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 ?.?. (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) 0.1804.

Step 1: Define the Poisson Process

The number of customers arriving is modeled by a Poisson process with a rate of \( \lambda = 4 \) customers per hour. The probability of observing \( k \) customers in one hour is given by the Poisson probability mass function (PMF): \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] where \( \lambda = 4 \), \( k \) is the number of customers, \( e \) is Euler's number (approximately 2.71828), and \( k! \) is the factorial of \( k \).

Step 2: Calculate the probability of no customer between 2 and 3 P.M. (Part a)

For part (a), we need to calculate the probability that \( k = 0 \) customers arrive between 2 and 3 P.M. Using the Poisson PMF, we find:\[ P(X = 0) = \frac{e^{-4} \times 4^0}{0!} = e^{-4} \approx 0.0183 \] Thus, the probability is about 0.0183.

Step 3: Calculate the probability of exactly two customers between 3 and 4 P.M. (Part b)

For part (b), we want the probability that exactly \( k = 2 \) customers arrive between 3 and 4 P.M. Using the Poisson PMF again:\[ P(X = 2) = \frac{e^{-4} \times 4^2}{2!} = \frac{16e^{-4}}{2} \approx 0.1465 \] Thus, the probability is about 0.1465.

Step 4: Calculate the probability of exactly two customers between 2 and 4 P.M. (Part c)

For part (c), we want the probability that exactly \( k = 2 \) customers arrive in two hours (from 2 to 4 P.M.).The process rate for two hours is \( \lambda = 8 \), so:\[ P(Y = 2) = \frac{e^{-8} \times 8^2}{2!} = \frac{64e^{-8}}{2} \approx 0.0148 \] Thus, the probability is about 0.0148.

Step 5: Calculate conditional probability of both customers arriving between 3 and 4 P.M. (Part d)

For part (d), we are given that exactly 2 customers arrive between 2 and 4 P.M., and we want the probability that both are in the interval 3 to 4 P.M. We use the conditional probability formula:\[ P(X = 2 \mid Y = 2) = \frac{P(X = 2 \text{ and } Y = 2)}{P(Y = 2)} = \frac{P(X = 2) \times P(X = 0)}{P(Y = 2)} \] where \(P(X=0)\) for the first hour is calculated in Step 2 as approximately 0.0183, and \(P(X=2)\) for the second hour is approximately 0.1465, and \(P(Y=2)\) is approximately 0.0148 as calculated in Step 4.Thus:\[ P(X = 2 \mid Y = 2) = \frac{0.1465 \times 0.0183}{0.0148} \approx 0.1804 \]Therefore, the probability is approximately 0.1804.

Key Concepts

Probability Mass Function

A Probability Mass Function (PMF) is a mathematical function that gives the probability of a discrete random variable being exactly equal to some value. When dealing with events like customer arrivals, modeled as a Poisson process, the PMF formula is incredibly useful for various calculations. For a Poisson distribution, the PMF is given by: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]

• \( \lambda \) represents the average rate of occurrence/event—in this case, the average number of customers arriving per hour.
• \( k \) is the specific number of events (or customers) we are interested in.
• \( e \) is Euler's number, approximately 2.71828, which is a constant.
• \( k! \) is the factorial of \( k \), which is the product of all positive integers up to \( k \).

To solve problems using PMF, it's essential to know the average rate \( \lambda \) and the number of events \( k \). This function helps determine probabilities for different scenarios, such as no customers in an hour, or exactly two customers arriving in a given time frame.

Conditional Probability

Conditional Probability is the probability of an event occurring given that another event has already occurred. It's crucial when events are dependent on one another, which is often the case in sequential processes. For example, in customer arrival modeling, we might need to determine the probability of specific customer arrivals, given certain prior arrivals. The formula for conditional probability is: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]

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

In our exercise, this concept is applied to find probabilities when customers arrive in different time slots but are in part dependent on one another through the overall arrival rate between two intervals.

Customer Arrival Modeling

Customer Arrival Modeling uses statistical tools to predict how customers will arrive over a period. The Poisson process is commonly utilized for such modeling due to its efficiency in dealing with random, independent occurrences over time, like customer arrivals. The key elements in customer arrival modeling include:

• The average arrival rate (\( \lambda \)), which guides the expected number of customers arriving.
• The time interval for which the prediction is being made.
• Using the Poisson PMF to determine probabilities of various arrival scenarios.

For instance, if you want to predict how many customers will arrive between 2 and 4 PM, you extend the average rate appropriately. In our exercise, the rate for a two-hour period is calculated as 8, assuming each hour independently has an average of 4 arrivals. This extension is critical for modeling longer time frames accurately.

Independent Events in Probability

In probability theory, independent events are those where the occurrence of one event does not affect the probability of another occurring. This concept is vital, especially in scenarios involving repetitive trials, like customer arrivals within different hours. Events are defined as independent if:

• The probability of both events occurring is the product of their individual probabilities.
• \( P(A \cap B) = P(A) \times P(B) \)

In the exercise, the number of customers arriving between 2 and 3 PM is independent of those arriving between 3 and 4 PM. This independence simplifies calculations significantly, especially when using conditional probabilities or extending the Poisson process over multiple intervals. Ensuring events are independent allows for straightforward multiplication of their individual probabilities to find joint probabilities.