Problem 54 - 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 five customers per hour. (a) Find the probability that exactly one 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 three 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 three customers arrive between 2 and 4 p.m., what is the probability that one arrives between 2 and 3 p.m. and two between 3 and 4 P.m.?

Step-by-Step Solution

Answer

(a) 0.0337, (b) 0.0842, (c) 0.0076, (d) 0.3737.

Step 1: Understand the Poisson Distribution

The Poisson distribution describes the probability of a given number of events happening in a fixed interval of time. The probability mass function (PMF) is given by \( P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \), where \( \lambda \) is the average rate, and \( k \) is the number of events. In this case, \( \lambda = 5 \) customers per 1-hour interval.

Step 2: Probability for Exactly One Customer (2-3 P.M.)

For part (a), between 2 and 3 P.M., we need \( P(X = 1) \) where \( \lambda = 5 \). Use the formula: \[ P(X = 1) = \frac{e^{-5} \cdot 5^1}{1!} = 5 \cdot e^{-5} \approx 0.0337 \]

Step 3: Probability for Exactly Two Customers (3-4 P.M.)

For part (b), between 3 and 4 P.M., we need \( P(X = 2) \) with \( \lambda = 5 \):\[ P(X = 2) = \frac{e^{-5} \cdot 5^2}{2!} = \frac{25 \cdot e^{-5}}{2} \approx 0.0842 \]

Step 4: Probability for Exactly Three Customers (2-4 P.M.)

For part (c), considering the period 2-4 P.M. as one interval with \( \lambda = 10 \) (since 5 per hour for 2 hours):\[ P(X = 3) = \frac{e^{-10} \cdot 10^3}{3!} = \frac{1000 \cdot e^{-10}}{6} \approx 0.0076 \]

Step 5: Conditional Probability (Part d)

For part (d), using the probability for exactly 1 customer from 2-3 P.M. and 2 customers from 3-4 P.M.:\[ P(X_{2-3}=1 \text{ and } X_{3-4}=2) = P(X_{2-3}=1) \cdot P(X_{3-4}=2) \approx 0.0337 \times 0.0842 = 0.00284 \]Then, the conditional probability is:\[ P(X_{2-3}=1 \text{ and } X_{3-4}=2 | X_{2-4} = 3) = \frac{P(X_{2-3}=1 \text{ and } X_{3-4}=2)}{P(X_{2-4} = 3)} \approx \frac{0.00284}{0.0076} \approx 0.3737 \]

Key Concepts

Probability Mass Function

When dealing with a Poisson distribution, the Probability Mass Function (PMF) is an essential concept. It provides a way to find the likelihood of a certain number of events occurring within a given time frame. In simpler terms, PMF helps us to understand how likely it is for a specific number of customers to arrive at the post office within one hour.

The PMF for a Poisson distribution is mathematically expressed as \( P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \). In this equation:

• \( P(X = k) \) represents the probability of observing exactly \( k \) events (like customer arrivals).
• \( \lambda \) is the average number of events per interval, also known as the expected rate. For example, if we expect 5 customers per hour, then \( \lambda = 5 \).
• \( e \) is the base of natural logarithms, approximately equal to 2.718.
• \( k! \) denotes the factorial of \( k \), which is the product of all positive integers less than or equal to \( k \).

Using the PMF, you can calculate how likely it is for exactly one customer to arrive within an hour, or for any other specific number of customers. This is crucial for scheduling and managing expectations in services like post offices.

Conditional Probability

Conditional probability helps us understand the likelihood of an event occurring, given that another event has already happened. It's like asking, "What are the odds of another event, knowing what we know now?" In the context of our Poisson distribution exercise, conditional probability is important when you want to predict customer arrivals over more complex time frames.

For instance, suppose we already know that exactly three customers arrived between 2 and 4 P.M. The question becomes: what are the chances that one customer arrived between 2 and 3 P.M., and two customers arrived between 3 and 4 P.M.?

This can be determined using the formula for conditional probability \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), where:

• \( P(A|B) \) is the probability of event A given that B has occurred.
• \( P(A \cap B) \) is the probability that both A and B occur.
• \( P(B) \) is the probability of event B.

By applying this formula, you can dynamically estimate occurrences, making conditional probability a powerful tool in statistics and decision-making.

Independent Events

In statistics, identifying independent events is crucial for accurately calculating probabilities. Two events are said to be independent if the occurrence of one does not influence the probability of the other occurring.

For the post office scenario, assume the number of customers arriving between 2 and 3 P.M. is independent of the number arriving between 3 and 4 P.M. This means that knowing how many came during the first hour gives no clue about the second hour.

Mathematically, events A and B are independent if \( P(A \cap B) = P(A) \cdot P(B) \). This independence simplifies calculations without needing to consider complex associations between the events.

Understanding independent events allows for clear-cut modeling in tasks such as planning and resource allocation, as what happens in one time frame is isolated from the others. This knowledge streamlines strategic decisions for industries relying on time-based customer interactions, like the postal service.