Explanation Given: The mice moves from A to B with a probability 0.3 and to C with a probability 0.4. The mice in compartment B moves to A and C with respective probabilities of 0.15 and 0.55. The mice in compartment C moves to A and B with respective probabilities 0.3 and 0.6, respectively. Explanation: The probabilites for the mice shifting from one compartment to another can be represented with the help of a transition matrix. Three compartments would correspond to a square transition matrix of order 3. Use the provided informtaion to set up the transition matrix as P . A B C P = A B C [ a 11 0.3 0.4 0.15 a 22 0.55 0.3 0.6 a 33 ] As the sum of all the entries of the transition matrix need to be 1, the unknown diagonal elements can be calcuated as a 11 + 0.3 + 0.4 = 1 a 11 + 0.7 = 1 a 11 = 1 − 0.7 = 0.3 0.15 + a 22 + 0.55 = 1 a 22 = 1 − 0.15 − 0.55 = 1 − 0.7 = 0.3 0.3 + 0.6 + a 33 = 1 a 33 = 1 − 0.3 − 0.6 = 1 − 0.9 = 0.1 This results in the following transition matrix A B C P = A B C [ 0.3 0.3 0.4 0.15 0.3 0.55 0.3 0.6 0.1 ] The long-range probabilities for the mice to be in each compartment can be computed as the equilibrium vector for the transition matrix P . A vector v is said to be an equilibrium vector for a transition matrix P , if the following condition holds: v P = v Let the vector v be defined as follows, so that the multiplication holds. [ v 1 v 2 v 3 ] Then, v P = v [ v 1 v 2 v 3 ] [ 0.3 0.3 0.4 0.15 0.3 0.55 0.3 0.6 0.1 ] = [ v 1 v 2 v 3 ] Simplify the left hand side to obtain [ 0.3 v 1 + 0.15 v 2 + 0.3 v 3 0.3 v 1 + 0.3 v 2 + 0.6 v 3 0.4 v 1 + 0.55 v 2 + 0.1 v 3 ] = [ v 1 v 2 v 3 ] This gives ride to the following system of equations 0.3 v 1 + 0.15 v 2 + 0.3 v 3 = v 1 0.3 v 1 + 0.3 v 2 + 0.6 v 3 = v 2 0.4 v 1 + 0.55 v 2 + 0.1 v 3 = v 3 This can be simplified to − 0.7 v 1 + 0.15 v 2 + 0.3 v 3 = 0 0.3 v 1 − 0.7 v 2 + 0.6 v 3 = 0 0.4 v 1 + 0.55 v 2 − 0.9 v 3 = 0 Also, use the property that v 1 , v 2 , and v 3 are entries of a probability vector so they should sum to 1.This gives v 1 + v 2 + v 3 = 1 The system involving these 4 equations can be solved with the help of Gauss Jordan method. The augmented matrix for the system is [ − 0.7 0.15 0.3 0 0.3 − 0.7 0.6 0 0.4 0.55 − 0.9 0 1 1 1 1 ] − 1 0

11th Edition

ISBN: 9780321979438

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 10.2, Problem 32E

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Chapter 10 Solutions

Finite Mathematics (11th Edition)

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The long-range probability of the mice being present in each compartment.

Solution Summary: The author explains how the long-range probability of mice moving from one compartment to another can be represented with the help of a transition matrix.

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