Math help probability

Question

An auto insurance company classifies its customers in three categories: poor, satisfactory, and preferred. Each year, 20% of those in the poor category are moved to satisfactory and 5% of those in the satisfactory category are moved to preferred. Also, 5% in the preferred category are moved to the satisfactory category, and 5% of those in the satisfactory category are moved to the poor category. Customers are never moved from poor to preferred, or conversely, in a single year. Assuming these percentages remain valid over a long period of time, how many customers can the company expect to have in each category in the long run?


Poor= % (round to the nearest tenth if necessary.)


Satisfactory= % (round to the nearest tenth if necessary.)


Preferred= % (round to the nearest tenth if necessary.)

Answer 1

This is an example of a Markov chain problem. Let x be the number of preferred drivers, let y be the number of satisfactory drivers, and let z be the number of poor drivers. Then we can set up a matrix and a vector this way.

Matrix A              vector u

0.95, 0.05, 0.00   x

0.05, 0.90, 0.20   y

0.00, 0.05, 0.80   z

 

After a year the vector of drivers will be A*u = u₂. After 2 years the vector of drivers will be A²*u = u_3, and so on. Since A is a Markov matrix, there exists a vector u such that A*u = u. Translating that fact in to a system of equations gives us this:

 

0.95x + 0.05y = x

0.05x + 0.90y + 0.20z = y

0.05y + 0.80z = z

 

Solving the first line for x gives us x = y

Substituting y for x in the second line gives us y = 4z. Letting z = 1 gives us the vector (4, 4, 1) or normalized (4/9, 4/9, 1/9). Given any distribution of drivers at time 0, the future distribution of drivers will approach the eigenvector (4/9, 4/9, 1/9). 

 

So after a long period of time 4/9 of the drivers will be preferred, 4/9 will be standard, and 1/9 will be poor.