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Chapter 4: Problem 17

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) Show that the limiting probabilities of the reversed Markov chain are the same as for the forward chain by showing that they satisfy the equations $$ \pi_{j}=\sum_{i} \pi_{i} Q_{u} $$ (b) Give an intuitive explanation for the result of part (a).

Consider a Markov chain with states \(0,1,2,3,4\). Suppose \(P_{0,4}=1\); and suppose that when the chain is in state \(l, i>0\), the next state is equally likely to be any of the states \(0,1, \ldots, i-1\). Find the limiting probabilities of this Markov chain.

For a series of dependent trials the probability of success on any trial is \((k+1) /(k+2)\) where \(k\) is equal to the number of successes on the previous two trials. Compute \(\lim _{n \rightarrow \infty} P\) success on the \(n\) th trialj.

Prove that if the number of states in a Markov chain is \(M\), and if state \(J\) can be reached from state \(i\), then it can be reached in \(M\) steps or less.

.Each day, one of \(n\) possible elements is requested, the \(i\) th one with probability \(P_{i}, i \geq 1, \Sigma_{1}^{n} P_{i}=1 .\) These elements are at all times arranged in an ordered list which is revised as follows: The element selected is moved to the front of the list with the relative positions of all the other elements remaining unchanged. Define the state at any time to be the list ordering at that time and note that there are \(n !\) possible states. (a) Argue that the preceding is a Markov chain. (b) For any state \(i_{1}, \ldots, i_{n}\) (which is a permutation of \(\left.1,2, \ldots, n\right)\), let \(\pi\left(i_{1}, \ldots, t_{n}\right)\) denote the limiting probability. In order for the state to be \(i_{1}, \ldots, i_{n}\), it is necessary for the last request to be for \(i_{1}\), the last non- \(i_{1}\) request for \(i_{2}\), the last non- \(i_{1}\) or \(i_{2}\) request for \(l_{3}\), and so on. Hence, it appears intuitive that Verify when \(n=3\) that the above are indeed the limiting probabilities.
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