Question

Three white balls and three black balls are distributed in two urns in such a way that each contain three balls. We say that the system is in state i, i=0,1,2,3, if the first urn contains i white balls. At each step, we draw one ball from each urn and place the ball drawn from the first urn into the second, and conversely with the ball from the second urn. Let Xn denote the state of the system after the nth step. Explain why {Xn, n=0 ,1, 2} is a Markov chain and calculate its transition probability matrix.

Answer 1

Answer:

Following are the answer to this question:

Step-by-step explanation:

In-state 0, it has urns of the 100% chance of shifting towards state 1 because a colored ball must be substituted.  

In the $state 1: \frac{1 \ white}{2 \ black} \ \ \ and \ \ \frac{2 \ white}{1 \ black}$

The probability to select White from both the probability to select Black from both are 2/9, therefore there are 4/9 possibilities to remain in State 1.  

It is probable which white from both the beginning is selected and black from the second, so that 1/9 probability of 0.  

The probability is 4/9 that the first black and the second white will be chosen and 4/9 possibility will be made to state 2.

In the $state 2: \frac{2 \ white}{1 \ black} \ \ \ and \ \ \frac{1 \ white}{2 \ black}$

This is essentially a state 1 mirror image because identical claims are used for reverse colors.  

In-State 3, the urns are 100% likely to revert to State 2.  

It is the representation of matrix M is, therefore:

$( ..0. ..1. ..0. ..0. )\\\\( \frac{1}{9} \frac{4}{9} \frac{4}{9}.. 0. )\\\\( ..0. \ \frac{4}{9} \ \frac{4}{9} \ \frac{1}{9})\\\\( ..0. ..0. ..1. ..0. )\\\\So, \\ X_n = M \times X_{n}-1 \\\\Or\\X_n = M^n \times X_0$