/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 An assembly consists of three me... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An assembly consists of three mechanical components. Suppose that the probabilities that the first, second, and third components meet specifications are \(0.95,0.98,\) and 0.99, respectively. 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
PMF is: P(X=0)=0.00001, P(X=1)=0.00167, P(X=2)=0.07663, P(X=3)=0.92229.

Step by step solution

01

Understanding the Problem

We have three components, with probabilities of meeting specifications as 0.95, 0.98, and 0.99 respectively. They are independent events. We want to find the probability mass function (PMF) for the number of components meeting specifications.
02

Define Random Variable

Let the random variable X represent the number of components meeting specifications. Therefore, X can take values 0, 1, 2, or 3.
03

Calculate P(X=0)

The probability that none of the components meet specifications is calculated by multiplying the probabilities that each component does not meet specifications. \[P(X=0) = (1-0.95) \times (1-0.98) \times (1-0.99)\]\[= 0.05 \times 0.02 \times 0.01 = 0.00001\]
04

Calculate P(X=1)

The probability that exactly one component meets specifications is calculated by considering each component individually meeting specifications while the others do not. Sum these probabilities:\[P(X=1) = (0.95 \times 0.02 \times 0.01) + (0.05 \times 0.98 \times 0.01) + (0.05 \times 0.02 \times 0.99)\]\[= 0.00019 + 0.00049 + 0.00099 = 0.00167\]
05

Calculate P(X=2)

The probability that exactly two components meet specifications is calculated by considering all pair combinations where two meet specifications and the other does not. Sum these probabilities:\[P(X=2) = (0.95 \times 0.98 \times 0.01) + (0.95 \times 0.02 \times 0.99) + (0.05 \times 0.98 \times 0.99)\]\[= 0.00931 + 0.01881 + 0.04851 = 0.07663\]
06

Calculate P(X=3)

The probability that all three components meet specifications is simply the product of their individual probabilities:\[P(X=3) = 0.95 \times 0.98 \times 0.99 = 0.92229\]
07

Verify Total Probability

Ensure the probabilities sum up to 1 to confirm the calculations are correct:\[P(X=0) + P(X=1) + P(X=2) + P(X=3) = 0.00001 + 0.00167 + 0.07663 + 0.92229 = 1.0\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Independent Events
In probability theory, independent events are events where the occurrence of one event does not affect the probability of another event occurring. This makes calculations simpler because you can multiply their probabilities directly. For example, in our exercise, we have three components, each having their own probability of meeting the specifications. They are considered independent.
  • The probability of one component meeting the specification does not change depending on how the other components perform.
  • This assumption enables us to use the multiplication rule to find the joint probabilities of multiple events occurring simultaneously.
The independence assumption simplifies calculations significantly when calculating joint probabilities in the real world, especially in reliability engineering scenarios.
Random Variable
A random variable is a fundamental concept in probability theory. It is a variable that takes on different values based on the outcome of a random phenomenon. In our exercise, we define a random variable, let's call it X, representing the number of components meeting the specifications.
  • Here, X can take on integer values: 0, 1, 2, or 3, where each number corresponds to the number of components fulfilling the specifications.
  • Each of these outcomes has a certain probability associated with it, allowing us to calculate probabilities for different scenarios.
Random variables are crucial for defining and working with probability distributions, whether they be for discrete or continuous events. They help us model real-world phenomena in a quantitative way.
Component Reliability
Component reliability refers to the probability that a component will perform its intended function under specified conditions for a specific period. In the context of the exercise, the reliability of each component is given as the probability that it meets the specifications, namely:
  • The first component has a reliability of 0.95,
  • The second component's reliability is 0.98,
  • And the third component's reliability is 0.99.
Understanding component reliability is essential in fields such as engineering and manufacturing since it directly relates to the performance and safety of larger systems made up of such components. Reliability assessments aid in quality assurance and designing systems that can tolerate failure of individual components while overall functionality remains unimpaired.
Probability Calculation
Probability calculation allows us to assess the likelihood of different potential outcomes. In the exercise, we calculate the probability mass function (PMF) for the random variable X. A PMF gives the probability that a discrete random variable is exactly equal to some value.
Sub-headlines:
**Calculating Probabilities**
The exercise involves calculating the probability for different values of X (0, 1, 2, and 3), representing how many components meet the specifications:
  • For no components meeting specifications, the probability is quite low at 0.00001.
  • For exactly one component meeting specifications, the likelihood slightly increases to 0.00167.
  • Two components reaching the standard lifts the probability significantly to 0.07663.
  • Finally, all components meeting specifications takes up the bulk of the probability at 0.92229.
These calculations use the relationships of independent probabilities to determine the likelihood of each scenario, ensuring the sum of all probabilities equals 1. Understanding this helps in planning and analyzing systems or events where outcomes are not immediately apparent.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let the random variable \(X\) be equally likely to assume any of the values \(1 / 8,1 / 4,\) or \(3 / 8 .\) Determine the mean and variance of \(X\).

A large bakery can produce rolls in lots of either \(0,\) \(1000,2000,\) or 3000 per day. The production cost per item is \(\$ 0.10 .\) The demand varies randomly according to the following distribution: $$ \begin{array}{lcccc} \text { Demand for rolls } & 0 & 1000 & 2000 & 3000 \\ \text { Probability of demand } & 0.3 & 0.2 & 0.3 & 0.2 \end{array} $$ Every roll for which there is a demand is sold for \(\$ 0.30 .\) Every roll for which there is no demand is sold in a secondary market for \(\$ 0.05 .\) How many rolls should the bakery produce each day to maximize the mean profit?

The number of content changes to a Web site follows a Poisson distribution with a mean of 0.25 per day. (a) What is the probability of two or more changes in a day? (b) What is the probability of no content changes in five days? (c) What is the probability of two or fewer changes in five days?

The number of views of a page on a Web site follows a Poisson distribution with a mean of 1.5 per minute. (a) What is the probability of no views in a minute? (b) What is the probability of two or fewer views in 10 minutes? (c) Does the answer to the previous part depend on whether the 10-minute period is an uninterrupted interval? Explain.

A utility company might offer electrical rates based on time-of-day consumption to decrease the peak demand in a day. Enough customers need to accept the plan for it to be successful. Suppose that among 50 major customers, 15 would accept the plan. The utility selects 10 major customers randomly (without replacement) to contact and promote the plan. (a) What is the probability that exactly two of the selected major customers accept the plan? (b) What is the probability that at least one of the selected major customers accepts the plan? (c) Instead of 15 customers, what is the minimum number of major customers that would need to accept the plan to meet the following objective? The probability that at least 1 selected major customer accepts the plan is greater than or equal to 0.95 .

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.