Construct the transition matrix based on the given probabilities.
Define states and transition probabilities.
Let state 1 represent the assembly line working correctly, and state 2 represent the assembly line not working correctly. The transition matrix \( P \) will have the following form:
\( P = \begin{array}{cc} \begin{bmatrix} P(1 \rightarrow 1) & P(1 \rightarrow 2) \\ P(2 \rightarrow 1) & P(2 \rightarrow 2) \end{bmatrix} \end{array} \)
- \( P(1 \rightarrow 1) \) is the probability that the line works correctly given it worked correctly the previous time, which is 0.6.
- \( P(1 \rightarrow 2) \) is the probability that the line does not work correctly given it worked correctly the previous time, which is \( 1 - 0.6 = 0.4 \).
- \( P(2 \rightarrow 1) \) is the probability that the line works correctly given it did not work correctly the previous time, which is 0.2.
- \( P(2 \rightarrow 2) \) is the probability that the line does not work correctly given it did not work correctly the previous time, which is \( 1 - 0.2 = 0.8 \).
Construct the transition matrix.
Therefore, the transition matrix is:
\( P = \begin{array}{cc} \begin{bmatrix} 0.6 & 0.4 \\ 0.2 & 0.8 \end{bmatrix} \end{array} \)
\(\boxed{[[0.6, 0.4], [0.2, 0.8]]}\)
Calculate the long-range probability that the assembly line will work correctly.
Set up the equations for the stationary distribution.
To find the long-range probability, we need to find the stationary distribution \( \pi = [\pi_1, \pi_2] \) such that \( \pi P = \pi \) and \( \pi_1 + \pi_2 = 1 \). This means:
\( \begin{bmatrix} \pi_1 & \pi_2 \end{bmatrix} \begin{bmatrix} 0.6 & 0.4 \\ 0.2 & 0.8 \end{bmatrix} = \begin{bmatrix} \pi_1 & \pi_2 \end{bmatrix} \)
This gives us the following equations:
- \( 0.6\pi_1 + 0.2\pi_2 = \pi_1 \)
- \( 0.4\pi_1 + 0.8\pi_2 = \pi_2 \)
- \( \pi_1 + \pi_2 = 1 \)
Solve the equations for the stationary distribution.
From the first equation:
\( 0.2\pi_2 = 0.4\pi_1 \)
\( \pi_2 = 2\pi_1 \)
Substituting this into the third equation:
\( \pi_1 + 2\pi_1 = 1 \)
\( 3\pi_1 = 1 \)
\( \pi_1 = \frac{1}{3} \)
Then, \( \pi_2 = 2 \cdot \frac{1}{3} = \frac{2}{3} \)
The long-range probability that the line will work correctly is \( \pi_1 = \frac{1}{3} \).
\(\boxed{\frac{1}{3}}\)
The transition matrix is:
\(\boxed{[[0.6, 0.4], [0.2, 0.8]]}\)
The long-range probability that the line will work correctly is:
\(\boxed{\frac{1}{3}}\)