Problem 217 - Applied Statistics and Probability for Engineers

Question

A square inch of carpeting contains 50 carpet fibers. The probability of a damaged fiber is \(0.0001 .\) Assume that the damaged fibers occur independently. (a) Approximate the probability of one or more damaged fibers in one square yard of carpeting. (b) Approximate the probability of four or more damaged fibers in one square yard of carpeting.

Step-by-Step Solution

Answer

(a) Approximately 0.9985. (b) Approximately 0.8522.

Step 1: Convert Units

First, we need to convert the area from square yards to square inches. Since 1 yard = 36 inches, 1 square yard is \[36^2 = 1296\] square inches. So, there are 1296 square inches in a square yard.

Step 2: Calculate Total Number of Fibers

Next, we calculate the total number of carpet fibers in a square yard. Given that each square inch contains 50 fibers, the total number of fibers in 1296 square inches is calculated as:\[1296 \times 50 = 64800\] fibers.

Step 3: Determine Expected Number of Damaged Fibers

The probability of a single fiber being damaged is 0.0001. To find the expected number of damaged fibers in the square yard, multiply the total number of fibers by this probability:\[64800 \times 0.0001 = 6.48\] Therefore, the expected number of damaged fibers is 6.48.

Step 4: Model the Problem with Poisson Distribution

Since we are dealing with a large number of trials (fibers) and a small probability, we use the Poisson distribution to approximate the probability of a certain number of damaged fibers.The parameter \(\lambda\) (mean of the Poisson distribution) is 6.48.

Step 5: Calculate Probability of One or More Damaged Fibers

The probability of one or more damaged fibers is 1 minus the probability of zero damaged fibers, given by the Poisson distribution:\[P(X \geq 1) = 1 - P(X = 0) = 1 - e^{-6.48} \cdot \frac{6.48^0}{0!} = 1 - e^{-6.48} \approx 0.9985\]Thus, the probability of one or more damaged fibers is approximately 0.9985.

Step 6: Calculate Probability of Four or More Damaged Fibers

Similarly, calculate the probability of four or more damaged fibers:\[P(X \geq 4) = 1 - (P(X=0) + P(X=1) + P(X=2) + P(X=3))\]Using the Poisson formula individually for each case:\[P(X=0) = e^{-6.48} \cdot \frac{6.48^0}{0!}\]\[P(X=1) = e^{-6.48} \cdot \frac{6.48^1}{1!}\]\[P(X=2) = e^{-6.48} \cdot \frac{6.48^2}{2!}\]\[P(X=3) = e^{-6.48} \cdot \frac{6.48^3}{3!}\]Finally, calculate and sum up these probabilities and subtract from 1. Performing this gives:\[P(X \geq 4) \approx 0.8522\]

Key Concepts

Poisson Distribution

The Poisson distribution is a vital tool in probability, especially when dealing with events that happen independently. It's often used to model the number of times an event occurs within a specific interval. For example, in this carpet fiber problem, we're examining how many fibers get damaged. When the probability of an individual event happening is low (like damaging a fiber) and you have a large number of trials (the total fibers), the Poisson distribution becomes handy. To use the Poisson distribution, you need a parameter \( \lambda \), which represents the average number of events (or damages, in this context) in the interval. In this problem, \( \lambda = 6.48 \), revealing damage occurs, on average, about 6.48 times in a square yard.

Expected Value

The expected value (`EX`) is simply the mean or average number of times an event should occur based on probabilities. For scenarios involving large sample sizes and small probabilities of success, like the damaged fibers, calculating the expected value can guide us on what to reasonably anticipate.In our case, the expected value of damaged fibers is found by multiplying the total fibers present by the probability of any single fiber being damaged. Thus, we have \[64800 \times 0.0001 = 6.48\]This tells us that, on average, we should expect about 6.48 damaged fibers in one square yard.

Probability Approximation

In probability, approximating allows us to make sense of complex situations. Using the Poisson distribution to approximate probabilities is common when dealing with many independent trials and low success probabilities. Steps involved in approximating:- Convert units to calculate total items or trials
- Compute the expected number of events
- Model with the Poisson distribution using \( \lambda \) as the expected number of eventsIn our exercise, the probabilities of specific scenarios (like 1 damaged fiber, or 4 or more) are approximated to make calculations feasible without direct enumeration of every outcome. This method simplifies the process significantly.

Independent Events

Independent events in probability mean that the occurrence of one event does not affect the occurrence of another. In the world of probability, the principle of independence simplifies calculations and applies to many real-world scenarios. Within our problem, damaged fibers are independent events. This suggests that if one fiber is damaged, it doesn't change the likelihood of another getting damaged. Independence is important, especially when: - Using models like the Poisson distribution, where you assume events occur randomly
- Making calculations simpler, since probabilities of independent events simply multiply Recognizing independence here helps underpin the assumption needed for Poisson distribution, making it a valid tool for this problem.