27 exam-style questions on Transition matrices; regular Markov chains; steady state — topic 4.19 of Statistics and Probability in the IB Mathematics: Applications and Interpretation syllabus — with answers and explanations. Samples below; the full set is free with an account.
A transition matrix T for a Markov chain has columns representing the current state. Which property must every column of T satisfy?
Answer: Its entries sum to 1
In a (column) stochastic transition matrix, each column gives the probabilities of moving from one state to all states, so its entries are non-negative and sum to 1.
A Markov chain has transition matrix T = [[0.8, 0.3], [0.2, 0.7]] with initial state vector s0 = [1, 0] (column vector). Calculate the state vector after one step, s1 = T·s0.
Answer: [0.8, 0.2]
T·[1,0]ᵀ selects the first column of T, giving [0.8, 0.2]ᵀ.
A regular Markov chain has transition matrix T = [[0.6, 0.5], [0.4, 0.5]]. Determine the steady-state (long-run) probability vector.
Answer: [5/9, 4/9]
Solving 0.6a + 0.5b = a with a + b = 1 gives 0.4a = 0.5b, so a = 5/9 and b = 4/9. Steady state ≈ [0.556, 0.444].
The long-run (steady-state) vector s of a regular Markov chain satisfies which equation?
Answer: Ts = s
The steady state is unchanged by another transition, so Ts = s; it is the eigenvector of T with eigenvalue 1, scaled so its entries sum to 1.
For a Markov chain with transition matrix T and initial state vector s0, the state vector after n steps is given by:
Answer: sn = Tⁿ·s0
Each step multiplies by T, so after n steps the state is sn = Tⁿ·s0.
A weather model treats each day as Sunny (S) or Rainy (R). P(S tomorrow | S today) = 0.9 and P(S tomorrow | R today) = 0.6. Determine the long-run proportion of Sunny days, correct to 2 decimal places.
Answer: 0.86
Steady state: s = 0.9s + 0.6(1 − s) ⇒ s = 0.6 + 0.3s ⇒ 0.7s = 0.6 ⇒ s = 6/7 ≈ 0.857 ≈ 0.86.
That was 6 of 27 Transition matrices; regular Markov chains; steady state questions.
Practise all 27 freeExplainAO3 · 1–3 marks
Give reasons or causes: say what happens AND why, linking each step with 'because' or 'so that'.
DescribeAO1 · 1–2 marks
Give a detailed account of what it is or how it happens — steps, features, pattern — without saying why.
StateAO1 · 1–2 marks
One word or one short sentence. No explanation, no working: the mark is for the fact itself.
27 Transition matrices; regular Markov chains; steady state questions — free with an account. Spaced repetition, streaks and full exam simulations included.
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