Math Problem Statement
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Part 1
A large group of mice is kept in a cage having compartments A, B, and C. Mice in compartment A move to B with probability
0.30.3
and to C with probability
0.50.5.
Mice in B move to A or C with probabilities
0.350.35
and
0.450.45,
respectively. Mice in C move to A or B with probabilities
0.30.3
and
0.30.3,
respectively. Find the long-range prediction for the fraction of mice in each of the compartments.
Question content area bottom
Part 1
The long-range prediction for the fraction of mice is
enter your response here
in compartment A,
enter your response here
in compartment B, and
enter your response here
in compartment C.
(Type integers or simplified fractions.)
Solution
This problem involves finding the steady-state distribution for a Markov chain, where the probabilities represent transitions between states (compartments A, B, and C). Let us solve this step by step:
Transition Matrix
The transition probabilities can be written as a matrix :
\begin{bmatrix} 0 & 0.35 & 0.3 \\ 0.3 & 0 & 0.3 \\ 0.5 & 0.45 & 0.4 \end{bmatrix}$$ Here: - Row $$i$$ corresponds to the compartment the mice are **leaving**. - Column $$j$$ corresponds to the compartment the mice are **entering**. The goal is to find the **steady-state vector** $$\pi = [\pi_A, \pi_B, \pi_C]$$, such that: $$\pi P = \pi \quad \text{and} \quad \pi_A + \pi_B + \pi_C = 1.$$ --- ### Solving the System The equation $$\pi P = \pi$$ translates to: $$\begin{aligned} \pi_A & = 0.35 \pi_B + 0.3 \pi_C, \\ \pi_B & = 0.3 \pi_A + 0.45 \pi_C, \\ \pi_C & = 0.5 \pi_A + 0.3 \pi_B.$$ Additionally, we know: $$\pi_A + \pi_B + \pi_C = 1.$$ This is a system of linear equations. Let’s solve this mathematically. --- ### Calculations I will now solve this system for $$\pi_A, \pi_B, \pi_C$$. numpy as np # Coefficient matrix for the system A = np.array([ [1, -0.35, -0.3], # From π_A = 0.35π_B + 0.3π_C [-0.3, 1, -0.45], # From π_B = 0.3π_A + 0.45π_C [-0.5, -0.3, 1], # From π_C = 0.5π_A + 0.3π_B [1, 1, 1] # Normalization condition: π_A + π_B + π_C = 1 ]) # Right-hand side vector b = np.array([0, 0, 0, 1]) # Solve the system using least squares to handle normalization steady_state = np.linalg.lstsq(A, b, rcond=None)[0] # Format results steady_stateThe steady-state distribution (long-range prediction) for the fraction of mice in each compartment is: - Compartment A: $$\pi_A = \frac{30861743}{100000000} \approx 0.3086$$ - Compartment B: $$\pi_B = \frac{32761961}{100000000} \approx 0.3276$$ - Compartment C: $$\pi_C = \frac{34037435}{100000000} \approx 0.3404$$ Thus, the mice will distribute approximately 30.86% in $$A$$, 32.76% in $$B$$, and 34.04% in $$C$$ in the long run. Would you like a detailed explanation of the calculations or have other questions? Here are some related queries you may explore: 1. What is the interpretation of a steady-state distribution? 2. How can transition matrices represent real-world systems beyond biology? 3. Can we simulate this Markov chain to verify the steady-state distribution? 4. What happens if the sum of transition probabilities in a row isn't 1? 5. How do we handle larger systems with more compartments? **Tip:** Steady-state distributions are independent of the initial state, as long as the Markov chain is irreducible and aperiodic!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability
Markov Chains
Steady-State Distribution
Linear Algebra
Formulas
Steady-state condition: πP = π
Normalization condition: π_A + π_B + π_C = 1
Theorems
Markov Chain Steady-State Theorem
Suitable Grade Level
Grades 10-12, College Level
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