Probability Of Defective Light Bulbs Calculation And Explanation

Jul 10, 2025 by Admin 65 views

Probability of Defective Light Bulbs in a Shipment

Let's delve into the probability problem concerning defective light bulbs. In this scenario, we have a factory producing light bulbs, and there's a known probability of 3% that any given bulb will be defective. The question we aim to answer is: What is the probability that exactly 2 out of a shipment of 30 light bulbs will be defective? This is a classic probability question that can be solved using the binomial probability formula. Before we jump into the calculations, let's understand why this is a binomial problem and the components involved.

Identifying a Binomial Problem

This problem fits the criteria for a binomial probability distribution because it meets the following conditions:

Fixed Number of Trials: We have a fixed number of trials, which is 30 (the number of light bulbs in the shipment).
Independent Trials: Each light bulb's condition (defective or not) is independent of the others. One bulb being defective doesn't influence the probability of another being defective.
Two Possible Outcomes: There are only two outcomes for each trial: the bulb is either defective or not defective. This is often referred to as success or failure.
Constant Probability of Success: The probability of a bulb being defective (success) is constant at 3% or 0.03 for each bulb.

Key Components of the Binomial Formula

To solve this problem, we will use the binomial probability formula. This formula helps us calculate the probability of getting exactly k successes in n trials, given the probability of success p on a single trial. The formula is as follows:

P(X = k) = {n race k} * p^k * (1 - p)^{(n - k)}

Where:

P(X = k) is the probability of exactly k successes in n trials
${n race k}$ is the binomial coefficient, which represents the number of ways to choose k successes from n trials. It's also known as "n choose k" and can be calculated as n! / (k!(n - k)!), where ! denotes factorial.
p is the probability of success on a single trial
n is the number of trials
k is the number of successes we want to find the probability for
(1 - p) is the probability of failure on a single trial

Now that we've established the context and the formula, let's identify the values for each component in our specific problem:

n (number of trials) = 30 (light bulbs)
k (number of successes) = 2 (defective bulbs)
p (probability of success) = 0.03 (probability of a bulb being defective)
(1 - p) (probability of failure) = 1 - 0.03 = 0.97 (probability of a bulb not being defective)

With these values in hand, we can now plug them into the binomial probability formula and calculate the probability of exactly 2 out of 30 light bulbs being defective.

Now, let's apply the binomial probability formula to our problem. We have all the necessary components, so it's just a matter of plugging in the values and performing the calculations.

Calculating the Binomial Coefficient

First, we need to calculate the binomial coefficient, ${n race k}$ , which represents the number of ways to choose 2 defective bulbs from a shipment of 30. Using the formula:

{n race k} = n! / (k!(n - k)!)

We get:

{30 race 2} = 30! / (2!(30 - 2)!) = 30! / (2! * 28!)

Calculating the factorials and simplifying:

{30 race 2} = (30 * 29 * 28!) / (2 * 1 * 28!) = (30 * 29) / 2 = 435

So, there are 435 different ways to choose 2 defective light bulbs from a shipment of 30.

Plugging Values into the Formula

Now we can plug all the values into the binomial probability formula:

P(X = 2) = {30 race 2} * (0.03)^2 * (0.97)^{(30 - 2)}

Substituting the values:

P(X = 2) = 435 * (0.03)^2 * (0.97)^{28}

Performing the Calculation

Let's break down the calculation step by step:

Calculate $(0.03)^2$ : This is simply 0.03 multiplied by itself, which equals 0.0009.
Calculate $(0.97)^{28}$ : This is 0.97 raised to the power of 28. This might require a calculator with exponential functions. The result is approximately 0.4198.
Multiply all the components: Now, we multiply the binomial coefficient, the probability of success squared, and the probability of failure raised to the power of 28:
$P(X = 2) = 435 * 0.0009 * 0.4198$ $P(X = 2) ≈ 0.1642$

Therefore, the probability of exactly 2 out of 30 light bulbs being defective is approximately 0.1642.

Our calculation shows that the probability of finding exactly 2 defective light bulbs in a shipment of 30, given a 3% defect rate, is approximately 0.1642. This means there's about a 16.42% chance of this occurring. Now, let's consider this result in the context of the problem and discuss its implications.

Comparing with Answer Choices

The original question provided answer choices. Let's see how our calculated probability compares to those:

A. 0.25
B. 0.0016
C. 0.0035
D. 0.167

Our calculated probability of 0.1642 is closest to option D, 0.167. This confirms that our calculations are accurate and that option D is the correct answer.

Implications of the Probability

Understanding this probability can be useful in several ways. For the factory, it helps in quality control. Knowing the probability of a certain number of defective bulbs in a shipment allows them to make informed decisions about quality assurance processes. For example, if the probability of 2 defective bulbs is considered too high, they might implement stricter quality control measures.

From a consumer perspective, this probability gives an idea of what to expect when purchasing light bulbs from this factory. While a 3% defect rate might seem low, the probability of getting exactly 2 defective bulbs in a shipment of 30 is not negligible. This highlights the importance of testing light bulbs upon purchase and having a warranty or return policy in case of defects.

Factors Affecting the Probability

It's also important to consider factors that might affect this probability. The 3% defect rate is an average, and the actual defect rate might vary depending on factors like the manufacturing process, the quality of materials, and the handling of the bulbs during shipping. If any of these factors change, the probability of finding 2 defective bulbs in a shipment could also change.

In conclusion, by applying the binomial probability formula, we were able to calculate the probability of finding exactly 2 defective light bulbs in a shipment of 30. This probability, approximately 0.1642, provides valuable information for both the manufacturer and the consumer, and understanding the factors that can influence this probability is crucial for making informed decisions.

In summary, the problem of calculating the probability of defective light bulbs in a shipment is a classic example of a binomial probability problem. By understanding the conditions for a binomial distribution and applying the binomial probability formula, we can accurately calculate the likelihood of specific outcomes. In this case, the probability of finding exactly 2 defective light bulbs in a shipment of 30, given a 3% defect rate, is approximately 0.1642, making option D (0.167) the correct answer.

This type of probability calculation has practical implications for quality control in manufacturing, consumer expectations, and risk assessment. It highlights the importance of understanding probability concepts in real-world scenarios. The binomial probability formula is a powerful tool for analyzing situations with binary outcomes and can be applied in various fields beyond manufacturing, such as healthcare, finance, and marketing.

Question: Light bulbs manufactured at a certain factory have a 3% probability of being defective. What is the probability that 2 out of a shipment of 30 will be defective?

Answer: D. 0.167