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Magazine Chapter 4: Problem 159
Emergency room bed availability. The mean number of patients admitted per day to the emergency room of a small hospital is \(2.7 .\) If, on a given day, there are only four beds available for new patients, what is the probability that the hospital will not have enough beds to accommodate its newly admitted patients?
Short Answer
Expert verified
The probability is approximately 13.9%.
Step by step solution
01
Understand the Problem
We are given that the mean number of patients admitted per day is 2.7, which suggests a Poisson distribution, as it deals with counting the number of events in a fixed interval. We need to find the probability that more than 4 patients are admitted since there are 4 beds available.
02
Define the Random Variable and Distribution
Let the random variable \( X \) represent the number of patients admitted on any given day. Since this is modeled by a Poisson distribution, \( X \sim \text{Poisson}(\lambda) \), where \( \lambda = 2.7 \).
03
Set up the Probability Expression
We need to calculate \( P(X > 4) \), which is the probability of the hospital admitting more than 4 patients in a day.
04
Use the Complement Rule
Using the complement rule, \( P(X > 4) = 1 - P(X \leq 4) \). This is because, in a probability distribution, all possible events sum to 1.
05
Calculate \( P(X \leq 4) \) Using the Poisson Formula
The probability mass function for a Poisson distribution is \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \] for \( k = 0, 1, 2, \ldots \). Calculate \( P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \).
06
Calculate Individual Probabilities
Calculate each probability: \( P(X = 0) = \frac{e^{-2.7} \times 2.7^0}{0!} \approx 0.067 \), \[ P(X = 1) = \frac{e^{-2.7} \times 2.7^1}{1!} \approx 0.181 \],\[ P(X = 2) = \frac{e^{-2.7} \times 2.7^2}{2!} \approx 0.244 \],\[ P(X = 3) = \frac{e^{-2.7} \times 2.7^3}{3!} \approx 0.220 \],\[ P(X = 4) = \frac{e^{-2.7} \times 2.7^4}{4!} \approx 0.149 \].
07
Sum the Probabilities
Sum the probabilities calculated: \[ P(X \leq 4) = 0.067 + 0.181 + 0.244 + 0.220 + 0.149 \approx 0.861. \]
08
Compute the Final Probability
Using the complement rule, find the probability that the hospital doesn't have enough beds: \[ P(X > 4) = 1 - 0.861 = 0.139. \] The probability is therefore approximately 0.139.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability Calculation
Probability calculation involves determining the likelihood of specific events occurring. In contexts like the emergency room scenario, this includes understanding how often an event, such as admitting more patients than available beds, may occur.
For calculating the probability of a Poisson random variable, we employ the Poisson probability mass function (PMF). This function helps us determine probabilities when the event count follows a Poisson distribution.
To find the probability that a hospital admits more patients than its bed count, we seek to calculate the likelihood of having more than four patient admissions. This requires first calculating the probability for each number of admissions up to four, then using these to find the probability of the complementary event鈥攖hat more than four patients are admitted.
For calculating the probability of a Poisson random variable, we employ the Poisson probability mass function (PMF). This function helps us determine probabilities when the event count follows a Poisson distribution.
To find the probability that a hospital admits more patients than its bed count, we seek to calculate the likelihood of having more than four patient admissions. This requires first calculating the probability for each number of admissions up to four, then using these to find the probability of the complementary event鈥攖hat more than four patients are admitted.
- The Poisson PMF formula: \[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]
- The probabilities: Sum the probabilities for all events up to four patients to find the cumulative probability, and then use the complementary event.
Complement Rule
The complement rule is a fundamental principle in probability, allowing us to calculate the probability of an event by considering everything else that can occur.
This is particularly useful when it's easier to calculate the probability of the complementary event. In probability, all possible outcomes sum up to one because they cover all potential scenarios.
In the emergency room problem, calculating the probability of admitting more than four patients involves using the complement rule. Rather than directly computing this probability, we calculate the probability of not exceeding four admissions, and subtract from one to find the probability of exceeding.
This is particularly useful when it's easier to calculate the probability of the complementary event. In probability, all possible outcomes sum up to one because they cover all potential scenarios.
In the emergency room problem, calculating the probability of admitting more than four patients involves using the complement rule. Rather than directly computing this probability, we calculate the probability of not exceeding four admissions, and subtract from one to find the probability of exceeding.
- Complement formula:\[ P(X > 4) = 1 - P(X \leq 4) \]
- This approach reduces complication by focusing first on straightforward calculations.
Random Variable Modeling
When dealing with events such as patient admissions in an emergency room, modeling with a random variable is essential. A random variable represents different outcomes of a random process.
In our scenario, the number of daily admissions can vary, thus we model it with a random variable \(X\). Drawing on the traits of a Poisson distribution, which is suited for counting events in a fixed interval with a known average rate, we assign \(\lambda = 2.7\) for the average admissions per day.
Utilizing the random variable in this way helps us predict outcomes and manage resources effectively. Understanding the distribution supporting the random variable allows administrators to anticipate different patient loads and prepare accordingly.
In our scenario, the number of daily admissions can vary, thus we model it with a random variable \(X\). Drawing on the traits of a Poisson distribution, which is suited for counting events in a fixed interval with a known average rate, we assign \(\lambda = 2.7\) for the average admissions per day.
Utilizing the random variable in this way helps us predict outcomes and manage resources effectively. Understanding the distribution supporting the random variable allows administrators to anticipate different patient loads and prepare accordingly.
- The Poisson distribution characteristics:
- Counts events over a fixed period
- Good for independent events, like arrivals
- Application: By modeling admissions, predictions about busy days and potential overflow can be made.
