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Chapter 4: Problem 24

Consider three urns, one colored red, one white, and one blue. The red urn contains 1 red and 4 blue balls; the white urn contains 3 white balls, 2 red balls, and 2 blue balls; the blue urn contains 4 white balls, 3 red balls, and 2 blue balls. At the initial stage, a ball is randomly selected from the red urn and then returned to that urn. At every subsequent stage, a ball is randomly selected from the urn whose color is the same as that of the ball previously selected and is then returned to that urn. In the long run, what proportion of the selected balls are red? What proportion are white? What proportion are blue?

Short Answer

Expert verified
In the long run, the proportion of selected balls will be approximately 22.9% red (\(\frac{27}{118}\)), 33.9% white (\(\frac{40}{118}\)), and 43.2% blue (\(\frac{51}{118}\)).

Step by step solution

01

Identify the Markov chain transition matrix

We need to find the transition matrix of the Markov chain. The transition matrix is a square matrix, where the entry at row i and column j represents the probability of going from state i to state j. In this problem, state i refers to choosing a ball from urn i. The transition matrix (P) will be of size 3x3, since there are 3 states (red, white, blue): \(P = \begin{pmatrix} P_{RR} & P_{RW} & P_{RB} \\ P_{WR} & P_{WW} & P_{WB} \\ P_{BR} & P_{BW} & P_{BB} \end{pmatrix}\) These probabilities are determined based on the contents of the urns.
02

Calculate the transition matrix elements

Calculate the probabilities for each state transition using the information from the problem: - \(P_{RR}\) (Red to Red): Probability of choosing a red ball from the red urn is 1 out of 5 balls, so it's 1/5 - \(P_{RW}\) (Red to White): There are no white balls in the red urn, so it's 0 - \(P_{RB}\) (Red to Blue): Probability of choosing a blue ball from the red urn is 4 out of 5 balls, so it's 4/5 - \(P_{WR}\) (White to Red): Probability of choosing a red ball from the white urn is 2 out of 7 balls, so it's 2/7 - \(P_{WW}\) (White to White): Probability of choosing a white ball from the white urn is 3 out of 7 balls, so it's 3/7 - \(P_{WB}\) (White to Blue): Probability of choosing a blue ball from the white urn is 2 out of 7 balls, so it's 2/7 - \(P_{BR}\) (Blue to Red): Probability of choosing a red ball from the blue urn is 3 out of 9 balls, so it's 1/3 - \(P_{BW}\) (Blue to White): Probability of choosing a white ball from the blue urn is 4 out of 9 balls, so it's 4/9 - \(P_{BB}\) (Blue to Blue): Probability of choosing a blue ball from the blue urn is 2 out of 9 balls, so it's 2/9 Putting these values in the transition matrix, we get: \(P = \begin{pmatrix} 1/5 & 0 & 4/5 \\ 2/7 & 3/7 & 2/7 \\ 1/3 & 4/9 & 2/9 \end{pmatrix}\)
03

Find the stationary distribution of the Markov chain

Now, we need to find the stationary distribution, which means solving the following equation for 蟺: \(\pi P = \pi\) Let 蟺 = \(( \pi_R, \pi_W, \pi_B)\), then the equation becomes: \( (\pi_R, \pi_W, \pi_B) \begin{pmatrix} 1/5 & 0 & 4/5 \\ 2/7 & 3/7 & 2/7 \\ 1/3 & 4/9 & 2/9 \end{pmatrix} = ( \pi_R, \pi_W, \pi_B)\) subject to the condition \(\pi_R + \pi_W + \pi_B = 1\) Solving this system of linear equations, we get: \(\pi_R = \frac{27}{118}\) \(\pi_W = \frac{40}{118}\) \(\pi_B = \frac{51}{118}\)
04

Proportion of selected balls

From the stationary distribution, we have the long-run proportion of selected balls for each color: - Red: \(\pi_R = \frac{27}{118} \approx 0.229\) - White: \(\pi_W = \frac{40}{118} \approx 0.339\) - Blue: \(\pi_B = \frac{51}{118} \approx 0.432\) In the long run, approximately 22.9% of the selected balls will be red, 33.9% white, and 43.2% blue.

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