91Ó°ÊÓ

Components coming off an assembly line are either free of defects \((\mathrm{S}\), for success \()\) or defective \((\mathrm{F}\), for failure). Suppose that \(70 \%\) of all such components are defect-free. Components are independently selected and tested one by one. Let \(y\) denote the number of components that must be tested until a defect-free component is obtained. a. What is the smallest possible \(y\) value, and what experimental outcome gives this \(y\) value? What is the second smallest \(y\) value, and what outcome gives rise to it? b. What is the set of all possible \(y\) values? c. Determine the probability of each of the five smallest \(y\) values. You should see a pattern that leads to a simple formula for \(p(y)\), the probability distribution of \(y\).

Step by step solution

01

Determine Smallest \(y\) Values

The smallest possible \(y\) value is 1, meaning that the first component tested is defect-free. This outcome occurs when \(S\) is achieved on the first trial. The second smallest \(y\) value is 2, which happens if the first component tested is defective (i.e., \(F\)) and the second one is defect-free (i.e., \(S\)).

02

Determine All Possible \(y\) Values

All possible \(y\) values range from 1 to infinity, as it could potentially take an infinite number of tests to find a defect-free component – but at least one trial is required. So, the set of all possible \(y\) values is \(y = \{1, 2, 3, ...\}\)

03

Calculate Probabilities of Smallest \(y\) Values

First calculate the probability for \(y = 1\). This just means the first component selected is defect-free, which happens with a 0.7 probability. So \(p(1) = 0.7\). Moving on to \(y = 2\), this means we first select a defective component (0.3 probability), and then select a good component (0.7 probability). These events are independent, hence we multiply the probabilities to get \(p(2) = 0.3 * 0.7 = 0.21\). Let's perform the same calculations for \(y\) values 3 to 5 and summarize all the results: \(p(3) = 0.3 * 0.3 * 0.7 = 0.063\), \(p(4) = 0.3^3 * 0.7 = 0.0189\), and \(p(5) = 0.3^4 * 0.7 = 0.00567\)

04

Identify Pattern and Probability Distribution Function

Looking at the calculation for \(p(y)\), we can observe a pattern, especially if we express each probability as a multiplication of probabilities. The general pattern is: \(p(y) = 0.3^{y - 1} * 0.7\). This is the probability distribution function of a geometric distribution.

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Key Concepts

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

Probability Distribution

A probability distribution describes how the probabilities are allocated to different possible outcomes of a random event. In the context of the geometric distribution, the probability distribution focuses on the number of trials needed until the first success is encountered.
For example, the probability distribution function of a geometric distribution is given by:

• The probability of success on each trial, denoted as \( p \).
• The probability of failure on each trial, \( (1 - p) \).

Where we let \( y \) denote the number of trials required to get the first success, the formula for the probability of \( y \) is \( p(y) = (1 - p)^{y - 1} \cdot p \).
This formula reflects the probability of experiencing a string of \( y-1 \) failures followed by a success, making it a simple yet powerful tool in determining outcomes.

Independent Events

Independent events are those whose occurrence or outcome does not affect the occurrence of other events.
In the context of our geometric distribution, if each component is tested individually and independently, the result of testing one component does not influence the others. This is crucial because:

• Each component has its own probability of being defect-free.
• The probabilities remain constant regardless of previous tests.

This independence allows us to multiply probabilities easily to find combined probabilities of sequential events, such as the occurrence of a defect and then a defect-free component. For instance, \( p(F \text{ then } S) = p(F) \cdot p(S) \).
Understanding independent events helps in identifying how actions or occurrences relate statistically in complex, sequential scenarios.

Defect-Free Components

Defect-free components play a key role in understanding probability in this exercise. When making or assembling products, ensuring that components are free of defects can significantly impact functionality and quality.

• A defect-free component follows the pre-determined probability of success.
• This definition aids in calculating the necessary steps to ensure that a satisfactory component is found.

Since our exercise sets the probability of a defect-free component at \( 0.7 \), one can gauge the efficiency and predictability of finding quality parts. This probability directly influences reliability forecasting and quality assurance practices.
Knowing this aids decisions in processes like batch testing, quality control, and production planning.

Mathematical Modeling

Mathematical modeling helps represent real-world phenomena through equations or expressions. In the context of this exercise, it allows us to predict how many tests are needed for defect-free components using probability and statistics.
Here, we apply the mathematical model of a geometric distribution to:

• Describe the outcome of sequential trials until success.
• Determine the distribution's probability formula \( p(y) = (1 - p)^{y-1} \cdot p \).

This model turns abstract probabilities into a format that can be easily computed and understood. Mathematical modeling in this scenario translates real-world testing into manageable data points, enabling predictive analysis and resource management within production. By doing so, companies can allocate their time and resources more efficiently to maintain or improve quality.