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Pulses arrive at a Geiger counter in accordance with a Poisson process at a rate of three arrivals per minute. Each particle arriving at the counter has a probability \(\frac{2}{3}\) of being recorded. Let \(X(t)\) denote the number of pulses recorded by time \(t\) minutes. (a) \(P[X(t)=0\\}=?\) (b) \(E[X(t)]=\) ?

Short Answer

Expert verified
(a) \(P[X(t)=0] = e^{-2t}\) (b) \(E[X(t)] = 2t\)

Step by step solution

01

Find the rate of recorded arrivals

The rate of the particles arriving at the Geiger counter is 3 per minute. Since the probability of a particle being recorded is \(\frac{2}{3}\), the rate of recorded arrivals is obtained by multiplying the arrival rate with the probability of recording a particle. Rate of recorded arrivals: \(3 \times \frac{2}{3} = 2\)
02

Calculate P[X(t)=0]

Now that we have the rate of recorded arrivals (2 per minute), we can use the Poisson distribution formula to find the probability that 0 pulses have been recorded by time t: \(P(X(t) = 0) = \frac{(\lambda t)^{k} e^{-\lambda t}}{k!}\) Here, \(\lambda = 2\) (rate of recorded arrivals) \(t =\) time in minutes \(k = 0\) (number of recorded arrivals) So, \(P[X(t)=0] = \frac{(2t)^{0} e^{-2t}}{0!}\) Simplifying: \(P[X(t)=0] = e^{-2t}\)
03

Calculate E[X(t)]

The expectation of a Poisson-distributed random variable is simply \(\lambda t\). So, \(E[X(t)] = 2t\) #Results# (a) The probability of no recorded pulses in \(t\) minutes is given by: \(P[X(t)=0] = e^{-2t}\) (b) The expected number of recorded pulses in \(t\) minutes is: \(E[X(t)] = 2t\)

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Key Concepts

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

Geiger counter
A Geiger counter is an essential instrument used for detecting and measuring ionizing radiation. It works by counting the number of particles, such as alpha particles, beta particles, and gamma rays, that pass through its sensor. Each particle is counted as a discrete pulse.
The Geiger counter is commonly used in various fields, including physics research, environmental monitoring, and nuclear industry operations. The device itself consists of a Geiger-Müller tube, which is filled with an inert gas. When a particle enters the tube, it ionizes the gas, creating a pulse of current that is amplified and counted by the electronics of the Geiger counter.
To contextualize it within the problem, pulses arrive at the Geiger counter following a Poisson process. This means the arrival of particles at the counter follows a specific statistical distribution that helps us calculate the likelihood of certain events, like the number of pulses recorded.
Poisson distribution
The Poisson distribution is a probability distribution that is used to model the number of times an event occurs within a specific interval. This interval could be related to time, space, or some other measure. The events in a Poisson process occur independently, and the number of events in each interval follows a Poisson distribution.
Mathematically, the Poisson distribution is represented by the formula:\[P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}\]Where:
  • \(P(X = k)\) is the probability of \(k\) events in the interval
  • \(\lambda\) is the average rate of events per interval
  • \(k\) is the number of occurrences
  • \(e\) is the base of the natural logarithm, approximately equal to 2.718
In the context of our exercise, pulses arriving at the Geiger counter follows a Poisson process with an adjusted arrival rate, accounting for the probability of a particle being recorded, which leads to a rate of \(\lambda = 2\) per minute. This is a critical step in calculating the probability of recording no pulses, \(P[X(t)=0]\), by applying the Poisson formula.
Probability of recording
The probability of recording refers to the chance that any given pulse arriving at the Geiger counter is actually recorded. In the exercise, each arriving particle has a probability of \(\frac{2}{3}\) to be recorded. This means that not all particles that enter the counter are counted.
The rate of recorded arrivals, also known as the effective arrival rate, can be calculated by multiplying the rate of arrival by the probability of recording. For our problem:
  • The original rate of pulses is 3 per minute
  • The probability of recording is \(\frac{2}{3}\)
Multiplying these values gives us:\[\text{Rate of recorded arrivals} = 3 \times \frac{2}{3} = 2\]Using this adjusted rate, we can determine the probability that zero pulses are recorded by time \(t\), using the formula for the Poisson distribution. This results in:\[P[X(t) = 0] = e^{-2t}\] Moreover, we can calculate the expected number of recorded pulses, \(E[X(t)]\), by multiplying the adjusted rate by time, yielding \(E[X(t)] = 2t\). These calculations involve understanding and utilizing the properties of the Poisson distribution combined with the given recording probability.

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Most popular questions from this chapter

Suppose that electrical shocks having random amplitudes occur at times distributed according to a Poisson process \(\\{N(t), t \geqslant 0]\) with rate \(\lambda\). Suppose that the amplitudes of the successive shocks are independent both of other amplitudes and of the arrival times of shocks, and also that the amplitudes have distribution \(F\) with mean \(\mu\). Suppose also that the amplitude of a shock decreases with time at an exponential 'rate \(\alpha\), meaning that an initial amplitude \(A\) will have value \(A e^{-\alpha x}\) after an additional time \(x\) has elapsed. Let \(A(t)\) denote the sum of all amplitudes at time \(t\). That is, $$ A(t)=\sum_{i=1}^{N(t)} A_{i} e^{-\alpha\left(t-S_{i}\right)} $$ where \(A_{i}\) and \(S_{l}\) are the initial amplitude and the arrival time of shock \(i\). (a) Find \(E[A(t)]\) by conditioning an \(N(t)\). (b) Without any computations, explain why \(A(t)\) has the same distribution as does \(D(t)\) of Example \(5.19 .\)

Events occur accórding to a Poisson process with rate \(\lambda\). Each time an event occurs, we must decide whether or not to stop, with our objective being to stop at the last event to occur prior to some specified time \(T\), where \(T>1 / \lambda\). That is, if an event occurs at time \(t, 0 \leqslant t \leqslant T\), and we decide to stop, then we win if there are no additional events by time \(T\), and we lose otherwise. If we do not stop when an event occurs and no. additional events occur by time \(T\), then we lose. Also, if no events occur by time \(T\), then we lose. Consider the strategy that stops at the first event to occur after some fixed time \(s, 0 \leqslant s \leqslant T\). (a) Using this strategy, what is the probability of winning? (b) What value of \(s\) maximizes the probability of winning? (c) Show that on?'s probability of winning when using the preceding strategy with the value of \(s\) specified in part (b) is \(1 / e\).

Teams 1 and 2 are playing a match. The teams score points according to independent Poisson processes with respective rates \(\bar{\lambda}_{1}\) and \(\lambda_{2}\). If the match ends when one of the teams has scored \(k\) more points than the other, find the probability that team 1 wins. Hint: Relate this to the gambler's ruin problem.

If \(X_{i}, i=1,2,3\), are independent exponential random variables with rates \(\lambda_{i}, i=1,2,3\), find (a) \(P\left\\{X_{1}

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent and identically distributed exponential random variables. Show that the probability that the largest of them is greater than the sum of the others is \(n / 2^{n-1}\). That is, if $$ M=\max _{j} X_{j} $$ then show $$ P\left\\{M>\sum_{i=1}^{n} X_{i}-M\right\\}=\frac{n}{2^{n-1}} $$ Hint: What is \(P\left[X_{1}>\sum_{i=2}^{n} X_{i}\right\\} ?\)

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