IB Mathematics AHL 4.19: Transition matrices AI HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
From the problem: If charged previous day, probability charges today is \(0.4\), so \(a = 0.4\).
Since rows sum to 1, \(b = 1 – 0.4 = 0.6\).
(i) \(\boxed{0.4}\)
(ii) \(\boxed{0.6}\)
(b)(i)
Initial state: charged on day 0 ⇒ \(v_0 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\).
To charge on all days 1 to 4 means sequence: C → C → C → C → C.
Probability = \(0.4^4 = 0.0256\).
\(\boxed{0.0256}\)
(b)(ii)
With \(p = 0.7\), \(M = \begin{pmatrix} 0.4 & 0.7 \\ 0.6 & 0.3 \end{pmatrix}\).
We need the first entry of \(v_4 = M^4 v_0\).
Compute \(M^4\) (using calculator):
\(M^4 \approx \begin{pmatrix} 0.5422 & 0.5341 \\ 0.4578 & 0.4659 \end{pmatrix}\).
Then \(v_4 = M^4 \begin{pmatrix} 1 \\ 0 \end{pmatrix} \approx \begin{pmatrix} 0.5422 \\ 0.4578 \end{pmatrix}\).
Probability charges on day 4 ≈ 0.542.
\(\boxed{0.542}\)
(c)
Let \(v = \begin{pmatrix} 1 \\ -1 \end{pmatrix}\).
Compute \(Mv = \begin{pmatrix} 0.4 & p \\ 0.6 & 1-p \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 0.4 – p \\ 0.6 – (1-p) \end{pmatrix} = \begin{pmatrix} 0.4 – p \\ p – 0.4 \end{pmatrix}\).
This equals \((0.4 – p) \begin{pmatrix} 1 \\ -1 \end{pmatrix}\).
Thus \(v\) is an eigenvector with eigenvalue \(\lambda = 0.4 – p\).
\(\boxed{\text{Eigenvalue } = 0.4 – p}\)
(d)
Steady state vector \(s = \begin{pmatrix} x \\ y \end{pmatrix}\) satisfies \(Ms = s\) and \(x + y = 1\).
From \(Ms = s\):
\(0.4x + py = x\) ⇒ \(py = 0.6x\) ⇒ \(y = \frac{0.6x}{p}\).
Using \(x + y = 1\):
\(x + \frac{0.6x}{p} = 1\) ⇒ \(x\left(1 + \frac{0.6}{p}\right) = 1\) ⇒ \(x = \frac{p}{p + 0.6}\).
\(\boxed{\frac{p}{p + 0.6}}\)
(e)
We require steady state probability \(\frac{p}{p + 0.6} \geq 0.6\).
\(\frac{p}{p + 0.6} \geq 0.6\) ⇒ \(p \geq 0.6(p + 0.6)\) ⇒ \(p \geq 0.6p + 0.36\) ⇒ \(0.4p \geq 0.36\) ⇒ \(p \geq 0.9\).
\(\boxed{p = 0.9}\)