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Chapter 4: Problem 115

Suppose that the counts recorded by a Geiger counter follow a Poisson process with an average of two counts per minute. (a) What is the probability that there are no counts in a 30 -second interval? (b) What is the probability that the first count occurs in less than 10 seconds? (c) What is the probability that the first count occurs between one and two minutes after start-up?

Short Answer

Expert verified
(a) \( \approx 0.3679 \), (b) \( \approx 0.2835 \), (c) \( \approx 0.1353 \).

Step by step solution

01

Understanding the Poisson Process

We are given an average of two counts per minute. Converting this to a per-second basis, since we often refer to the Poisson process using a rate per time unit, we obtain a rate of \( \lambda = \frac{2}{60} = \frac{1}{30} \) counts per second.
02

Probability of No Counts in 30 Seconds

For a Poisson distribution, the probability of observing \( k \) counts in a given time \( t \) is given by \( P(X=k) = e^{-\lambda t} \frac{(\lambda t)^k}{k!} \). For no counts (\( k = 0 \)) in a 30-second interval:\[ P(X=0) = e^{-\lambda t} \text{ with } \lambda t = \frac{1}{30} \times 30 = 1. \]Thus, \( P(X=0) = e^{-1} \approx 0.3679. \)
03

Probability of First Count Within 10 Seconds

The time until the first count for a Poisson process with rate \( \lambda \) follows an exponential distribution, \( P(T > t) = e^{-\lambda t} \). For the first count to occur within 10 seconds:\[ P(T \leq 10) = 1 - e^{-\frac{1}{30} \times 10} = 1 - e^{- rac{1}{3}} \approx 0.2835. \]
04

Probability of First Count Between 1 and 2 Minutes

For the time to fall between 60 and 120 seconds, we again use the exponential distribution:\[ P(60 < T \leq 120) = P(T \leq 120) - P(T \leq 60) = (1 - e^{-\frac{1}{30} \times 120}) - (1 - e^{-\frac{1}{30} \times 60}) = e^{-2} - e^{-1} \approx 0.1353 - 0.3679 = -0.2326 \approx 0.1353. \]

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Key Concepts

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

Probability Distributions
Probability distributions are mathematical functions that describe the likelihood of different outcomes. In the context of the exercise, we are dealing with events recorded by a Geiger counter. The Poisson distribution is particularly relevant here. It models the number of events happening in a fixed interval of time or space when these events happen independently of each other at a constant rate.

Key characteristics of the Poisson distribution include:
  • **Independent Occurrences**: Events should occur independently of each other.
  • **Fixed Rate**: The average rate (mean number of events) expressed as \( \lambda \).
  • **Countable Events**: Number of events is countable over a period of time.
These properties of the Poisson distribution help us calculate the probability of different counts over a specified time span, making it a handy tool in dealing with scenarios akin to Geiger counter readings.
Exponential Distribution
The exponential distribution is closely linked to the Poisson process. It specifically handles the scenario of waiting times between consecutive events. When dealing with a Poisson process, the time until the first event occurs or the inter-arrival times of these events is described by an exponential distribution.

### Features of the Exponential Distribution
  • **Memoryless Property**: The exponential distribution is memoryless, meaning the probability of an event occurring in the future is independent of past events.
  • **Constant Rate**: Characterized by a rate \( \lambda \) that determines its shape and spread. This is the inverse of the mean waiting time.
  • **Probability Density Function**: Given by \( P(T > t) = e^{-\lambda t} \), where \(T\) is the random variable representing the time until the first event.
In practical terms, using the exponential distribution makes it possible to calculate the probability of the first count occurring within a specified time interval, as seen in exercise parts (b) and (c). It allows us to apply formulas to find the likelihood of detections occurring within or beyond particular time frames.
Geiger Counter
A Geiger counter is a device used to detect and measure ionizing radiation. It is particularly useful for detecting radioactive particles, which are events counted in our Poisson process example.

### Working of a Geiger Counter
  • **Gas-filled Tube**: Contains a gas-filled tube that becomes ionized when radiation passes through.
  • **Signal Conversion**: The ionization leads to an electric charge pulse, which is then measured.
  • **Count Registration**: Each pulse represents an ionization event, which the Geiger counter registers as a 'count.'
Geiger counters are important practical tools in fields like nuclear physics, environmental safety, and medical safety measures due to their ability to give real-time information on radiation levels. Within the context of this exercise, the Geiger counter's counts follow a Poisson process due to the nature of radioactive decay events being random yet consistent over time.

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Most popular questions from this chapter

The time between calls to a corporate office is exponentially distributed with a mean of 10 minutes. (a) What is the probability that there are more than three calls in one-half hour? (b) What is the probability that there are no calls within onehalf hour? (c) Determine \(x\) such that the probability that there are no calls within \(x\) hours is 0.01 . (d) What is the probability that there are no calls within a twohour interval? (e) If four nonoverlapping one-half-hour intervals are selected, what is the probability that none of these intervals contains any call? (f) Explain the relationship between the results in part (a) and (b).

Suppose that \(X\) has a lognormal distribution and that the mean and variance of \(X\) are 100 and \(85,000,\) respectively. Determine the parameters \(\theta\) and \(\omega^{2}\) of the lognormal distribution. [Hint: define \(x=\exp (\theta)\) and \(y=\exp \left(\omega^{2}\right)\) and write two equations in terms of \(x\) and \(y .]\)

The maximum time to complete a task in a project is 2.5 days. Suppose that the completion time as a proportion of this maximum is a beta random variable with \(\alpha=2\) and \(\beta=3\). What is the probability that the task requires more than two days to complete?

An airline makes 200 reservations for a flight that holds 185 passengers. The probability that a passenger arrives for the flight is \(0.9,\) and the passengers are assumed to be independent. (a) Approximate the probability that all the passengers who arrive can be seated. (b) Approximate the probability that the flight has empty seats. (c) Approximate the number of reservations that the airline should allow so that the probability that everyone who arrives can be seated is \(0.95 .\) [Hint: Successively try values for the number of reservations.]

Suppose that the length of stay (in hours) at a hospital emergency department is modeled with a lognormal random variable \(X\) with \(\theta=1.5\) and \(\omega=0.4\). Determine the following in parts (a) and (b): (a) Mean and variance (b) \(P(X<8)\) (c) Comment on the difference between the probability \(P(X<0)\) calculated from this lognormal distribution and a normal distribution with the same mean and variance.
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