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An assembly consists of two mechanical components. Suppose that the probabilities that the first and second components meet specifications are 0.95 and \(0.98 .\) Assume that the components are independent. Determine the probability mass function of the number of components in the assembly that meet specifications.

Short Answer

Expert verified
The PMF is: \( P(X=0) = 0.0011 \), \( P(X=1) = 0.068 \), \( P(X=2) = 0.931 \).

Step by step solution

01

Define the Problem

We need to find the probability mass function (PMF) of the number of components out of two that meet the required specifications. Let the random variable \( X \) represent the number of components meeting specifications. \( X \) can take values 0, 1, or 2.
02

Calculate Probabilities

For an individual component, the probability that the first component meets specifications is 0.95, and for the second component, it is 0.98. They are independent events, which allows us to find joint probabilities by multiplying the individual probabilities.
03

Probability for 0 Components Meeting Specifications

Compute the probability that neither component meets specifications. This is given by the product of the probabilities of each component failing: \( P(X=0) = (1-0.95) \times (1-0.98) = 0.05 \times 0.02 = 0.0011 \).
04

Probability for 1 Component Meeting Specifications

There are two scenarios here: either the first component meets specifications and the second does not, or vice versa. Compute each and sum the probabilities:\[ P(X=1) = (0.95 \times 0.02) + (0.05 \times 0.98) = 0.019 + 0.049 = 0.068 \]
05

Probability for 2 Components Meeting Specifications

Compute the probability that both components meet specifications. This is given by:\[ P(X=2) = 0.95 \times 0.98 = 0.931 \]
06

Summarize the PMF

Combine the results from previous steps to form the probability mass function for \( X \):- \( P(X=0) = 0.0011 \)- \( P(X=1) = 0.068 \)- \( P(X=2) = 0.931 \)

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Key Concepts

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

Independent Events
In probability theory, events are considered independent if the occurrence of one event does not affect the occurrence of another event. In the context of the problem, the independence of the two mechanical components means that how the first component performs does not change the probabilities related to the second component.
This concept is crucial because it allows us to calculate the probability of combined events by simply multiplying the probabilities of individual events. In this exercise, since the components are independent, the probability that both meet the specifications is calculated by multiplying their probabilities. Often problems will specify independence, and knowing this tells you that you don't need to worry about interactions or influences between the events.
  • If two events, A and B, are independent, then: \[ P(A \text{ and } B) = P(A) \times P(B) \]
  • This simplifies complex probability problems significantly.
Probability of Failure
The probability of failure is a fundamental aspect of probability calculations, especially when dealing with reliability. It is simply the opposite of the probability of success. For instance, if the probability that a component meets specifications is 0.95, the probability that it fails to meet specifications is 1 minus this probability, or 0.05.
In this exercise, both components have a specified probability of meeting specifications. Knowing the chance of failure helps in calculating scenarios where one or both components may not perform as expected. For events that are independent, these probabilities of failure also remain independent.
This concept is useful because:
  • It highlights the areas where attention is needed for improvement.
  • It is used to calculate joint probabilities of multiple failures.
It significantly impacts reliability assessments where it's critical to know the chances of failures happening independently.
Random Variables
Random variables are a core concept in statistics and probability, representing numerical outcomes of random phenomena. In our scenario, the random variable, denoted as \( X \), symbolizes the number of components in the assembly that meet specifications. Here, \( X \) can take on the values 0, 1, or 2 based on how many components meet the specifications.
Creating a probability mass function (PMF) for a random variable means listing all possible outcomes and their probabilities. PMFs provide a complete picture of the probability distribution for discrete random variables, helping in understanding and predicting outcomes:
  • The PMF assigns probabilities to each possible value \( X \) can assume, ensuring the total probability sums to 1.
  • In this example, the PMF shows us how likely we are to have 0, 1, or 2 components meeting specifications.
The process of defining a random variable and its PMF is instrumental in mapping out the expectations and variabilities in any given scenario.

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Most popular questions from this chapter

Suppose that the number of customers who enter a bank in an hour is a Poisson random variable, and suppose that \(P(X=0)=0.05 .\) Determine the mean and variance of \(X .\)

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This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 100 customer orders to fill. Each order requires one component part that is purchased from a supplier. However, typically, \(2 \%\) of the components are identified as defective, and the components can be assumed to be independent. (a) If the manufacturer stocks 100 components, what is the probability that the 100 orders can be filled without reordering components? (b) If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components? (c) If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

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