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Chapter 4: Problem 52

A taxi driver provides service in two zones of a city. Fares picked up in zone \(A\) will have destinations in zone \(A\) with probability \(.6\) or in zone \(B\) with probability .4. Fares picked up in zone \(B\) will have destinations in zone A with probability . 3 or in zone \(B\) wih probability .7. The driver's expected profit for a trip entirely in zone \(A\) is 6 ; for a trip entirely in zone \(B\) is 8 ; and for a trip that involves both zones is 12 . Find the taxi driver's average profit per trip.

Short Answer

Expert verified
The taxi driver's average profit per trip is \(8.8\).

Step by step solution

01

1. Calculate the probability of each type of trip

Assuming the driver picks up an equal amount of fares in each zone, we have the probabilities as follows: - Zone A to Zone A: \(0.5 \times 0.6 = 0.3\) - Zone A to Zone B: \(0.5 \times 0.4 = 0.2\) - Zone B to Zone A: \(0.5 \times 0.3 = 0.15\) - Zone B to Zone B: \(0.5 \times 0.7 = 0.35\)
02

2. Calculate the expected profit for each type of trip

We multiply the probability of each type of trip by the corresponding expected profit, as follows: - Profit A to A: \(0.3 \times 6 = 1.8\) - Profit A to B: \(0.2 \times 12 = 2.4\) - Profit B to A: \(0.15 \times 12 = 1.8\) - Profit B to B: \(0.35 \times 8 = 2.8\)
03

3. Calculate the average profit per trip

Now, we sum the expected profits for each type of trip to find the average profit per trip: Average profit = Profit A to A + Profit A to B + Profit B to A + Profit B to B Average profit = \(1.8 + 2.4 + 1.8 + 2.8\) Average profit = \(8.8\) So, the taxi driver's average profit per trip is \(8.8\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Models
When we talk about probability models in the context of daily activities like a taxi service, we're referring to a mathematical representation of various outcomes and their likelihood of occurring. In our exercise, the model is used to determine the chances of a taxi picking up a fare in one zone and dropping off in the same or a different one. It breaks down a complex scenario into simpler, quantifiable probabilities, such as fares from zone A to zone A or zone B to zone B.

The strength of probability models lies in their ability to predict outcomes in situations of uncertainty. For the taxi driver, this means anticipating where their fares will lead them throughout a workday based on the probabilities of picking up passengers in different zones. By understanding these models, individuals and businesses can make informed decisions and strategize efficiently in stochastic (random) environments.
Expected Value
The concept of expected value is fundamental in understanding long-term outcomes in probabilistic scenarios. It provides a way to quantify the average outcome of a random event, assuming the event is repeated many times. In the taxi driver's case, it is a weighted average of the profits from all possible trips they can make, with the weights being the probabilities of each trip occurring.

Mathematically, the expected value is calculated by multiplying each outcome by its probability and adding all these products together. It's important to note that while expected value offers a 'big picture' analysis, it does not guarantee outcomes in the short term. So a taxi driver may not earn the exact expected profit on each trip, but over many trips, the average should approximate the expected value.
Profit Analysis
Profit analysis is a key concept for any business venture, including a taxi service. It involves determining the potential gains from various business activities or choices. By assessing the expected profits for trips within and between zones A and B, the driver can better understand which trips are more lucrative and how their choices impact overall profitability.

This analysis becomes even more insightful when considering probability models and expected values. The probability-adjusted expected profits give a clearer picture of potential earnings over time. For instance, although a trip from zone B to zone A has the same expected profit as a trip within zone A, its lower probability of occurring makes it contribute less to the average profit per trip.

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Most popular questions from this chapter

It follows from Theorem \(4.2\) that for a time reversible Markov chain $$ P_{i j} P_{j k} P_{k i}=P_{i k} P_{k j} P_{j i}, \quad \text { for all } i, j, k $$ It turns out that if the state space is finite and \(P_{i j}>0\) for all \(i, j\), then the preceding is also a sufficient condition for time reversibility. [That is, in this case, we need only check Equation (4.26) for paths from \(i\) to \(i\) that have only two intermediate states.] Prove this. Hint: Fix \(i\) and show that the equations $$ \pi_{j} P_{j k}=\pi_{k} P_{k j} $$ are satisfied by \(\pi_{j}=c P_{i j} / P_{j i}\), where \(c\) is chosen so that \(\sum_{j} \pi_{j}=1\).

Trials are performed in sequence. If the last two trials were successes, then the next trial is a success with probability \(0.8\); otherwise the next trial is a success with probability 0.5. In the long run, what proportion of trials are successes?

For a time reversible Markov chain, argue that the rate at which transitions from \(i\) to \(j\) to \(k\) occur must equal the rate at which transitions from \(k\) to \(j\) to \(i\) occur.

Consider the Markov chain \(\left\\{X_{n}, n \geqslant 0\right\\}\) with states \(0,1,2\), whose transition probability matrix is $$ \mathbf{P}=\left[\begin{array}{lll} 0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0 \\ 1 & 0 & 0 \end{array}\right] $$ Let \(f(0)=0, f(1) \neq f(2)=1\). If \(Y_{n}=f\left(X_{n}\right)\), is \(\left\\{Y_{n}, n \geqslant 0\right\\}\) a Markov chain?

Three white and three black balls are distributed in two urns in such a way that each contains 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 \(X_{n}\) denote the state of the system after the \(n\) th step. Explain why \(\left\\{X_{n}, n=0,1,2, \ldots\right\\}\) is a Markov chain and calculate its transition probability matrix.
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