Problem 53 - Calculus for Biology and Medicine

Question

Suppose the number of customers per hour arriving at the post office is a Poisson process with an average of four customers per hour. (a) Find the probability that no customer arrives between 2 and 3 P.M. (b) Find the probability that exactly two customers arrive between 3 and 4 P.M. (c) Assuming that the number of customers arriving between 2 and 3 P.M. is independent of the number of customers arriving between 3 and 4 P.M., find the probability that exactly two customers arrive between 2 and 4 P.M. (d) Assume that the number of customers arriving between 2 and 3 P.M. is independent of the number of customers arriving between 3 and 4 P.M. Given that exactly two customers arrive between 2 and 4 P.M., what is the probability that both arrive between 3 and 4 P.M.?

Step-by-Step Solution

Answer

(a) 0.0183; (b) 0.1465; (c) 0.0148; (d) 0.1805.

Step 1: Define the Poisson Distribution

A Poisson process with an average rate of 4 customers per hour means that the number of customers arriving in a period follows a Poisson distribution with parameter \( \lambda = 4 \). The probability of observing \( k \) events (customers) in a time period is given by the formula: \[P(X = k) = \frac{\lambda^k \cdot e^{-\lambda}}{k!}\] where \( \lambda \) is the average number of events (4), \( k \) is the number of events for which we want the probability, and \( e \) is approximately 2.71828.

Step 2: (a) Probability that No Customer Arrives between 2 and 3 P.M.

To find the probability of no customer arriving between 2 and 3 P.M., set \( k = 0 \). Using the Poisson probability formula: \[P(X = 0) = \frac{4^0 \cdot e^{-4}}{0!} = e^{-4} \approx 0.0183\]So, the probability that no customer arrives is approximately 0.0183.

Step 3: (b) Probability of Exactly Two Customers between 3 and 4 P.M.

For exactly two customers arriving between 3 and 4 P.M., use \( k = 2 \): \[P(X = 2) = \frac{4^2 \cdot e^{-4}}{2!} = \frac{16 \cdot e^{-4}}{2} = 8 \cdot e^{-4} \approx 0.1465\]Hence, the probability that exactly two customers arrive is approximately 0.1465.

Step 4: Understand Independence between 2-3 P.M. and 3-4 P.M.

Given that the arrivals in each hour are independent of each other, the combined probability distribution can be considered separately for each hour, then combined for independent events.

Step 5: (c) Probability of Exactly Two Customers between 2 and 4 P.M.

Between 2 and 4 P.M. is a 2-hour period, so \( \lambda = 8 \) (since it's 4 per hour): \[P(Y = 2) = \frac{8^2 \cdot e^{-8}}{2!} = \frac{64 \cdot e^{-8}}{2} = 32 \cdot e^{-8} \approx 0.0148\]Thus, the probability of exactly two customers arriving in the two-hour period is approximately 0.0148.

Step 6: (d) Given Exactly Two Arrive, Probability Both are 3-4 P.M.

Consider that a total of 2 customers arrive between 2 and 4 P.M. Use the conditional probability formula. The probability that both arrive in the second hour is found using: \[\frac{P(2 \text{ in } 3-4 \text{ P.M.}) \cdot P(0 \text{ in } 2-3 \text{ P.M.})}{P(2 \text{ in } 2-4 \text{ P.M.})} = \frac{0.1465 \times 0.0183}{0.0148} \approx 0.1805\]Thus, given that exactly two arrive between 2 and 4 P.M., the probability both arrive between 3 and 4 P.M. is approximately 0.1805.

Key Concepts

Probability Distribution

A probability distribution is a mathematical function that provides the probabilities of different outcomes in an experiment. It’s a way to represent the likelihood of various outcomes quantitatively. When we talk about the Poisson distribution, it’s a specific type of probability distribution. It’s used for counting the number of events that happen in a fixed interval of time or space.
For example, if you want to calculate how many customers might visit a post office in an hour, the Poisson distribution is helpful. A key characteristic of the Poisson distribution is that it’s defined by a single parameter, the average rate (denoted by \( \lambda \)). This parameter represents the average number of events occurring in a time period.
The mathematical expression for the Poisson probability distribution is:

• \( P(X = k) = \frac{\lambda^k \cdot e^{-\lambda}}{k!} \)

This formula helps to find the probability of observing \(k\) events.
Let’s say you are dealing with a post office scenario where, on average, 4 customers visit per hour. Using the Poisson distribution, you can calculate the probability of observing exactly 0, 2, or any number of customers.

Independent Events

In probability, independent events are events whose outcomes do not affect each other. Understanding independence is crucial when dealing with multiple events occurring in a sequence of time intervals, like the customers arriving in different hours at a post office.
For example, when the time period is broken into several intervals, such as 2-3 P.M. and 3-4 P.M., the events occurring in each interval can be considered independent if the occurrence in one interval doesn't change the probability outcome of the other. This principle means that the number of customers arriving in one hour doesn’t influence the number of arrivals in the following hour.
In the context of the exercise, knowing that the customers arriving between these two different periods are independent allows us to simplify calculations. We can calculate probabilities for each period separately and then use these probabilities to find combined probabilities for the entire period, leveraging the notion of product of probabilities.

Conditional Probability

Conditional probability is a measure of the probability of an event occurring given that another event has already occurred. It helps us to refine our predictions based on known outcomes.
When calculating probabilities for events under specified conditions, conditional probability becomes an essential tool. For instance, given exactly two customers arrive between 2 and 4 P.M., you might want to find the probability that both customers arrive between 3 and 4 P.M.
The formula for computing conditional probability is:

• \( P(A | B) = \frac{P(A \cap B)}{P(B)} \)

Where \( P(A | B) \) is the probability of event \( A \) occurring given that \( B \) is true, and \( A \cap B \) represents the joint probability of both \( A \) and \( B \) occurring.
Using this concept, we determine the probability of specific distributions given known constraints, refining our understanding of outcomes based on preset conditions.