5. Accidents on a certain stretch of I-10 occur according to a Poisson process with an average of 3 every 2 hours. Find the probability that more than one accident occur during a particular 1-hour period."
5. Accidents on a certain stretch of I-10 occur according to a Poisson process with an average of 3 every 2 hours. Find the probability that more than one accident occur during a particular 1-hour period."
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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![**Problem Statement:**
Accidents on a certain stretch of I-10 occur according to a Poisson process with an average of 3 every 2 hours. Find the probability that more than one accident occurs during a particular 1-hour period.
**Explanation:**
This problem involves understanding and applying the Poisson distribution, which is commonly used to model the number of events occurring within a fixed interval of time or space. The key elements of this problem are:
- **Average Rate of Occurrence (λ):** Accidents occur with an average of 3 every 2 hours.
- **Adjusted Rate for 1 Hour:** Since we need to find the probability for a 1-hour period, we adjust the rate to 1.5 accidents per hour (half of 3).
- **Objective:** Calculate the probability of more than one accident in 1 hour using the adjusted rate.
**Steps to Solve:**
To solve, you can use the formula for the Poisson probability mass function (PMF):
\[ P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \]
where:
- \( X \) is a random variable representing the number of accidents.
- \( \lambda \) is the average rate of occurrence (1.5 in this case).
- \( k \) is the number of occurrences (number of accidents).
For more than one accident (i.e., \( P(X > 1) \)), calculate:
\[ P(X > 1) = 1 - (P(X = 0) + P(X = 1)) \]
1. **Calculate \( P(X = 0) \):**
\[ P(X = 0) = \frac{e^{-1.5} \cdot 1.5^0}{0!} = e^{-1.5} \]
2. **Calculate \( P(X = 1) \):**
\[ P(X = 1) = \frac{e^{-1.5} \cdot 1.5^1}{1!} = 1.5 \cdot e^{-1.5} \]
3. **Calculate \( P(X > 1) \):**
\[ P(X > 1) = 1 - (P(X = 0) + P(X = 1)) = 1 - (e^{-1.5} + 1.5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ff7519c-27e7-46e8-b0a7-228922513039%2F8a8802a8-a45f-46af-afa5-722172b5d366%2F2wp0lov_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Accidents on a certain stretch of I-10 occur according to a Poisson process with an average of 3 every 2 hours. Find the probability that more than one accident occurs during a particular 1-hour period.
**Explanation:**
This problem involves understanding and applying the Poisson distribution, which is commonly used to model the number of events occurring within a fixed interval of time or space. The key elements of this problem are:
- **Average Rate of Occurrence (λ):** Accidents occur with an average of 3 every 2 hours.
- **Adjusted Rate for 1 Hour:** Since we need to find the probability for a 1-hour period, we adjust the rate to 1.5 accidents per hour (half of 3).
- **Objective:** Calculate the probability of more than one accident in 1 hour using the adjusted rate.
**Steps to Solve:**
To solve, you can use the formula for the Poisson probability mass function (PMF):
\[ P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!} \]
where:
- \( X \) is a random variable representing the number of accidents.
- \( \lambda \) is the average rate of occurrence (1.5 in this case).
- \( k \) is the number of occurrences (number of accidents).
For more than one accident (i.e., \( P(X > 1) \)), calculate:
\[ P(X > 1) = 1 - (P(X = 0) + P(X = 1)) \]
1. **Calculate \( P(X = 0) \):**
\[ P(X = 0) = \frac{e^{-1.5} \cdot 1.5^0}{0!} = e^{-1.5} \]
2. **Calculate \( P(X = 1) \):**
\[ P(X = 1) = \frac{e^{-1.5} \cdot 1.5^1}{1!} = 1.5 \cdot e^{-1.5} \]
3. **Calculate \( P(X > 1) \):**
\[ P(X > 1) = 1 - (P(X = 0) + P(X = 1)) = 1 - (e^{-1.5} + 1.5
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