Question

Defective Computer Chips A piece of electronic equipment contains six computer chips, two of which are defective. Three computer chips are randomly chosen for inspection, and the number of defective chips is recorded. Find the probability distribution for \(x\), the number of defective computer chips. Compare your results with the answers obtained in Exercise \(4.90 .\)

Short answer

Answer: The probability distribution for the number of defective computer chips, x, is given as follows: P(x) = {0.2 if x = 0, 0.6 if x = 1, 0.2 if x = 2}

Step-by-step solution

Identify the possible outcomes

There can be a total of 3 possible outcomes in this case. They are: - x = 0: No defective chips found (all 3 chips are non-defective) - x = 1: Only one defective chip found (2 chips are non-defective) - x = 2: Two defective chips found (1 chip is non-defective)

Calculate the total number of ways to choose 3 chips

The total number of ways to choose 3 chips from 6 chips, which is called combinations, can be represented as \(\binom{6}{3}\). We can find the combinations using the formula: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) In our case, n = 6 and k = 3. \(\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6\times5\times4}{3\times2\times1} = 20\) There are a total of 20 ways to choose 3 computer chips for inspection.

Calculate the probability for each outcome

For each outcome (x = 0, x = 1, and x = 2), we will calculate the number of successful events and divide them by the total number of events (20). a) When x = 0 (No defective chips), we choose 3 non-defective chips out of the 4 non-defective chips (6-2): \(\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4\times3\times2}{3\times2} = 4\) So, there are 4 successful events for x = 0. To find the probability: \(P(x=0) = \frac{4}{20} = 0.2\) b) When x = 1 (One defective chip), we choose 1 defective chip out of the 2 defective chips and 2 non-defective chips out of the 4 non-defective chips: \(\binom{2}{1} \times \binom{4}{2} = \frac{2!}{1!(2-1)!} \times \frac{4!}{2!(4-2)!} = \frac{2!}{1!1!} \times \frac{4\times3}{2\times1} = 2 \times 6 = 12\) So, there are 12 successful events for x = 1. To find the probability: \(P(x=1) = \frac{12}{20} = 0.6\) c) When x = 2 (Two defective chips), we choose 2 defective chips out of the 2 defective chips and 1 non-defective chip out of the 4 non-defective chips: \(\binom{2}{2} \times \binom{4}{1} = \frac{2!}{2!(2-2)!} \times \frac{4!}{1!(4-1)!} = \frac{2!}{2!0!} \times \frac{4!}{1!3!} = 1 \times 4 = 4\) So, there are 4 successful events for x = 2. To find the probability: \(P(x=2) = \frac{4}{20} = 0.2\)

Write the probability distribution

Now that we have all the probabilities for each outcome, we can write the probability distribution: P(x) = {0.2 if x = 0, 0.6 if x = 1, 0.2 if x = 2}

Key concepts

Combinatorics

Combinatorics is a branch of mathematics dealing with combinations, permutations, and the counting of objects. In the context of probability, combinatorics helps determine how many different ways you can choose or arrange items. This is crucial for calculating probabilities.

A combination is a way of selecting items from a larger pool, where the order does not matter. For example, choosing 3 chips from a selection of 6 involves combinations because the order in which you choose the chips isn't important.

We denote the number of combinations of selecting k items from n items as \(\binom{n}{k}\), which is calculated with the formula:

\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

Here, \(n!\) denotes factorials, which are the product of all positive integers up to n. Using this concept, you can calculate how many different groups of 3 chips can be selected from the total of 6.

Understanding combinatorics is vital in probability distribution because it aids in finding both the total number of possible outcomes and the successful outcomes.

Defective Items

Defective items in a probability problem often refer to objects in a sample space that do not meet certain criteria or are faulty in some way. In our example with computer chips, two out of the six chips are considered defective.

When addressing this kind of probability exercise, the goal is often to determine the probability of selecting a certain number of defective items within a sample.

Identifying the total number of defective items and understanding how they are distributed becomes the basis for creating probability models.

• For instance, calculating how likely it is to select zero, one, or two defective chips from the sample helps build a probability distribution.
• The strategy often involves enumerating all possible ways of selecting defective as well as non-defective items in the group chosen for inspection.

Analyzing defective items provides a practical approach to exploring real-world probabilities and helps in understanding larger concepts in quality control and failure rates.

Probability Calculation

Probability calculation involves determining the likelihood of occurrence for each possible outcome in a defined sample space. It results in a probability distribution, which in this context, indicates how probable it is to select a certain number of defective computer chips.

To compute probability, use the formula:

\[ P(A) = \frac{\text{Number of successful outcomes for the event A}}{\text{Total number of possible outcomes}} \]

This basic probability principle says the probability of an event is the ratio of favorable cases to the total number of cases possible.

For the case of computer chips:

• When \(x = 0\), the probability of picking 0 defective chips is calculated by the ratio of the number of ways to select non-defective chips to the total number of possible selections.
• Similarly, compute the probability for \(x = 1\) and \(x = 2\) using combinations to count successful outcomes of each scenario.

After computing individual event probabilities, you sum them up as part of the overall probability distribution, which tells you the probabilities for different numbers of defective chips being drawn.