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) 0.0671, (d) 0.532

Step 1: Explain Poisson Distribution

The Poisson distribution is used to calculate the probability of a given number of events happening in a fixed interval of time. The formula is \( P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \), where \( \lambda \) is the average rate of success (5 in this case), \( k \) is the number of occurrences we want to find the probability for, and \( e \) is approximately 2.71828.

Step 2: Calculate Probability for Part (a)

To find the probability that exactly one customer arrives between 2 and 3 P.M., we use \( \lambda = 5 \) and \( k = 1 \). The formula becomes \( P(X = 1) = \frac{e^{-5} \cdot 5^1}{1!} = 5e^{-5} \). Using a calculator, we find \( P(X = 1) \approx 0.0337 \).

Step 3: Calculate Probability for Part (b)

To find the probability that exactly two customers arrive between 3 and 4 P.M., use \( \lambda = 5 \) and \( k = 2 \). The formula becomes \( P(X = 2) = \frac{e^{-5} \cdot 5^2}{2!} = \frac{25e^{-5}}{2} \). Simplifying gives \( P(X = 2) \approx 0.0842 \).

Step 4: Combine Events for Part (c)

The probability of exactly three customers arriving between 2 and 4 P.M. is a sum of independent events: one customer in the first hour and two in the second, or vice versa, or all three in one hour. We calculate \( P(1,2) = P(1)*P(2) + P(2)*P(1) + P(3,0) + P(0,3)\). However, the simplest calculation comes from directly using \( k=3 \) for the 2-hour interval, \( \lambda = 10 \).\[ P(X = 3) = \frac{e^{-10} \cdot 10^3}{3!} = \frac{1000e^{-10}}{6} \approx 0.0671 \].

Step 5: Conditional Probability for Part (d)

Given that exactly three total customers arrive between 2 and 4 P.M., we use the probabilities calculated in Step 2 and Step 3. The conditional probability is \( P(1,2|3) = \frac{P(1)*P(2)}{P(3)} \). From the previous steps, \( P(1,2) = 0.0357 \) and \( P(3) = 0.0671 \). Therefore, \( P(1,2|3) = \frac{0.0357}{0.0671} \approx 0.532 \).

Key Concepts

Probability Calculation

In the context of the Poisson distribution, probability calculation plays a crucial role in determining the likelihood of a specific number of events occurring within a set period of time. The Poisson distribution is particularly useful because it models situations where events occur independently of each other, with a known average rate. The key formula for calculating probability in a Poisson distribution is:\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]Here, \( \lambda \) represents the average number of occurrences (also called the distribution's parameter or "rate"), \( k \) denotes the number of events you want to find the probability for, and \( e \) is the base of natural logarithms, approximately equal to 2.71828. This formula lets you determine the probability of observing \( k \) number of events in a fixed interval, given the average rate. You simply plug in the values for \( \lambda \) and \( k \), carry out the calculations, and find the probability.For instance, if the post office experiences an average of five customers per hour, you can calculate the probability of exactly one customer or exactly two customers using the above formula by substituting \( \lambda = 5 \) and \( k = 1 \) or \( k = 2 \), respectively.

Independent Events

Independent events are a cornerstone concept when dealing with probabilities. In mathematical terms, two events are considered independent if the occurrence of one does not affect the probability of the other occurring. In the case of the post office example, we assume that the number of customers arriving between 2 and 3 P.M. is independent of the number between 3 and 4 P.M.Mathematically, if two events \( A \) and \( B \) are independent, the probability of both events occurring together is just the product of their individual probabilities:\[ P(A \cap B) = P(A) million P(B) \]In our exercise, this means we can calculate the combined probability of customers arriving in separate hourly intervals independently. Thus, if the probability of one customer arriving between 2 and 3 P.M. is \( P(1) \) and the probability of two customers arriving between 3 and 4 P.M. is \( P(2) \), the combined probability for these independent events would simply be multiplied together.This property is particularly handy in complex problems as it simplifies the calculation of probabilities when dealing with more than one event.

Conditional Probability

Conditional probability is the probability of an event occurring given that another event has already occurred. This involves understanding the dependence of one event on the occurrence of another, contrasting with independent events which do not influence each other.The formula for conditional probability is expressed as follows:\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]This formula indicates the probability of event \( A \) happening given that event \( B \) has happened. In the problem provided, this pertains to finding the likelihood of one customer arriving between 2 and 3 P.M. while two arrive between 3 and 4 P.M., given three customers arrive in total between 2 and 4 P.M.Using values calculated previously, this conditional probability is determined by dividing the joint probability of both events by the total probability of the observed events happening (i.e., exactly three customers arriving within the two-hour interval). This provides insight into how specific scenarios affect the likelihood of certain outcomes in a random event context. For example, knowing that three customers arrived, the interest lies in the probability distribution between the intervals, using conditional probability.