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Many manufacturers have quality control programs that include inspection of incoming materials for defects. Suppose a computer manufacturer receives circuit boards in batches of five. Two boards are selected from each batch for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair \((1,2)\) represents the selection of boards 1 and 2 for inspection. a. List the ten different possible outcomes. b. Suppose that boards 1 and 2 are the only defective boards in a batch. Two boards are to be chosen at random. Define \(X\) to be the number of defective boards observed among those inspected. Find the probability distribution of \(X\). c. Let \(F(x)\) denote the cdf of \(X\). First determine \(F(0)=\) \(P(X \leq 0), F(1)\), and \(F(2)\); then obtain \(F(x)\) for all other \(x\)

Short Answer

Expert verified
The probability distribution of \(X\) is \(P(X=0)=\frac{3}{10}, P(X=1)=\frac{6}{10}, P(X=2)=\frac{1}{10}\). The cdf \(F(x)\) is 0 for \(x < 0\), \(\frac{3}{10}\) for \(x=0\), \(\frac{9}{10}\) for \(x=1\), and 1 for \(x \geq 2\).

Step by step solution

01

List All Possible Outcomes

The task is to select any two boards for inspection out of a total of five. We can list all combinations using pairs: \((1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), (4,5)\). These are the ten different possible combinations.
02

Define the Random Variable X

The random variable \(X\) represents the number of defective boards among the two selected. Since boards 1 and 2 are the defective ones, the possible values of \(X\) can be 0, 1, or 2.
03

Calculate Probability Distribution of X

We calculate the probability for each value of \(X\). - \(X = 0\) (no defective boards): Pairs \((3,4), (3,5), (4,5)\). Probability \(= \frac{3}{10}\).- \(X = 1\) (one defective board): Pairs \((1,3), (1,4), (1,5), (2,3), (2,4), (2,5)\). Probability \(= \frac{6}{10}\).- \(X = 2\) (two defective boards): Pair \((1,2)\). Probability \(= \frac{1}{10}\).
04

Determine Cumulative Distribution Function F(x)

The cdf \(F(x)\) represents the cumulative probability \(P(X \leq x)\):- \(F(0) = P(X \leq 0) = \frac{3}{10}\).- \(F(1) = P(X \leq 1) = \frac{9}{10}\).- \(F(2) = P(X \leq 2) = 1\).
05

CDF for Other Values of x

Since the random variable \(X\) represents the number of defective boards and ranges from 0 to 2, the cdf \(F(x)\) will be 0 for \(x < 0\) and 1 for \(x > 2\). Thus, \(F(x) = 0\) for \(x < 0\) and \(F(x) = 1\) for \(x > 2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cumulative Distribution Function
The cumulative distribution function, abbreviated as CDF and denoted as \(F(x)\), is a very important tool in probability and statistics. It gives us a way to describe the probability that a random variable \(X\) is less than or equal to a particular value \(x\).

In the exercise provided, we calculated \(F(0)\), \(F(1)\), and \(F(2)\) for our random variable \(X\), which represents the number of defective circuit boards chosen. These values give us the probabilities of choosing 0, 1, or 2 defective boards respectively.
- \(F(0) = \frac{3}{10}\), which means there is a 30% chance of selecting 0 defective boards.- \(F(1) = \frac{9}{10}\), meaning there is a 90% chance of selecting 1 or fewer defective boards.- \(F(2) = 1\), which means there is a 100% chance of selecting 2 or fewer (since it's not possible to select more than 2 defective boards in this context).

The CDF always starts from 0 and goes up to 1. It is useful because once we have this function, we can easily calculate the probability of \(X\) being in any interval by subtracting \(F(x)\) values corresponding to the limits of the interval. For example, \(P(0 < X \leq 1) = F(1) - F(0) = \frac{9}{10} - \frac{3}{10}\).
Random Variables
A random variable is a mathematical function that maps outcomes of a random process to numbers. In simple terms, it is a way to quantify uncertainty. In our exercise, the random variable \(X\) represents the number of defective boards selected in a random draw of two boards.

This random variable \(X\) can take on values 0, 1, or 2, because there are only two defective boards, and we are selecting two boards in each draw. Describing \(X\) helps us to form a probability distribution, detailing how the probabilities are distributed for each possible outcome.
- \(X = 0\): This outcome means no defective boards were chosen.- \(X = 1\): Exactly one defective board was chosen.- \(X = 2\): Both defective boards were chosen.

A good understanding of random variables is crucial, as they form the foundation for more complex concepts in statistics like probability distributions and various statistical modeling techniques.
Combination and Permutations
Combination and permutations are two fundamental principles in combinatorics, which is the field of mathematics dealing with counting, arrangement, and combination of objects. They help us understand different ways to select or arrange objects or events.

In this context, we use combinations because the order in which the boards are selected does not matter. We simply want to count distinct groups of boards. Mathematically, combinations are represented by \(\binom{n}{k}\), where \(n\) is the total number of items, and \(k\) is the number of items to choose.

For example:
  • When selecting 2 boards from a batch of 5, we calculate the combinations as \(\binom{5}{2} = 10\). This means there are 10 different ways to choose 2 boards out of 5, which we listed as \((1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), (4,5)\).
Permutations, unlike combinations, do consider the order of selection, and while they're not used in this exercise, knowing the difference is important for solving a variety of problems.

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