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Automobiles arrive at a vehicle equipment inspection station according to a Poisson process with rate \(\alpha=10\) per hour. Suppose that with probability \(.5\) an arriving vehicle will have no equipment violations. a. What is the probability that exactly ten arrive during the hour and all ten have no violations? b. For any fixed \(y \geq 10\), what is the probability that \(y\) arrive during the hour, of which ten have no violations? c. What is the probability that ten "no-violation" cars arrive during the next hour? [Hint: Sum the probabilities in part (b) from \(y=10\) to \(\left.x_{-}\right]\)

Short Answer

Expert verified
a. 0.000122 b. Solve using the formula in Step 4. c. 0.000122

Step by step solution

01

Calculate the probability of exactly 10 arrivals

The number of arrivals in an hour follows a Poisson distribution with rate \(\alpha=10\). The probability of exactly 10 arrivals, denoted as \( P(Y = 10) \), is given by the formula for the Poisson distribution: \[P(Y = 10) = \frac{e^{-10} \cdot 10^{10}}{10!}\]Calculating this gives approximately \(0.1251\).
02

Calculate the probability all 10 have no violations

Given that 10 cars arrive, each car has a 0.5 probability of having no violations. The probability that all 10 cars have no violations is \((0.5)^{10}\). \[(0.5)^{10} = 0.00097656\]
03

Calculate probability all 10 arrive with no violations for part (a)

Multiply the probability of exactly 10 arrivals by the probability that all 10 cars have no violations:\[0.1251 \times 0.00097656 = 0.000122\]
04

Probability calculation in part (b)

For \(y \geq 10\), the probability that exactly \(y\) cars arrive and 10 have no violations involves choosing 10 from \(y\), with all having no violations:\[P(\text{10 no violations and } y \text{ arrive}) = P(Y = y) \cdot \binom{y}{10} \cdot (0.5)^{10} \cdot (0.5)^{y-10}\]Compute \(P(Y = y)\) using the Poisson formula, then multiply by the above expression.
05

Calculate the probability for part (c)

For part (c), calculate the probability that 10 "no-violation" cars arrive by summing the probabilities from \(y = 10\) to infinity:\[P(\text{10 no-violation cars}) = \sum_{y=10}^{\infty} P(\text{10 no-violations and } y \text{ arrive})\]Using the formula from step 4, this turns into:\[\sum_{y=10}^{\infty} \frac{e^{-10} \cdot 10^y}{y!} \cdot \binom{y}{10} \cdot (0.5)^y\]Recognizing this as a negative binomial situation, summation using negative binomial properties gives:\[\frac{e^{-10} \cdot 10^{10}}{10!} = 0.000122\]

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Key Concepts

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

Probability Calculations
Probability calculations are the foundation of understanding events that might happen. They provide a way to quantify the likelihood of different outcomes.
In problems such as these, where we're exploring car arrivals and checking for violations, it's essential to set up the problem with the right probability models. For instance, the Poisson distribution is crucial because it's used for events happening independently over time, such as the arrival of cars at a station.
To find the probability of a certain number of cars arriving, Poisson's formula is used:
  • If \( \lambda \) is the average number per interval (10 cars per hour in this case), then the probability of exactly \( k \) events (arrivals) is:\[ P(X = k) = \frac{e^{-\lambda} \cdot \lambda^{k}}{k!} \]
  • For instance, if we want to know the probability of exactly 10 cars arriving, we’d plug 10 into this formula.
Understanding these calculations allows us to not only interpret critical data but make informed decisions based on probability.
Vehicle Inspection
Vehicle inspections are an everyday application of probability and help in studying patterns in vehicle behavior such as equipment failures. Imagine that each arriving vehicle is like a new coin toss with probabilities being assigned based on the likelihood of an event, such as having or not having violations.
For this scenario, when a car arrives at a vehicle inspection station, it might or might not have equipment violations with a probability of 0.5 for no violations. This means for each car, there's an equal chance of being clear or having faults. This probability helps us determine the chance of having all 10 cars arriving without violations, which needs multiplying the individual probabilities for each car:
  • For example, the probability that all 10 cars have no violations: \[ (0.5)^{10} \]
By examining these probabilities carefully, we can evaluate and improve our vehicle inspection standards, predicting and addressing potential issues efficiently.
Negative Binomial Distribution
The Negative Binomial Distribution is a probability distribution that applies when we count the number of successes before a specified number of failures occur. It's a generalization of the geometric distribution, allowing more than one failure before stopping.
In this exercise, by considering the situation where exactly 10 cars without violations arrive, we apply this concept to compute the chance of this happening if more than ten cars may arrive in total.
  • In mathematical terms for this scenario, if \( y \) cars arrive (where \( y \ge 10 \)), computing the probability involves choosing which among the \( y \) had no violations and multiplying by the probabilities of occurrences:\[ \binom{y}{10} \cdot (0.5)^{10} \cdot (0.5)^{y-10} \]
  • This formula takes the familiar binomial coefficient to choose 10 from \( y \), and multiplies the independent probabilities of each outcome accordingly.
This distribution is particularly useful as it animates real-world scenarios where conditions must meet precise configurations for success and is instrumental in fields like risk management.
No Violation Probability
In probability theory, especially with practical applications like vehicles inspection, the concept of "No Violation Probability" involves calculating the occurrence of all cars having no violations in a specific timeframe.
After determining the individual probabilities of cars arriving without violations, which we've seen when multiplying the probabilities using Poisson and further extended through the Negative Binomial model, we can solve more complex questions like the ones in our task.
For instance, when summing probabilities from a minimum number of arrivals (in this case, 10 "no-violation" arrivals) to infinity, we're considering the
  • total probability of all scenarios where at least a certain condition holds
. The sum, which uses properties of the negative binomial distribution, turns out to be equivalent to the Poisson probability for exactly the specified number of arrivals. This remarkable intersection between different probabilistic models is a valued insight in statistical mathematics.

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Most popular questions from this chapter

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