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Chapter 4: Problem 25

Each morning an individual leaves his house and goes for a run. He is equally likely to leave either from his front or back door. Upon leaving the house, he chooses a pair of running shoes (or goes running barefoot if there are no shoes at the door from which he departed). On his return he is equally likely to enter, and leave his running shoes, either by the front or back door. If he owns a total of \(k\) pairs of running shoes, what proportion of the time does he run barefooted?

Short Answer

Expert verified
The individual never runs barefooted, as there must always be running shoes present at either the front door or the back door. The probability of running barefoot is 0.

Step by step solution

01

Represent the total number of running shoes and distribution of running shoes

Let k be the total number of running shoes. And let x be the number of running shoes present at the front door and y be the number of running shoes present at the back door. Therefore, \(k=x+y\) as all running shoes are distributed between the front and back doors.
02

Calculate the probability of selecting a door

Since the individual is equally likely to leave from the front or the back door, the probability of selecting a door can be denoted as: - Probability of choosing the front door: \(P(F) = \frac{1}{2}\) - Probability of choosing the back door: \(P(B) = \frac{1}{2}\)
03

Calculate the probabilities of finding no running shoes at the selected door

The probability of finding no running shoes at the front door with x running shoes is: \(P(\text{No Shoes} | F) = \frac{0}{x} = 0\) The probability of finding no running shoes at the back door with y running shoes is: \(P(\text{No Shoes} | B) = \frac{0}{y} = 0\)
04

Calculate the overall probability of running barefoot

We will now combine the probability of choosing a door with the probability of finding no running shoes at that chosen door. For this, we will use the total probability theorem: \(P(\text{No Shoes}) = P(\text{No Shoes} | F) \times P(F) + P(\text{No Shoes} | B) \times P(B)\) Plugging in the numbers, we get: \(P(\text{No Shoes}) = 0 \times \frac{1}{2} + 0 \times \frac{1}{2}\) \(P(\text{No Shoes}) = 0\) The answer is 0, meaning that he never runs barefooted, as there must always be running shoes present at either the front door or the back door.

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Most popular questions from this chapter

A group of \(n\) processors is arranged in an ordered list. When a job arrives, the first processor in line attempts it; if it is unsuccessful, then the next in line tries it; if it too is unsuccessful, then the next in line tries it, and so on. When the job is successfully processed or after all processors have been unsuccessful, the job leaves the system. At this point we are allowed to reorder the processors, and a new job appears. Suppose that we use the one- closer reordering rule, which moves the processor that was successful one closer to the front of the line by interchanging its position with the one in front of it. If all processors were unsuccessful (or if the processor in the first position was successful), then the ordering remains the same. Suppose that each time processor \(i\) attempts a job then, independently of anything else, it is successful with probability \(p_{i}\). (a) Define an appropriate Markov chain to analyze this model. (b) Show that this Markov chain is time reversible. (c) Find the long-run probabilities.

Specify the classes of the following Markov chains, and determine whether they are transient or recurrent: $$\mathbf{P}_{1}=\left\|\begin{array}{lll} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0 \end{array} \mid, \quad \mathbf{P}_{2}=\right\| \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \|$$ $$\mathbf{P}_{3}=\left\|\begin{array}{|ccccc|} \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ \frac{1}{4} & \frac{1}{2} & \frac{1}{4} & 0 & 0 \\ \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \end{array} \mid, \quad \mathbf{P}_{4}=\right\| \begin{array}{ccccc} \frac{1}{4} & \frac{3}{4} & 0 & 0 & 0 \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & \frac{1}{3} & \frac{2}{3} & 0 \\ 1 & 0 & 0 & 0 & 0 \end{array} \|$$

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?

\(M\) balls are initially distributed among \(m\) urns. At each stage one of the balls is selected at random, taken from whichever urn it is in, and then placed, at random, in one of the other \(M-1\) urns. Consider the Markov chain whose state at any time is the vector \(\left(n_{1}, \ldots, n_{m}\right)\) where \(n_{i}\) denotes the number of balls in urn \(i\). Guess at the limiting probabilities for this Markov chain and then verify your guess and show at the same time that the Markov chain is time reversible.

Consider a process \(\left\\{X_{n}, n=0,1, \ldots\right\\}\), which takes on the values 0,1 , or 2 . Suppose $$ \begin{aligned} &P\left\\{X_{n+1}=j \mid X_{n}=i, X_{n-1}=i_{n-1}, \ldots, X_{0}=i_{0}\right\\} \\ &\quad=\left\\{\begin{array}{ll} P_{i j}^{\mathrm{I}}, & \text { when } n \text { is even } \\ P_{i i}^{\mathrm{II}}, & \text { when } n \text { is odd } \end{array}\right. \end{aligned} $$ where \(\sum_{j=0}^{2} P_{i j}^{\mathrm{I}}=\sum_{j=0}^{2} P_{i j}^{\mathrm{II}}=1, i=0,1,2 .\) Is \(\left\\{X_{n}, n \geqslant 0\right\\}\) a Markov chain? If not, then show how, by enlarging the state space, we may transform it into a Markov chain.
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