Problem 1
Question
Customers arrive at a bank at a Poisson rate \(\lambda\) Suppose that two customers arrived during the first hour. What is the probability that (a) both arrived during the first 20 minutes? (b) at least one arrived during the first 20 minutes?
Step-by-Step Solution
Verified Answer
The probability that both customers arrived within the first 20 minutes is approximately 7.33%, and the probability that at least one customer arrived within the first 20 minutes is approximately 48.66%.
1Step 1: Calculate the number of arrivals in each time interval
First, let us calculate the expected number of arrivals in each time interval given the Poisson rate 𝜆. Since we are given that there are two arrivals in the first hour (60 minutes), we can use this to find the expected number of arrivals in the first 20 minutes and the last 40 minutes:
Expected number of arrivals in the first 20 minutes: \(\frac{1}{3} * 2 = \frac{2}{3}\)
Expected number of arrivals in the last 40 minutes: \(\frac{2}{3} * 2 = \frac{4}{3}\)
2Step 2: Calculate the probability for scenario (a)
For scenario (a), we want to know the probability that both customers arrived within the first 20 minutes. In this case, X = 2, k = 2, and t = 20/60.
Using the Poisson probability mass function:
\(P(X = 2) = \frac{e^{-\frac{2}{3} (\frac{1}{3})} (\frac{2}{3} \frac{1}{3})^2}{2!} \approx 0.073262\)
So, the probability that both customers arrived within the first 20 minutes is approximately 7.33%.
3Step 3: Calculate the probability for scenario (b)
For scenario (b), we want to know the probability that at least one customer arrived within the first 20 minutes. We can calculate this by finding the probability that neither customer arrived within the first 20 minutes (in the last 40 minutes) and then subtracting that probability from 1.
First, we will calculate the probability that neither customer arrived within the first 20 minutes: X = 0, k = 0, and t = 20/60.
Using the Poisson probability mass function:
\(P(X = 0) = \frac{e^{-\frac{2}{3} (\frac{1}{3})} (\frac{2}{3} \frac{1}{3})^0}{0!} \approx 0.513417\)
Now, we subtract this probability from 1 to find the probability that at least one customer arrived within the first 20 minutes:
\(1 - P(X = 0) \approx 1 - 0.513417 \approx 0.486583\)
So, the probability that at least one customer arrived within the first 20 minutes is approximately 48.66%.
Key Concepts
Poisson ProcessProbability Mass FunctionExpected Number of Arrivals
Poisson Process
The Poisson process is a statistical model that describes a series of events occurring independently and at a constant average rate. It is widely used in various fields to model random events such as the number of customers arriving at a bank, as demonstrated in our exercise.
In the context of the exercise, customers arriving at a bank is a classic example of a Poisson process. To break it down, the key features of this process include events (customers arriving) happening independently of each other, and the average rate (symbolized by \( \lambda \)) being constant. In essence, the Poisson process gives us a framework for predicting how likely it is that a certain number of events will occur during a fixed period.
Diving into our specific problem, which involves 20-minute intervals, we can apply the principles of the Poisson process to calculate the probability of different scenarios of customer arrivals within these time frames.
In the context of the exercise, customers arriving at a bank is a classic example of a Poisson process. To break it down, the key features of this process include events (customers arriving) happening independently of each other, and the average rate (symbolized by \( \lambda \)) being constant. In essence, the Poisson process gives us a framework for predicting how likely it is that a certain number of events will occur during a fixed period.
Diving into our specific problem, which involves 20-minute intervals, we can apply the principles of the Poisson process to calculate the probability of different scenarios of customer arrivals within these time frames.
Probability Mass Function
The probability mass function (PMF) is a function that gives the probability that a discrete random variable is exactly equal to some value. Essentially, it describes the distribution of probability for a set of possible outcomes.
For the Poisson distribution, which is a discrete probability distribution, the PMF is particularly important. It is expressed using the formula:\[ P(X = k) = \frac{e^{-\lambda t} (\lambda t)^k}{k!} \]where \( P(X = k) \) represents the probability of observing \( k \) events in a fixed interval, \( \lambda \) is the average rate of occurrence, \( t \) is the length of the time interval, and \( k! \) denotes the factorial of \( k \).
When we apply the PMF to the exercise problem, we are able to calculate the precise probabilities for scenarios (a) and (b) given the expected number of arrivals within the specific time intervals.
For the Poisson distribution, which is a discrete probability distribution, the PMF is particularly important. It is expressed using the formula:\[ P(X = k) = \frac{e^{-\lambda t} (\lambda t)^k}{k!} \]where \( P(X = k) \) represents the probability of observing \( k \) events in a fixed interval, \( \lambda \) is the average rate of occurrence, \( t \) is the length of the time interval, and \( k! \) denotes the factorial of \( k \).
When we apply the PMF to the exercise problem, we are able to calculate the precise probabilities for scenarios (a) and (b) given the expected number of arrivals within the specific time intervals.
Expected Number of Arrivals
The expected number of arrivals in a Poisson process is a measure of the average number of events we can anticipate over a certain period. It is calculated by multiplying the rate of occurrence \( \lambda \) by the length of the time interval \( t \) being considered.
In the exercise, we first determine the expected number of arrivals for different segments of the hour (20 minutes and 40 minutes) using this concept. For instance, if two customers are expected in an hour (\( \lambda = 2 \) per hour), then the expected number for the first 20 minutes (\( t = 1/3 \) of an hour) is \( \lambda * t = 2/3 \) customers.
Understanding this concept is crucial as it sets the stage for computing probabilities in the Poisson distribution. It allows us to define our PMF parameters accurately to solve for the likelihood of specific arrival scenarios in the given time frames.
In the exercise, we first determine the expected number of arrivals for different segments of the hour (20 minutes and 40 minutes) using this concept. For instance, if two customers are expected in an hour (\( \lambda = 2 \) per hour), then the expected number for the first 20 minutes (\( t = 1/3 \) of an hour) is \( \lambda * t = 2/3 \) customers.
Understanding this concept is crucial as it sets the stage for computing probabilities in the Poisson distribution. It allows us to define our PMF parameters accurately to solve for the likelihood of specific arrival scenarios in the given time frames.
Other exercises in this chapter
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