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A concrete beam may fail either by shear \((S)\) or flexure \((F)\). Suppose that three failed beams are randomly selected and the type of failure is determined for each one. Let \(X=\) the number of beams among the three selected that failed by shear. List each outcome in the sample space along with the associated value of \(X\).

Short Answer

Expert verified
The sample space is: \ 1. \( (S, S, S) \): \( X = 3 \) \ 2. \( (S, S, F) \), (S, F, S), and (F, S, S): \( X = 2 \) \ 3. \( (S, F, F) \), (F, S, F), and (F, F, S): \( X = 1 \) \ 4. \( (F, F, F) \): \( X = 0 \).

Step by step solution

01

Understanding the Problem

We are given three concrete beams and each one can fail either by shear \( (S) \) or flexure \( (F) \). We need to determine the number of beams that failed by shear (denoted as \( X \)) for each possible outcome.
02

Identifying Possible Outcomes

Since each beam can independently fail by shear or flexure, we have 2 choices per beam. For 3 beams, this gives us \( 2^3 = 8 \) different outcomes.
03

Listing Sample Space

We list all possible outcomes for the failures of the beams: 1. \( (S, S, S) \)2. \( (S, S, F) \)3. \( (S, F, S) \)4. \( (S, F, F) \)5. \( (F, S, S) \)6. \( (F, S, F) \)7. \( (F, F, S) \)8. \( (F, F, F) \).
04

Determining the Value of X for Each Outcome

For each outcome, count the number of beams that failed by shear, which gives us the value of \( X \): - \( (S, S, S) \): \( X = 3 \)- \( (S, S, F) \): \( X = 2 \)- \( (S, F, S) \): \( X = 2 \)- \( (S, F, F) \): \( X = 1 \)- \( (F, S, S) \): \( X = 2 \)- \( (F, S, F) \): \( X = 1 \)- \( (F, F, S) \): \( X = 1 \)- \( (F, F, F) \): \( X = 0 \).
05

Listing Outcomes with Associated X Values

We can summarize the sample space along with the associated value of \( X \) as follows:1. \( (S, S, S) \): \( X = 3 \)2. \( (S, S, F) \): \( X = 2 \)3. \( (S, F, S) \): \( X = 2 \)4. \( (S, F, F) \): \( X = 1 \)5. \( (F, S, S) \): \( X = 2 \)6. \( (F, S, F) \): \( X = 1 \)7. \( (F, F, S) \): \( X = 1 \)8. \( (F, F, F) \): \( X = 0 \).

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Key Concepts

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

Random Variable
A **Random Variable** is a fundamental concept in probability theory. It's a numerical outcome of a random process. Here, the random variable is denoted as \(X\), which represents the number of concrete beams failed by shear in our experiment.

In simple terms, a random variable like \(X\) takes on different possible values, each associated with a particular outcome of a random experiment. For example, an outcome such as \((S, S, F)\), where two beams fail by shear, corresponds to \(X = 2\).

This concept helps in quantifying variations and expressing probabilities. By analyzing \(X\), we can understand how often a certain number of failures might occur if the process is repeated many times. Understanding random variables is crucial for interpreting data and making sense of unpredictable real-world happenings.
Sample Space
The **Sample Space** is the set of all possible outcomes of a random process. In our exercise, we're observing the outcomes of beam failures, which can either be "S" for shear failure or "F" for flexure failure.

The sample space contains every combination of failure modes over three beams. As each beam can independently fail by shear or flexure, the total number of outcomes is \(2^3 = 8\). These are listed as follows:
  • (S, S, S)
  • (S, S, F)
  • (S, F, S)
  • (S, F, F)
  • (F, S, S)
  • (F, S, F)
  • (F, F, S)
  • (F, F, F)
Each outcome is an element of the sample space, representing a complete account of what could happen when assessing the failure of all three beams.
Failure Modes
In probability contexts like this one, understanding **Failure Modes** is key to interpreting data. Failure modes refer to the ways in which a process or product can fail. Here, we focus on two types of beam failure: shear \((S)\) and flexure \((F)\).

This distinction is crucial because it informs engineers and decision-makers about the most likely and critical areas of failure. By knowing the type of failure mode, better designs and preventive measures can be implemented to avoid similar issues in the future.

Analyzing these failures through the study of different possible outcomes also helps in developing strategies to enhance the strength and reliability of materials, in this case, concrete beams.
Combinatorial Analysis
**Combinatorial Analysis** deals with counting, arranging, and finding patterns in a set of objects. It plays a vital role in probability problems where we need to explore all possible outcomes, as we saw in this beam failure exercise.

Consider the eight possible configurations of beam failures. Combinatorial analysis, in this instance, helps us systematically count and list these configurations, ensuring every possibility is considered. For three beams, the analysis involves the combination of two failure types \((S, F)\) leading to \(2^3 = 8\) outcomes.

This approach is foundational to solving problems in probability theory, enabling a structured assessment of all potential outcomes. It offers insights into the frequency or likelihood of events, which is essential for making informed predictions and decisions.

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Most popular questions from this chapter

Consider a disease whose presence can be identified by carrying out a blood test. Let \(p\) denote the probability that a randomly selected individual has the disease. Suppose \(n\) individuals are independently selected for testing. One way to proceed is to carry out a separate test on each of the \(n\) blood samples. A potentially more economical approach, group testing, was introduced during World War II to identify syphilitic men among army inductees. First, take a part of each blood sample, combine these specimens, and carry out a single test. If no one has the disease, the result will be negative, and only the one test is required. If at least one individual is diseased, the test on the combined sample will yield a positive result, in which case the \(n\) individual tests are then carried out. If \(p=.1\) and \(n=3\), what is the expected number of tests using this procedure? What is the expected number when \(n=5\) ? [The article "Random Multiple-Access Communication and Group Testing" (IEEE Trans. on Commun., 1984: 769-774) applied these ideas to a communication system in which the dichotomy was active/idle user rather than diseased/nondiseased.]

Consider a collection \(A_{1}, \ldots, A_{k}\) of mutually exclusive and exhaustive events, and a random variable \(X\) whose distribution depends on which of the \(A_{i}\) 's occurs (e.g., a commuter might select one of three possible routes from home to work, with \(X\) representing the commute time). Let \(E\left(X \mid A_{i}\right)\) denote the expected value of \(X\) given that the event \(A_{i}\) occurs. Then it can be shown that \(E(X)=\) \(\Sigma E\left(X \mid A_{i}\right) \cdot P\left(A_{i}\right)\), the weighted average of the individual "conditional expectations" where the weights are the probabilities of the partitioning events. a. The expected duration of a voice call to a particular telephone number is 3 minutes, whereas the expected duration of a data call to that same number is 1 minute. If \(75 \%\) of all calls are voice calls, what is the expected duration of the next call? b. A deli sells three different types of chocolate chip cookies. The number of chocolate chips in a type \(i\) cookie has a Poisson distribution with parameter \(\lambda_{i}=i+1\) \((i=1,2,3)\). If \(20 \%\) of all customers purchasing a chocolate chip cookie select the first type, \(50 \%\) choose the second type, and the remaining \(30 \%\) opt for the third type, what is the expected number of chips in a cookie purchased by the next customer?

An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let \(X=\) the number of months between successive payments. The cdf of \(X\) is as follows: $$ F(x)= \begin{cases}0 & x<1 \\ .30 & 1 \leq x<3 \\ .40 & 3 \leq x<4 \\ .45 & 4 \leq x<6 \\ .60 & 6 \leq x<12 \\ 1 & 12 \leq x\end{cases} $$ a. What is the pmf of \(X\) ? b. Using just the cdf, compute \(P(3 \leq X \leq 6)\) and \(P(4 \leq X)\).

Of all customers purchasing automatic garage-door openers, \(75 \%\) purchase a chain-driven model. Let \(X=\) the number among the next 15 purchasers who select the chain-driven model. a. What is the pmf of \(X\) ? b. Compute \(P(X>10)\). c. Compute \(P(6 \leq X \leq 10)\). d. Compute \(\mu\) and \(\sigma^{2}\). e. If the store currently has in stock 10 chaindriven models and 8 shaft- driven models, what is the probability that the requests of these 15 customers can all be met from existing stock?

A family decides to have children until it has three children of the same gender. Assuming \(P(B)=P(G)=.5\), what is the pmf of \(X=\) the number of children in the family?

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