Problem 26 - Algebra and Trigonometry

Question

Quality Control An assembly line that manufactures fuses for automotive use is checked every hour to ensure the quality of the finished product. Ten fuses are selected randomly, and if any one of the ten is found to be defective, the process is halted and the machines are recalibrated. Suppose that at a certain time 5\(\%\) of the fuses being produced are actually defective. What is the probability that the assembly line is halted at that hour's quality check?

Step-by-Step Solution

Answer

The probability that the line is halted is approximately 0.4013.

Step 1: Define the Problem

We need to find the probability that at least one fuse among the 10 selected is defective, given that there's a 5% chance of any individual fuse being defective.

Step 2: Calculate the Probability of a Fuse Being Non-defective

The probability of a single fuse being non-defective is the complement of being defective. So, if a fuse has a 5% chance of being defective, then it has a 95% chance of being non-defective. This is calculated as: \[ P( ext{non-defective}) = 1 - 0.05 = 0.95 \]

Step 3: Calculate the Probability of All Fuses Being Non-defective

Assuming each test is independent, the probability that all 10 fuses in the hour's sample are non-defective can be calculated by raising the probability of a single fuse being non-defective to the power of 10: \[ P( ext{all non-defective}) = 0.95^{10} \]

Step 4: Calculate the Probability of the Line Being Halted

The assembly line is halted if at least one fuse is defective. This probability is the complement of all fuses being non-defective. Thus, it's calculated by:\[ P( ext{at least one defective}) = 1 - P( ext{all non-defective}) = 1 - 0.95^{10} \]

Step 5: Compute the Final Probability

Now, calculate the numerical value for the probability:\[ P( ext{at least one defective}) = 1 - 0.95^{10} \approx 1 - 0.5987 \approx 0.4013 \]

Key Concepts

Quality Control

Quality control is an essential process used in various industries to ensure the products manufactured meet certain standards. In our given scenario, an assembly line producing automotive fuses is subjected to regular checks every hour. This quality control measure ensures that only high-quality and functional fuses reach the consumers.

Here's how the process works in the context of our exercise:

• Every hour, a sample of 10 fuses is randomly selected from the production line.
• If any single fuse in this sample is found to be defective, the entire assembly process is paused.
• The machines responsible for producing the fuses are then recalibrated to fix any potential issues.

These regular checks help maintain the manufacturer's reputation by preventing defective products from reaching the market and causing dissatisfaction or vehicle malfunction.

Complement Rule

The complement rule is a foundational concept in probability, which proves incredibly useful in a variety of scenarios, especially when calculating probabilities of at least one event occurring.

Let's explore how this rule is applied in our exercise:

• The probability of an event happening (like at least one defective fuse being found) is often easier to calculate using its complement.
• The complement of an event is everything not included in the event. For example, for "at least one fuse being defective," the complement is "all fuses being non-defective."

Using the complement rule, we find our desired probability with this formula:
\[ P(\text{at least one defective}) = 1 - P(\text{all non-defective}) \]

By calculating the probability that all 10 fuses are non-defective (which is found to be roughly 0.5987), we subtract this value from 1 to discover that there is approximately a 0.4013 probability of finding at least one defective fuse.

Independent Events

In probability, when we say that events are independent, we mean that the outcome of one event does not influence the outcome of another.

In relation to the exercise at hand:

• The probability of each individual fuse being non-defective is 0.95, regardless of what happens with the other fuses in the sample.
• When calculating the probability of all 10 fuses being non-defective, we assume these probabilities are independent. Thus, we multiply the probability of one non-defective fuse by itself for each fuse:

\[ P(\text{all non-defective}) = 0.95^{10} \] The reliance on the independence of events simplifies the calculation process significantly, allowing us to use power rules to compute probabilities more efficiently. This underscores the importance of understanding how independent events work, especially in highly controlled environments like manufacturing.