Estimating Probability Of Fewer Than 2 Defective Light Bulbs In A Large Group

Probability is a fascinating and crucial concept in mathematics, especially when dealing with scenarios involving chance and uncertainty. In this article, we will delve into a practical application of probability by exploring a scenario involving defective light bulbs. Specifically, we will address the problem of estimating the probability that there are fewer than two defective light bulbs in a large group, given a certain defect rate. This problem showcases how probability theory can be used to make predictions and assessments in real-world situations, such as quality control and risk management. The core of our analysis will revolve around using the Poisson distribution as an approximation for the binomial distribution, a technique that is particularly useful when dealing with rare events in large populations. This approach allows us to simplify the calculations and arrive at a reasonable estimate of the desired probability.

Defective light bulbs can be a significant concern for manufacturers and consumers alike. Imagine a scenario where it is known that 1 out of every 44,500 light bulbs produced is defective. Now, consider a large group of 177,500 light bulbs. Our objective is to estimate the likelihood that this group contains fewer than two defective light bulbs. This problem is a classic example of a situation where we can apply probability theory to make an informed estimation. The challenge lies in the fact that we are dealing with a relatively rare event (a defective light bulb) in a large population (177,500 light bulbs). This makes it computationally intensive to calculate the exact probability using the binomial distribution. Therefore, we will explore an alternative approach using the Poisson distribution, which provides a convenient and accurate approximation in such cases. This problem not only tests our understanding of probability concepts but also demonstrates the practical application of statistical methods in quality control and reliability analysis.

To tackle the problem of estimating the probability of fewer than two defective light bulbs, we will employ a powerful technique known as the Poisson approximation to the binomial distribution. This method is particularly effective when we are dealing with rare events occurring in a large number of trials, which perfectly describes our scenario. The binomial distribution, in its essence, calculates the probability of obtaining a specific number of successes (in our case, defective light bulbs) in a fixed number of independent trials (the 177,500 light bulbs), with each trial having the same probability of success (the defect rate of 1 in 44,500). However, when the number of trials is large and the probability of success is small, the binomial distribution can become computationally cumbersome. This is where the Poisson approximation comes to the rescue. The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event. In our context, the "events" are defective light bulbs, and the "fixed interval" is the group of 177,500 light bulbs. The key advantage of using the Poisson distribution is its simplicity and ease of calculation, especially when compared to the binomial distribution for large numbers of trials. By approximating the binomial distribution with the Poisson distribution, we can significantly reduce the computational complexity while still obtaining a highly accurate estimate of the probability.

Step 1: Calculate the Mean (λ)

The mean (λ) is a crucial parameter in the Poisson distribution, representing the average number of events (defective light bulbs) we expect to observe in our group. To calculate this mean, we multiply the number of trials (light bulbs) by the probability of success (defect rate). In our case, we have 177,500 light bulbs, and the probability of a light bulb being defective is 1 in 44,500. Therefore, the mean (λ) is calculated as follows:

λ = (Number of light bulbs) × (Probability of being defective) λ = 177,500 × (1 / 44,500) λ = 4

This result tells us that, on average, we expect to find 4 defective light bulbs in a group of 177,500, given the defect rate of 1 in 44,500. The mean (λ) serves as the central value around which the Poisson distribution is centered, and it plays a vital role in determining the probabilities of different numbers of defective light bulbs.

Step 2: Poisson Probability Formula

Now that we have calculated the mean (λ), we can utilize the Poisson probability formula to determine the probability of observing a specific number of defective light bulbs. The Poisson probability formula is given by:

P(x) = (e^(-λ) * λ^x) / x!

Where:

• P(x) is the probability of observing x events (defective light bulbs).
• e is the base of the natural logarithm (approximately 2.71828).
• λ is the mean number of events (4 in our case).
• x is the number of events we are interested in.
• x! is the factorial of x.

This formula allows us to calculate the probability of any specific number of defective light bulbs, from 0 to any positive integer. In our problem, we are interested in the probability of having fewer than two defective light bulbs, which means we need to calculate the probabilities for x = 0 and x = 1 and then add them together. This will give us the overall probability of having either zero or one defective light bulb in the group.

Step 3: Calculate P(0) and P(1)

To determine the probability of having fewer than two defective light bulbs, we need to calculate P(0), the probability of having zero defective light bulbs, and P(1), the probability of having one defective light bulb. We will use the Poisson probability formula we introduced earlier:

P(x) = (e^(-λ) * λ^x) / x!

For P(0), we substitute x = 0 and λ = 4:

P(0) = (e^(-4) * 4^0) / 0!

Since any number raised to the power of 0 is 1, and 0! is defined as 1, we have:

P(0) = e^(-4)

P(0) ≈ 0.0183

This means that there is approximately a 1.83% chance of finding no defective light bulbs in the group.

Next, we calculate P(1), the probability of having one defective light bulb, by substituting x = 1 and λ = 4:

P(1) = (e^(-4) * 4^1) / 1!

P(1) = (e^(-4) * 4) / 1

P(1) = 4 * e^(-4)

P(1) ≈ 4 * 0.0183

P(1) ≈ 0.0733

This indicates that there is roughly a 7.33% chance of finding exactly one defective light bulb in the group. These individual probabilities are crucial components in determining the overall probability of having fewer than two defective light bulbs.

Step 4: Calculate the Final Probability

To find the final probability of having fewer than two defective light bulbs, we simply add the probabilities of having zero defective light bulbs (P(0)) and one defective light bulb (P(1)). This is because these two events are mutually exclusive; you cannot have both zero and one defective light bulb at the same time. Therefore, the probability of either event occurring is the sum of their individual probabilities:

P(fewer than 2 defective) = P(0) + P(1)

We previously calculated P(0) ≈ 0.0183 and P(1) ≈ 0.0733. Now, we add these probabilities:

P(fewer than 2 defective) ≈ 0.0183 + 0.0733

P(fewer than 2 defective) ≈ 0.0916

Therefore, the estimated probability of having fewer than two defective light bulbs in a group of 177,500 is approximately 0.0916, or 9.16%. This result provides a quantitative measure of the likelihood of encountering a small number of defective light bulbs in the batch, given the specified defect rate. This type of analysis is invaluable in quality control processes, allowing manufacturers to assess the risk of defective products and make informed decisions about production and inspection procedures.

Our analysis has revealed that the estimated probability of finding fewer than two defective light bulbs in a group of 177,500, given a defect rate of 1 in 44,500, is approximately 9.16%. This result is a testament to the power of the Poisson approximation in simplifying complex probability calculations. By using the Poisson distribution, we were able to efficiently estimate the probability of a rare event (a defective light bulb) occurring in a large population. The calculated probability of 9.16% suggests that it is relatively unlikely to find fewer than two defective light bulbs in such a large group. Conversely, this implies that there is a higher probability of finding two or more defective light bulbs. This information is crucial for manufacturers as it provides insights into the expected quality of their products. A probability of 9.16% for fewer than two defective items indicates a need for stringent quality control measures to minimize the occurrence of defects. The Poisson approximation not only simplifies calculations but also provides a valuable tool for decision-making in various fields, including manufacturing, healthcare, and finance, where assessing the probability of rare events is paramount.

In conclusion, this article has demonstrated the application of probability theory, specifically the Poisson approximation, to estimate the likelihood of finding fewer than two defective light bulbs in a large group. We successfully navigated the problem by first calculating the mean number of defective light bulbs (λ) and then utilizing the Poisson probability formula to determine the probabilities of having zero and one defective light bulb. By summing these probabilities, we arrived at an estimated probability of 9.16% for having fewer than two defective light bulbs in the group. This result underscores the effectiveness of the Poisson distribution in approximating binomial probabilities when dealing with rare events in large populations. The methodology we employed can be extended to a wide range of real-world scenarios where estimating the probability of rare events is crucial, such as in quality control, risk management, and reliability analysis. The ability to make such estimations is invaluable for making informed decisions and implementing effective strategies to mitigate potential risks. Furthermore, this exercise highlights the importance of understanding and applying probability concepts in practical settings, showcasing the versatility and power of mathematics in solving real-world problems.