EXM.1.AHL.TZ0.17

pestleMathematicsAIHLPaper 1EXM· ahl-4-19-transition-matrices-markov-chainssource ↗

Sue sometimes goes out for lunch. If she goes out for lunch on a particular day then the probability that she will go out for lunch on the following day is 0.4. If she does not go out for lunch on a particular day then the probability she will go out for lunch on the following day is 0.3.

Write down the transition matrix for this Markov chain.

[2]

We know that she went out for lunch on a particular Sunday, find the probability that she went out for lunch on the following Tuesday.

[2]

Find the steady state probability vector for this Markov chain.

[3]

Markscheme / solution

$\left( {\begin{array}{*{20}{c}} {0.4}&{0.3} \\ {0.6}&{0.7} \end{array}} \right)$ M1A1

[2 marks]

${\left( {\begin{array}{*{20}{c}} {0.4}&{0.3} \\ {0.6}&{0.7} \end{array}} \right)^2}\left( {\begin{array}{*{20}{c}} 1 \\ 0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {0.34} \\ {0.66} \end{array}} \right)$ M1

So probability is 0.34 A1

[2 marks]

$\left( {\begin{array}{*{20}{c}} {0.4}&{0.3} \\ {0.6}&{0.7} \end{array}} \right)\left( {\begin{array}{*{20}{c}} p \\ {1 - p} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} p \\ {1 - p} \end{array}} \right) \Rightarrow 0.4p + 0.3\left( {1 - p} \right) = p \Rightarrow p = \frac{1}{3}$ M1A1

So vector is $\left( {\begin{array}{*{20}{c}} {\tfrac{1}{3}} \\ {\tfrac{2}{3}} \end{array}} \right)$ A1

[or by investigating high powers of the transition matrix]

[3 marks]

Examiners’ report

[N/A]