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Date May Example question Marks available 2 Reference code EXM.1.AHL.TZ0.17
Level Additional Higher Level Paper Paper 1 Time zone Time zone 0
Command term Find Question number 17 Adapted from N/A

Question

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]
a.

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]
b.

Find the steady state probability vector for this Markov chain.

[3]
c.

Markscheme

( 0.4 0.3 0.6 0.7 )       M1A1

[2 marks]

a.

( 0.4 0.3 0.6 0.7 ) 2 ( 1 0 ) = ( 0.34 0.66 )      M1

So probability is 0.34       A1

[2 marks]

b.

( 0.4 0.3 0.6 0.7 ) ( p 1 p ) = ( p 1 p ) 0.4 p + 0.3 ( 1 p ) = p p = 1 3     M1A1

So vector is  ( 1 3 2 3 )        A1

[or by investigating high powers of the transition matrix]

[3 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 4—Statistics and probability » AHL 4.19—Transition matrices – Markov chains
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Topic 4—Statistics and probability

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