Question

Improperly wired control panels were mistakenly installed on two of eight large automated machine tools. It is uncertain which of the machine tools have the defective panels, and a sample of four tools is randomly chosen for inspection. What is the probability that the sample will include no defective panels? Both defective panels?

Short answer

Answer: The probability of selecting a sample with no defective panels is approximately 0.2143, and the probability of selecting a sample with both defective panels is also approximately 0.2143.

Step-by-step solution

Calculate the total number of possible combinations

To calculate the total number of possible combinations, we can use the combination formula: C(n, r) = n! / (r!(n-r)!) where C(n, r) is the number of ways to choose r items from a set of n items, n! is the factorial of n (n*(n-1)*(n-2)*...*1), and r! is the factorial of r. In this case, we have 8 machine tools and we want to choose a sample of 4, so: C(8, 4) = 8! / (4!(8-4)!) Calculating the factorials: 8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40320 4! = 4 * 3 * 2 * 1 = 24 Now substitute the factorials back into the combination formula: C(8, 4) = 40320 / (24 * 24) = 70 So, there are a total of 70 possible combinations when selecting 4 tools out of 8.

Calculate the probability of no defective panels

To calculate the probability of no defective panels, we need to find the number of ways to choose 4 tools from the 6 non-defective ones. C(6, 4) = 6! / (4!(6-4)!) Calculating the factorials: 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720 4! = 24 (already calculated in Step 1) Substitute the factorials back into the combination formula: C(6, 4) = 720 / (24 * 2) = 15 Now, to find the probability of no defective panels, divide the number of combinations with no defective panels (15) by the total number of possible combinations (70): Probability(No Defective) = 15 / 70 = 3 / 14 ≈ 0.2143 (rounded to 4 decimal places)

Calculate the probability of both defective panels

To calculate the probability of both defective panels being included, we need to find the number of ways to choose 2 tools from the remaining 6 non-defective tools. C(6, 2) = 6! / (2!(6-2)!) Calculating the factorials: 2! = 2*1 = 2 4! = 24 (already calculated in Step 1) Substitute the factorials back into the combination formula: C(6, 2) = 720 / (2*24) = 15 Since the two defective panels must be included in our sample of 4, the total number of combinations that include both defective panels is 1 * 1 * 15, as each combination will consist of both defective panels and two of the six non-defective panels. The probability of selecting a sample with both defective panels can then be calculated by dividing the number of combinations including both defective panels (15) by the total number of possible combinations (70): Probability(Both Defective) = 15 / 70 = 3 / 14 ≈ 0.2143 (rounded to 4 decimal places) In conclusion, the probability of the sample including no defective panels is approximately 0.2143, and the probability of the sample including both defective panels is also approximately 0.2143.

Key concepts

Combinations

In probability theory, combinations are used to determine the total number of ways a certain number of items can be chosen from a larger set. A combination is distinct from a permutation because the order in which the items are selected does not matter.
To calculate the number of combinations, we use the formula: \[ C(n, r) = \frac{n!}{r! (n-r)!} \]Here, \( n \) represents the total number of items to choose from, and \( r \) is the number of items to select. The combination formula is handy in various scenarios such as lottery draws, team selections, and our current example of selecting machine tools. Applying it to our exercise, we calculate how many ways we can select 4 tools from a set of 8, resulting in a variety of combinations that allow us to understand the probability involved.

Factorial

A factorial, denoted by the exclamation mark \(!\), is a mathematical function that multiplies a series of descending natural numbers down to 1. For example, the factorial of 4, or \(4!\), is equal to \(4 \times 3 \times 2 \times 1 = 24\).
The factorial is crucial in calculating combinations as it facilitates determining how many ways a particular number of items can be arranged or ordered, even when order isn't essential.
This is why in our solution, when calculating \( C(8, 4) \) or \( C(6, 4) \), we use factorials to simplify the process of finding how many potential combinations exist and thus, interpret the probabilities accurately.

Defective Panels

In the context of the probability problem, defective panels represent an undesirable condition—panels that were improperly wired and installed on the machine tools. With two defective panels in a group of eight, identifying these becomes essential because it impacts the functionality of the tools.
When considering probability, knowing how many defective items are present helps make calculated decisions about risks. By determining the likelihood of selecting defective panels, we can assess how samples may or may not expose the faults.
In this problem, understanding the arrangement of defective versus non-defective panels allows us to apply probability theory effectively. This knowledge is crucial in quality assurance processes, where maintaining operational standards relies on identifying and mitigating defective components.

Sample Selection

Sample selection plays a key role in determining probabilities, especially in quality control scenarios like the exercise we are dealing with. Here, we choose a smaller group (sample) from a larger population (all the machine tools) to infer characteristics of the entire set.
Two crucial aspects govern effective sample selection:

• Randomness: Ensures that each item has an equal chance of being selected, reducing bias.
• Size: A larger sample can often provide better reliability, but it must be manageable.

In our example, by randomly selecting 4 out of 8 tools, we get a sample capable of reflecting the likelihood of containing defective panels. This random sampling technique aids in statistically estimating the presence of defects without inspecting every single tool, making the process both efficient and practical.