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The number of cars driving past a parking area in a one-minute time interval has a Poisson distribution with mean \(\lambda\). The probability that any individual driver actually wants to park his or her car is \(p .\) Assume that individuals decide whether to park independently of one another. a. If one parking place is available and it will take you one minute to reach the parking area, what is the probability that a space will still be available when you reach the lot? (Assume that no one leaves the lot during the one- minute interval.) b. Let \(W\) denote the number of drivers who wish to park during a one-minute interval. Derive the probability distribution of \(W\)

Short Answer

Expert verified
a. Probability is \( e^{-\lambda p} \). b. \( W \sim \text{Poisson}(\lambda p) \).

Step by step solution

01

Define the Scenario (Part a)

We are considering a Poisson distribution where the average rate of cars passing is \( \lambda \) per minute, and each car has a probability \( p \) of wanting to park. The goal is to determine the probability that no one parks in the next minute when there is one parking space available.
02

Define the Poisson Probability (Part a)

A Poisson random variable \( X \) with mean \( \lambda \times p \) is used to model the number of cars wishing to park in one minute. The probability mass function is \( P(X=k) = \frac{{(\lambda \times p)^k \cdot e^{-\lambda \times p}}}{k!} \). We specifically want \( P(X=0) \), meaning no cars want to park.
03

Calculate Probability for No Cars Wanting to Park (Part a)

The probability that no cars want to park is given by \( P(X=0) = \frac{{(\lambda \times p)^0 \cdot e^{-\lambda \times p}}}{0!} = e^{-\lambda \times p} \). This is the probability that the parking space remains available.
04

Solution Summary (Part a)

The parking space will be available at the shed if no cars wish to park. Therefore, \( e^{-\lambda p} \) is the probability that the space is available when you arrive.
05

Define Driver's Wish to Park (Part b)

The number of drivers who wish to park, \( W \), could be seen as a Binomial distribution since each driver decides independently with probability \( p \). Given a Poisson arrival process and independent parking decisions, \( W \) is a sum of Bernoulli trials.
06

Derive the Distribution of W (Part b)

\( W \) follows a Poisson distribution with parameter \( \lambda \times p \). This is derived from aggregating independent homogeneous Poisson processes with thinning, where the event occurs with probability \( p \). Thus, \( W \sim \text{Poisson}(\lambda \times p) \).
07

Solution Summary (Part b)

The number of drivers who wish to park follows a Poisson distribution with mean \( \lambda p \), represented as \( W \sim \text{Poisson}(\lambda p) \).

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Key Concepts

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

Probability
Probability is a measure of how likely an event is to occur. It is quantified between 0 (impossible event) and 1 (certain event). In the context of this exercise, probability is used to determine how likely it is for certain events to occur, such as a driver wanting to park in a given minute.

Probability helps in predicting outcomes by providing a numerical value for the chance of an event. For example, if the probability that any driver wants to park is \( p \), then the probability that a driver does not want to park is \( 1-p \).
  • The sum of probabilities of all possible outcomes of an event always equals 1.
  • In scenarios with multiple independent trials, probabilities can be multiplied for combined events.
Understanding probability is key to predicting outcomes more accurately and making informed decisions in uncertain situations.
Binomial Distribution
The binomial distribution models the number of successes in a fixed number of trials, where each trial has two possible outcomes: success or failure. It is characterized by two parameters: \( n \) (the number of trials) and \( p \) (the probability of success in each trial).

In this exercise, the decision of each driver to park is modeled initially by a binomial process because each driver independently decides to park or not. However, as the number of trials (cars arriving) is governed by a Poisson distribution, we eventually use a Poisson distribution for simplicity.
  • A binomial distribution becomes a Poisson distribution in situations where the number of trials is large, the success probability is small, and the mean is \( np \).
  • A typical scenario for employing the binomial distribution is when calculating the probability of a certain number of successes over a repeated and fixed number of independent trials.
While binomial distribution might sound complicated, it is just a way to manage situations where outcomes happen in series, like drivers choosing to park.
Random Variable
A random variable is a variable whose possible values are numerical outcomes of a random phenomenon. In our exercise, the random variable is the number of cars trying to park, symbolized as \( X \) or \( W \).

This concept is crucial because it allows us to convert real-world random occurrences into a form that we can analyze mathematically. For example:
  • A discrete random variable, like \( W \), has a countable number of possible values, such as 0, 1, 2, etc.
  • Probability distributions, like Poisson or Binomial, describe how the probabilities are distributed over the possible values of a random variable.
Understanding random variables helps turn the randomness of everyday events into something we can systematically work through and predict, like determining parking-related probabilities.
Independent Events
Independent events are those whose outcomes do not affect each other. In this problem, whether one driver decides to park is independent of another driver's decision. This independence means the probability of each driver choosing to park stays constant, regardless of others’ choices.

This concept is fundamental in probabilistic calculations because it influences how probabilities combine. If two events are independent, the probability of both occurring is the product of their individual probabilities. For instance, if the probability of each driver\( p \) is independent, then for no one to park, it will be \( e^{-\lambda p} \), derived from multiple independent decisions.
  • Independence is a key assumption in many probability models, including binomial and Poisson distributions.
  • Knowing that events are independent allows us to simplify probability calculations significantly.
Recognizing and applying the concept of independence enables more intuitive and tractable solutions, exactly like the scenario where multiple drivers independently decide whether to park or not.

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