A2 June 2022 Q3

EdexcelCurrent spec8 marksRecurrence Relations

Figure 1: a frog on lily pad A, with lily pads B and C below — Figure 1

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.

(a) Explain, with reference to the model, why \(p_1 = 0\) (1)

The probability \(p_n\) satisfies the recurrence relation

\[p_{n+1} = \frac{1}{2}\left(1 - p_n\right) \qquad n \geqslant 1 \quad \text{where } p_1 = 0\]

(b) Prove by induction that, for \(n \geqslant 1\)\[p_n = \frac{2}{3}\left(-\frac{1}{2}\right)^n + \frac{1}{3}\] (6)

(c) Use the result in part (b) to explain why, in the long term, the probability that the frog is on lily pad A is \(\dfrac{1}{3}\) (1)

Mark scheme (a)

Scheme	Marks	AO
Implies moves from A Has to hop to different lily pad to A Will not hop on the 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	2.4
	(1)

Notes

B1: Explains has to move from A

Mark scheme (b)

Scheme	Marks	AO
\(p_1 = \frac{2}{3}\left(-\frac{1}{2}\right)^1 + \left(\frac{1}{3}\right) = 0\) Minimum needed \(p_1 = \frac{2}{3}\left(-\frac{1}{2}\right) + \left(\frac{1}{3}\right) = 0\)	B1	2.1
Assume result is true for \(n = k\) Or \(p_k = \frac{2}{3}\left(-\frac{1}{2}\right)^k + \left(\frac{1}{3}\right)\)	M1	2.4
Finds an 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	1.1b
\(p_{k+1} = \dfrac{1}{3} + \dfrac{2}{3}\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)^k\) or \(\dfrac{1}{2} - \dfrac{1}{6} + \dfrac{2}{3}\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)^k\) \(p_{k+1} = \dfrac{1}{3} - \dfrac{1}{3} \times -2\left(-\frac{1}{2}\right)^{k+1}\) or \(\dfrac{1}{2} - \dfrac{1}{6} - \dfrac{1}{3} \times -2\left(-\frac{1}{2}\right)^{k+1}\)	M1	2.1
\(p_{k+1} = \frac{2}{3}\left(-\frac{1}{2}\right)^{k+1} + \left(\frac{1}{3}\right)\)	A1	1.1b
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	2.2a
	(6)

Notes

B1: Shows the statement is true for \(n = 1\). Needs to show that \(p_1 = 0\) and conclusion true for \(n = 1\), this statement can be recovered in their conclusion if says e.g. true for \(n = 1\)

M1: Makes a statement that assumes the result is true for \(n = k\). Assume (true for) \(n = k\) is sufficient. This mark may be recovered in their conclusion if they say e.g. if true for \(n = k\) then …etc

M1: Finds an expression for \(p_{k+1}\) using the recurrence relation and substitutes in for \(p_k\)

M1: Shows a correct intermediate stage \(p_{k+1} = \frac{1}{3} + \frac{2}{3}\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)^k\)

Condone \(\frac{1}{3} - \frac{2}{3}\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)^k\)

A1: \(p_{k+1} = \frac{2}{3}\left(-\frac{1}{2}\right)^{k+1} + \left(\frac{1}{3}\right)\) cso

A1: Correct complete conclusion. This mark is dependent on previous four marks, the B mark is not required. It is gained by conveying the ideas of all underlined points either at the end of their solution or as a narrative in their solution.

Note Using \(p_{k+1} = \frac{1}{2} - \frac{1}{2}p_k = \frac{1}{2} - \frac{1}{2}\left[\frac{2}{3}\left(-\frac{1}{2}\right)^k + \left(\frac{1}{3}\right)\right] = \frac{1}{2} + \frac{2}{3}\left(-\frac{1}{2}\right)^{k+1} - \left(\frac{1}{6}\right)\) scores 2^nd M1 3^rd M1

Scheme	Marks	AO
As \(n \to \infty,\ \left(-\frac{1}{2}\right)^n \to 0 \therefore p_n \to \dfrac{1}{3}\)	B1	3.4
	(1)
	(8 marks)