Question

A piece of electronic equipment contains six computer chips, two of which are defective. Three chips are selected at random, removed from the piece of equipment, and inspected. Let \(x\) equal the number of defectives observed, where \(x=0,1,\) or 2 . Find the probability distribution for \(x\) Express the results graphically as a probability histogram.

Short answer

Based on the given problem, the probability distribution for the number of defective chips observed (x=0, 1, or 2) is as follows: ``` x | Probability ---|------------ 0 | 1/5 1 | 3/5 2 | 1/5 ``` The histogram represents this distribution: ``` _______________________ | | | | | | 1/5| | | 3/5 | x=0 | x=1 | x=2 |___________|___________| ```

Step-by-step solution

Calculate the total number of possible outcomes

There are 6 chips in total, and we are selecting 3, the total number of possible outcomes without considering the order can be calculated using combinations as \({6 \choose 3}= \frac{6!}{3!3!} = 20\) outcomes.

Calculate the probability for x=0 (no defective chips)

In this case, all 3 chips selected are good. There are a total of 4 good chips, so we have a combination of \({4 \choose 3}\) possible outcomes this: \({4 \choose 3}= \frac{4!}{3!1!} = 4\) outcomes. Thus, the probability for x=0 is \(\frac{4}{20} = \frac{1}{5}\).

Calculate the probability for x=1 (one defective chip)

In this case, 1 chip selected is defective and the other 2 are good. There are \({2 \choose 1}\) combinations for the defective chip and \({4 \choose 2}\) combinations for the good chips. So, the number of possible outcomes is \({2 \choose 1} {4 \choose 2} = (2) \frac{4!}{2!2!} = 2(6) = 12\) outcomes. Therefore, the probability for x=1 is \(\frac{12}{20} = \frac{3}{5}\).

Calculate the probability for x=2 (two defective chips)

In this scenario, both defective chips are chosen and there is 1 good chip. There are \({2 \choose 2}\) combinations for the defective chips and \({4 \choose 1}\) combinations for the good chip. So, the number of possible outcomes is \({2 \choose 2} {4 \choose 1} = (1)(4) = 4\) outcomes. The probability for x=2 is \(\frac{4}{20} = \frac{1}{5}\).

Represent graphically as a probability histogram

Below is the probability distribution for x and corresponding probability histogram chart for the given data: ``` x | Probability ---|------------ 0 | 1/5 1 | 3/5 2 | 1/5 ``` Histogram: ``` _______________________ | | | | | | 1/5| | | 3/5 | x=0 | x=1 | x=2 |___________|___________| ```

Key concepts

Combinatorics

Combinatorics is a branch of mathematics that focuses on counting, arrangement, and combination possibilities within a finite set. In the context of the original exercise, combinatorics helps us determine how many ways we can select a subset of chips from a larger set. This process is essential to solve probability problems.

To understand how combinatorics works, consider a formula known as the combination formula, denoted as \({n \choose r}\). It calculates the number of ways to choose \(r\) objects from a total of \(n\) objects without considering order. Essentially, it tells us how many ways we can form a group. The formula is given by:

• \[{n \choose r} = \frac{n!}{r!(n-r)!}\]

In our exercise, we applied this formula several times:

• First, to find the total number of ways to pick any 3 chips out of 6, we calculated \({6 \choose 3} = 20\).
• Then, for each case (0, 1, or 2 defective chips), we calculated the combinations separately as necessary, such as \({4 \choose 3}\) for 0 defective chips, and \({2 \choose 2}{4 \choose 1}\) for 2 defective chips, and so on.

Combinatorics simplifies the process of solving probability problems by providing a structured approach to counting possibilities.

Probability Histogram

A probability histogram is a graphical representation of the probability distribution of a discrete random variable. This type of histogram helps in visualizing the distribution, making it easier to understand at a glance how probabilities are assigned to different outcomes.

In our exercise, the process involved calculating the probability distribution for \(x\), the number of defective chips selected. Once we had the probabilities for 0, 1, and 2 defective chips, we represented them graphically in a histogram:

• Each bar's height corresponds to the probability of each outcome. For instance, the probability of selecting 1 defective chip being \(\frac{3}{5}\) resulted in the tallest bar.
• The histogram shows all possibilities together, providing a clear visual of the likelihood of each possible outcome.

Creating such a histogram helps in quickly comparing different probabilities and provides an intuitive sense of the situation. Unlike numerical tables, histograms make it easier to notice trends and make deductions about the probability distributions at a glance.

Random Selection

Random selection is a process where sample items are chosen in such a way that each item has an equal chance of being selected. This concept ensures that results are unbiased and representative of the larger set.

In the context of the original exercise, chips are selected randomly, which means the selection of each chip does not depend on the selection of others. This impacts the probability distribution because all outcomes must consider this fair chance of selection.

• This randomness affects the number of ways chips can be drawn and the resulting probabilities, ensuring no particular chip or outcome is favored over another.
• It emphasizes the calculation of each combination without bias, making events like selecting a defective or a non-defective chip equally probable when using the combinatorics approach.

Random selection is a cornerstone of fair probability assessments, assuring that statistical calculations reflect true variability and uncertainty inherent in real-world scenarios.