must convey ideas of all underlined points
(6)

Part (c)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
As $n\to\infty,\ \left(-\frac{1}{2}\right)^n\to 0\ \therefore p_n\to\frac{1}{3}$	B1	See scheme
(1)
(8 marks)

## Question 3:

### Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Implies moves from A: has to hop to different lily pad to A / will not hop on same pad / they can't stay on lily pad A / goes to B or C but not A / can't be where it started / can't stay on A | B1 | Explains has to move from A |
| | **(1)** | |

### Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| $p_1=\frac{2}{3}\left(-\frac{1}{2}\right)^1+\left(\frac{1}{3}\right)=0$; minimum needed $p_1=\frac{2}{3}\left(-\frac{1}{2}\right)+\left(\frac{1}{3}\right)=0$ | B1 | Shows statement true for $n=1$. Needs to show $p_1=0$ and conclusion true for $n=1$ |
| Assume result is true for $n=k$: $p_k=\frac{2}{3}\left(-\frac{1}{2}\right)^k+\left(\frac{1}{3}\right)$ | M1 | Makes statement assuming result is true for $n=k$ |
| Finds expression for $p_{k+1}=\frac{1}{2}\left(1-\left[\frac{2}{3}\left(-\frac{1}{2}\right)^k+\left(\frac{1}{3}\right)\right]\right)$ | M1 | Finds expression for $p_{k+1}$ using recurrence relation and substitutes in $p_k$ |
| $p_{k+1}=\frac{1}{3}+\frac{2}{3}\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)^k$ or $\frac{1}{2}-\frac{1}{6}-\frac{2}{3}\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)^k$ $p_{k+1}=\frac{1}{3}-\frac{1}{3}\times-2\left(-\frac{1}{2}\right)^{k+1}$ or $\frac{1}{2}-\frac{1}{6}-\frac{1}{3}\times-2\left(-\frac{1}{2}\right)^{k+1}$ | M1 | Shows correct intermediate stage $p_{k+1}=\frac{1}{3}+\frac{2}{3}\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)^k$; condone $\frac{1}{3}-\frac{2}{3}\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)^k$ |
| $p_{k+1}=\frac{2}{3}\left(-\frac{1}{2}\right)^{k+1}+\left(\frac{1}{3}\right)$ | A1 | $p_{k+1}=\frac{2}{3}\left(-\frac{1}{2}\right)^{k+1}+\left(\frac{1}{3}\right)$ cso |
| If true for $n=k$ then it is true for $n=k+1$ and as it is true for $n=1$, the statement is true for all $n$ (allow 'for all values') | A1 | Correct complete conclusion, dependent on previous four marks. Must convey ideas of **all** underlined points |
| | **(6)** | |

### Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| As $n\to\infty,\ \left(-\frac{1}{2}\right)^n\to 0\ \therefore p_n\to\frac{1}{3}$ | B1 | See scheme |
| | **(1)** | |
| | **(8 marks)** | |

Show LaTeX source

3.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{9516df6d-0e85-45d8-afb0-281c80450159-08_321_615_294_726}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

There are three lily pads on a pond. A frog hops repeatedly from one lily pad to another.\\
The frog starts on lily pad A, as shown in Figure 1.\\
In a model, the frog hops from its position on one lily pad to either of the other two lily pads with equal probability.

Let $p _ { n }$ be the probability that the frog is on lily pad A after $n$ hops.
\begin{enumerate}[label=(\alph*)]
\item Explain, with reference to the model, why $p _ { 1 } = 0$

The probability $p _ { n }$ satisfies the recurrence relation

$$p _ { n + 1 } = \frac { 1 } { 2 } \left( 1 - p _ { n } \right) \quad n \geqslant 1 \quad \text { where } p _ { 1 } = 0$$
\item Prove by induction that, for $n \geqslant 1$

$$p _ { n } = \frac { 2 } { 3 } \left( - \frac { 1 } { 2 } \right) ^ { n } + \frac { 1 } { 3 }$$
\item Use the result in part (b) to explain why, in the long term, the probability that the frog is on lily pad A is $\frac { 1 } { 3 }$
\end{enumerate}

\hfill \mbox{\textit{Edexcel FP2 2022 Q3 [8]}}

This paper (10 questions)

View full paper

Q1 6 Q2 8 Q3 8 Q4 7 Q5 6 Q6 6 Q7 8 Q8 7 Q9 7 Q10 12