# Markov Chains - A small town has only two dry cleaners, Johnson and NorthClean

215 points

A small town has only two dry cleaners, Johnson and NorthClean. Each customer brings in one batch of clothes per week to one of these cleaners. The following transition matrix shows the probabilities that a customer initially using a particular dry cleaner will use a particular dry cleaner the next week.

$\begin{bmatrix}0.2 & 0.8\\0.65 & 0.35\end{bmatrix}$

Row 1: Johnson

Row 2: NorthClean

Col 1: Johnson

Col 2: NorthClean

Answer parts (a) through (e) based on this transition matrix.

(a) Find the probability that a customer initially bringing a batch of clothes to Johnson also brings a batch of clothes to Johnson aﬁer 1 week.

(b) Find the probability that a customer initially bringing a batch of clothes to Johnson also brings a batch of clothes to Johnson after 2 weeks.

(c) Find the probability that a customer initially bringing a batch of clothes to Johnson also brings a batch of clothes to Johnson aﬁer 3 weeks.

(d) Find the probability that a customer initially bringing a batch of clothes to Johnson also brings a batch of clothes to Johnson after 4 weeks.

(e) Find the probability that a customer initially bringing a batch of clothes to NorthClean brings a batch of clothes to Johnson after 2 weeks.

Markov Chains
homework

100 points

Crap

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9.9k points

part (a) In this question rows represent which cleaner was used the first week, and columns represent the second week. Since we are interested in finding the probability that a person who uses Johnson the first week also uses it the second week we need to examine the transition matrix,

$$A=\begin{bmatrix}0.2 & 0.8\\0.65 & 0.35\end{bmatrix}$$

We are looking at row 1 and column 1 which is 0.2

for part (b) we need to find $A^2$ by matrix multiplication $AA=A^2$

$$\begin{bmatrix}0.2 & 0.8\\0.65 & 0.35\end{bmatrix} \begin{bmatrix}0.2 & 0.8\\0.65 & 0.35\end{bmatrix}$$

$$= \begin{bmatrix}0.56 & 0.44\\0.3575 & 0.6425\end{bmatrix}$$

again the probability that a customer who uses Johnson initially will return and use them on the second week means we examine row 1 and column 1, so the probability is 0.56

for parts (c) and (d) compute $A^3$ and $A^4$ respectively and proceed in the same manner as in parts (a) and (b).

for part (e) again use $A^2$ which was already computed, but this time examine row 2 and column 1 which is 0.65

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