IB Maths AI HL · Unit 4: Statistics and Probability
IB Maths AI HL Markov Chains and Transition Matrices Questions
Exam-style IB Maths AI HL markov chains and transition matrices questions with worked solutions. Start with the 3 fully worked examples below — each is solved step by step the way an IB examiner expects — then try the practice questions and check your working against the mark scheme.
- 13 questions
- Paper 1: 13
- 4 easy
- 4 medium
- 5 hard
- 3 worked examples
Practise Markov Chains and Transition Matrices questions →
AI HL formula booklet
What you need to know
Weather forecasting, consumer brand-switching, disease spread. HL AI's Markov questions ask for the steady-state distribution — the left eigenvector of the transition matrix. Markov chains and transition matrices overview →
What's examined in AI HL markov chains and transition matrices
The question bank covers these markov chains and transition matrices question types (number of questions in brackets):
- Steady State Analysis (7)
- Future State Probabilities (5)
- Constructing Transition Matrices (1)
Key formulas
- Transition matrix (Markov chain)
- \(\vec{s}_{n+1} = T\,\vec{s}_n\)
In the same notation as the IB formula booklet. All AI HL formulas →
Markov Chains and Transition Matrices worked examples
Worked example 1: Constructing a Transition Matrix · easy
The transition between two states, A and B, is represented by a directed network. The probability of moving from A to B is $0.4$, and A to A is $0.6$. The probability of moving from B to A is $0.15$, and B to B is $0.85$. Write down the $2 \times 2$ transition matrix $T$ for this Markov chain, ensuring the states are mapped alphabetically.
1. Recall the structure of a transition matrix: columns represent the 'initial state' (From) and rows represent the 'final state' (To).
2. Construct the first column (From A): Top row (To A) is $0.6$, bottom row (To B) is $0.4$. Note that $0.6 + 0.4 = 1$.
3. Construct the second column (From B): Top row (To A) is $0.15$, bottom row (To B) is $0.85$. Note that $0.15 + 0.85 = 1$.
4. Write the final matrix: $T = \begin{pmatrix} 0.6 & 0.15 \\ 0.4 & 0.85 \end{pmatrix}$.
Examiner tip: In the IB AI syllabus, transition matrices are set up so that the columns sum to $1$. If you set it up such that the rows sum to $1$, you will get incorrect state vectors upon multiplication.
Worked example 2: Finding the State Vector after $n$ Transitions · medium
Using the transition matrix $T = \begin{pmatrix} 0.6 & 0.15 \\ 0.4 & 0.85 \end{pmatrix}$, assume the system initially starts in state A with $100\%$ certainty. Find the probability that the system is in state B after exactly 2 transitions.
1. Define the initial state vector $S_0$. Since the system is definitively in state A, $S_0 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$.
2. Apply the transition rule for $n$ steps: $S_n = T^n S_0$.
3. Set up the calculation for $S_2$: $\begin{pmatrix} 0.6 & 0.15 \\ 0.4 & 0.85 \end{pmatrix}^2 \begin{pmatrix} 1 \\ 0 \end{pmatrix}$.
4. Compute $T^2$ using the GDC: $\begin{pmatrix} 0.42 & 0.2175 \\ 0.58 & 0.7825 \end{pmatrix}$.
5. Multiply $T^2$ by $S_0$: $\begin{pmatrix} 0.42 & 0.2175 \\ 0.58 & 0.7825 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0.42 \\ 0.58 \end{pmatrix}$.
6. The bottom element represents state B. The probability is $0.58$.
Examiner tip: You can simply look at the first column of $T^2$ to find the probabilities after 2 steps originating entirely from State A. The matrix power $T^n$ contains all paths of length $n$.
Worked example 3: Calculating Steady State Probabilities · hard
For the transition matrix $T = \begin{pmatrix} 0.6 & 0.15 \\ 0.4 & 0.85 \end{pmatrix}$, find the exact steady state probability vector $\mathbf{s} = \begin{pmatrix} x \\ y \end{pmatrix}$.
1. State the steady state condition: multiplying the transition matrix by the steady state vector yields the exact same vector ($T\mathbf{s} = \mathbf{s}$).
2. Set up the matrix equation: $\begin{pmatrix} 0.6 & 0.15 \\ 0.4 & 0.85 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$.
3. Extract the top row into a linear equation: $0.6x + 0.15y = x$.
4. Simplify: $0.15y = 0.4x \implies 15y = 40x \implies 3y = 8x$.
5. Apply the probabilistic constraint that the states must sum to $1$: $x + y = 1 \implies y = 1 - x$.
6. Substitute and solve: $3(1 - x) = 8x \implies 3 - 3x = 8x \implies 11x = 3 \implies x = \frac{3}{11}$.
7. Consequently, $y = \frac{8}{11}$. The exact steady state vector is $\begin{pmatrix} 3/11 \\ 8/11 \end{pmatrix}$.
Examiner tip: If you try to solve the system using both rows of the matrix equation, you will get infinite solutions because the equations are linearly dependent. You must always swap out one row for the sum constraint $x+y=1$.
Try these IB Maths AI HL markov chains and transition matrices questions
Three questions from the bank, easiest first. Mark schemes and AI marking of your written working are in the practice area.
Question 1 · easy · 4 marks · Paper 1
A system is modelled by the transition matrix \(T = \begin{pmatrix} 0.7 & 0.2 \\ 0.3 & 0.8 \end{pmatrix}\).
The initial state of the system is given by the column vector \(s_0 = \begin{pmatrix} 100 \\ 50 \end{pmatrix}\).
Calculate the exact state vector after one transition, \(s_1\).
Question 2 · medium · 5 marks · Paper 1
A car rental company operates in two cities, City X and City Y.
- A car rented in City X has a \(70\%\) chance of being returned to City X, and a \(30\%\) chance of being returned to City Y.
- A car rented in City Y has a \(90\%\) chance of being returned to City Y, and a \(10\%\) chance of being returned to City X.
Currently, there are \(400\) cars in City X and \(200\) cars in City Y.
Using your GDC, find the number of cars expected to be in City X after 5 rental cycles (transitions). Round your answer to the nearest whole car.
Question 3 · hard · 7 marks · Paper 1
Consider the transition matrix \(T = \begin{pmatrix} 0.8 & 0.4 \\ 0.2 & 0.6 \end{pmatrix}\).
In linear algebra, a scalar \(\lambda\) is an eigenvalue of a matrix \(T\) if \(\det(T - \lambda I) = 0\), where \(I\) is the identity matrix.
Prove algebraically that \(\lambda = 1\) is an eigenvalue of \(T\), and find the other eigenvalue.
All 13 markov chains and transition matrices questions with mark schemes →
FAQ
How many IB Maths AI HL markov chains and transition matrices questions are there?
There are 13 exam-style markov chains and transition matrices questions in the AI HL question bank (Paper 1: 13), graded 4 easy, 4 medium, 5 hard. Every question has a full IB-style mark scheme (M, A and R marks).
Is markov chains and transition matrices on Paper 1 or Paper 2?
In the question bank these questions are set as Paper 1 questions.
Where can I get the mark schemes?
Open the AI HL Unit 4 practice page: every question has a step-by-step IB-style mark scheme, and you can photograph your working for instant AI marking. The worked examples on this page are free.
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