Question

Customers arrive at a bank at a Poisson rate \(\lambda .\) Suppose 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?

Short answer

The probabilities for the given scenarios are: (a) Both customers arrived during the first 20 minutes: \(P(\text{both in 20 min} | \text{2 in 1 hour}) = \frac{1}{9}\) (b) At least one customer arrived during the first 20 minutes: \(P(\text{at least 1 in 20 min} | \text{2 in 1 hour}) = \frac{5}{9}\)

Step-by-step solution

(Step 1: Define the parameter for the Poisson distribution)

Since we know that the customers arrive at a Poisson rate of \(\lambda\), we can represent the number of arrivals in a certain time interval as a Poisson distribution with parameter \(\lambda t\), where \(t\) denotes the length of the interval. In this exercise, we are given that two customers arrived during the first hour. So, we have a Poisson distribution with parameter \(\lambda \cdot 1\) hour.

(Step 2: Find the probability that both customers arrived during the first 20 minutes)

To find the probability that both customers arrived during the first 20 minutes, we need to define a new Poisson distribution for the first 20 minutes, with parameter \(\lambda \cdot \frac{1}{3}\) (since 20 minutes = \(\frac{1}{3}\) hour). First, we note that the events "both customers arrive during the first 20 minutes" and "both customers arrive after the first 20 minutes" are mutually exclusive and cover all possibilities. Then, we need to find the probability that two customers arrived during the first 20 minutes, which can be calculated as: \(P(\text{both in 20 min}) = \frac{(\lambda \cdot \frac{1}{3})^2 \cdot e^{-(\lambda \cdot \frac{1}{3})}}{2!} \) Since we know that two customers arrived during the first hour, the probability of both customers arrived during the first 20 minutes is: \(P(\text{both in 20 min} | \text{2 in 1 hour}) = \frac{P(\text{both in 20 min} \cap \text{2 in 1 hour})}{P(\text{2 in 1 hour})}\) Using the properties of Poisson distribution and the fact that the number of arrivals in disjoint time intervals is independent: \(P(\text{both in 20 min} | \text{2 in 1 hour}) = \frac{P(\text{2 in 20 min})}{P(\text{2 in 1 hour})}\) This probability can be calculated as: \(P(\text{both in 20 min} | \text{2 in 1 hour}) = \frac{(\lambda \cdot \frac{1}{3})^2 \cdot e^{-(\lambda \cdot \frac{1}{3})}/2!}{(\lambda \cdot 1)^2 \cdot e^{-(\lambda \cdot 1)}/2!} \) Simplify the expression: \(P(\text{both in 20 min} | \text{2 in 1 hour}) = \frac{1}{3^2} = \frac{1}{9} \)

(Step 3: Find the probability that at least one customer arrived during the first 20 minutes)

To find the probability that at least one customer arrived during the first 20 minutes, we'll use the fact that this is equivalent to the complement of both customers arriving after the first 20 minutes. The probability of both customers arriving after the first 20 minutes can be calculated in the same manner as in Step 2, but now considering a Poisson distribution for the last 40 minutes, with parameter \(\lambda \cdot \frac{2}{3}\): \(P(\text{at least 1 in 20 min} | \text{2 in 1 hour}) = 1 - P(\text{both after 20 min} | \text{2 in 1 hour})\) \(P(\text{at least 1 in 20 min} | \text{2 in 1 hour}) = 1 - \frac{(\lambda \cdot \frac{2}{3})^2 \cdot e^{-(\lambda \cdot \frac{2}{3})}/2!}{(\lambda \cdot 1)^2 \cdot e^{-(\lambda \cdot 1)}/2!} \) Simplify the expression: \(P(\text{at least 1 in 20 min} | \text{2 in 1 hour}) = 1 - \frac{2^2}{3^2} = 1 - \frac{4}{9} = \frac{5}{9} \) So, the probability that: (a) both customers arrived during the first 20 minutes is \(\frac{1}{9}\) (b) at least one customer arrived during the first 20 minutes is \(\frac{5}{9}\)

Key concepts

Probability Theory

Probability theory is a branch of mathematics concerned with the analysis of random phenomena. The central objects of probability theory are random variables, stochastic processes, and events: in our exercise, the event is customers arriving at a bank. Probability can range from 0 (impossibility) to 1 (certainty), representing the likelihood of occurrence of a given event.

When applying probability theory to the Poisson distribution scenario provided in the exercise, we realize that the distribution gives us a way to calculate the likelihood of a certain number of events happening within a fixed time interval or space area, given a constant average rate of occurrence. The Poisson distribution is especially useful for modeling events with a low probability of occurrence within a large population or time frame.

For instance, when customers arrive at a bank at a certain rate, we can use the Poisson distribution to predict the probability of a specific number of customers arriving in any given period of time. This helps in assessing the resources needed to handle customer traffic efficiently.

Event Independence

Event independence in probability theory means that the occurrence of one event does not affect the occurrence of another. This is a key assumption when working with the Poisson distribution, as it allows us to treat the number of events in disjoint or non-overlapping intervals as independent occurrences.

The exercise demonstrates event independence by considering the arrival of customers in separate time intervals. It's essential to note that the probability of customers arriving in the first 20 minutes is independent of the customers arriving in the next 40 minutes. This independence is crucial when we calculate the probability of both customers arriving during the first 20 minutes and the probability of at least one customer arriving in the same timeframe.

Without event independence, the calculations would become significantly more complex, as each event would be influenced by the occurrence of the others, necessitating a different approach than the Poisson distribution provides.

Mathematical Statistics

Mathematical statistics is the application of mathematics to the field of statistics, involving the collection, analysis, interpretation, and presentation of masses of numerical data. In our exercise, statistics help in making informed predictions about customer behavior at the bank.

Interpreting Poisson Distribution

In the context of the exercise, mathematical statistics allows for the interpretation of the Poisson distribution's results. For example, the probability of both customers arriving in the first 20 minutes was calculated as \(\frac{1}{9}\), which gives a quantitative measure of that event's likelihood. Similarly, understanding that the probability of at least one customer arriving during this period is \(\frac{5}{9}\) uses statistical techniques to predict outcomes in specific scenarios.

Decision-Making

These probabilities assist bank managers in making decisions, such as how many tellers are needed during different time intervals to manage the expected number of customers. Mathematical statistics thus aid in converting raw data into useful information for practical decision-making and resource allocation.