/*! 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 99 A \(k\)-out-of-n system is one t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(k\)-out-of-n system is one that will function if and only if at least \(k\) of the \(n\) individual components in the system function. If individual components function independently of one another, each with probability \(.9\), what is the probability that a 3 -out-of-5 system functions?

Short Answer

Expert verified
The probability is 0.99144.

Step by step solution

01

Understanding the Problem

We need to find the probability that at least 3 out of 5 components are functioning, where each functions independently with a probability of 0.9.
02

Define the Random Variable

Let X be a binomial random variable representing the number of functioning components. Here, X follows a binomial distribution with parameters n = 5 (components) and p = 0.9 (success probability for each component).
03

Set Up the Desired Probability

We need to find the probability P(X ≥ 3), which represents the system working when at least 3 components function.
04

Calculate P(X = 3)

The probability of exactly 3 components functioning is given by the binomial probability formula: \[ P(X = 3) = \binom{5}{3} (0.9)^3 (0.1)^2 \] Calculate this value: \[ P(X = 3) = 10 \times 0.729 \times 0.01 = 0.0729 \]
05

Calculate P(X = 4)

The probability of exactly 4 components functioning is: \[ P(X = 4) = \binom{5}{4} (0.9)^4 (0.1)^1 \] Calculate this value: \[ P(X = 4) = 5 \times 0.6561 \times 0.1 = 0.32805 \]
06

Calculate P(X = 5)

The probability of all 5 components functioning is: \[ P(X = 5) = \binom{5}{5} (0.9)^5 (0.1)^0 \] Calculate this value:\[ P(X = 5) = 1 \times 0.59049 \times 1 = 0.59049 \]
07

Combine Probabilities for At Least 3 Components

Sum the probabilities calculated for X = 3, 4, and 5 to find P(X ≥ 3): \[ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) \] Calculate this sum: \[ P(X \geq 3) = 0.0729 + 0.32805 + 0.59049 = 0.99144 \]
08

Conclusion

The probability that the system functions is 0.99144.

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.

k-out-of-n system
A "k-out-of-n system" is a concept often encountered in reliability engineering and probability theory. In this system, we have a total of \( n \) components. For the system to function properly, at least \( k \) of these components must be working. This setup is useful to model real-world scenarios where redundancy is built into systems to improve their reliability.
For example, if we're dealing with a 3-out-of-5 system, then even if two components fail, the system can still function because the other three are operational. This principle underscores the importance of designing systems that can withstand failures and still perform as expected, thereby increasing the robustness of the system.
Knowing how to calculate the probability of such a system functioning is crucial, and this is where probability concepts like the binomial distribution come into play.
binomial distribution
The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent Bernoulli trials. Each trial results in a "success" or "failure" and has the same probability of success, denoted by \( p \). The number of trials is represented by \( n \).
In the context of our problem involving the 3-out-of-5 system, the binomial distribution helps us calculate the probability of having at least 3 successes (functioning components) out of 5, with each component having a 0.9 probability of success.
This distribution is vital in scenarios where there are a fixed number of trials, and each trial is independent. The key parameters of a binomial distribution are the number of trials \( n \) and the probability of success \( p \). This makes it particularly useful in systems like our \( k \)-out-of-\( n \) system discussed earlier.
independent events
Independent events are events where the outcome of one event does not affect the outcome of another. In probability, the concept of independent events is fundamental because it allows for simplification when calculating probabilities.
For example, in the problem provided, each component in the system functions independently, meaning the functioning or failure of one component does not influence the functioning of another. This independence assumption allows us to use the binomial distribution, as each trial is independent of the others.
When dealing with binomial probabilities, the fact that events are independent means you multiply the probabilities of individual successes or failures together. Understanding this independence simplifies calculations and provides insight into the reliability and behavior of systems.
binomial probability formula
The binomial probability formula is used to find the probability of achieving exactly \( k \) successes in \( n \) independent trials, with each success having a probability \( p \). The formula is:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
In this formula, \( \binom{n}{k} \) is a binomial coefficient, calculating the number of ways \( k \) successes can occur. \( p^k \) is the probability of having \( k \) successes, and \( (1-p)^{n-k} \) accounts for the failures within the \( n \) trials.
To solve the exercise, we use this formula to find the probabilities of exactly 3, 4, and 5 successes. By calculating each and summing them, we determine the probability of the 3-out-of-5 system functioning. Understanding how to apply this formula is critical in analyzing and predicting the behavior of systems under similar conditions.

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

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter \(\lambda=20\) (suggested in the article "Dynamic Ride Sharing: Theory and Practice," J. of Transp. Engr., 1997: 308-312). What is the probability that the number of drivers will a. Be at most 10 ? b. Exceed 20 ? c. Be between 10 and 20 , inclusive? Be strictly between 10 and 20 ? d. Be within 2 standard deviations of the mean value?

Show that the cdf \(F(x)\) is a nondecreasing function; that is, \(x_{1}

The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a rate parameter of five per hour. a. What is the probability that exactly four arrivals occur during a particular hour? b. What is the probability that at least four people arrive during a particular hour? c. How many people do you expect to arrive during a 45 min period?

A library subscribes to two different weekly news magazines, each of which is supposed to arrive in Wednesday's mail. In actuality, each one may arrive on Wednesday, Thursday, Friday, or Saturday. Suppose the two arrive independently of one another, and for each one \(P(\) Wed. \()=.3\), \(P(\) Thurs. \()=.4, P(\) Fri. \()=.2\), and \(P(\) Sat. \()=.1\). Let \(Y=\) the number of days beyond Wednesday that it takes for both magazines to arrive (so possible \(Y\) values are \(0,1,2\), or 3 ). Compute the pmf of \(Y\). [Hint: There are 16 possible outcomes; \(Y(W, W)=0, Y(F, T h)=2\), and so on.] 20\. Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the probability that any particular couple or individual arrives late is .4 (a couple will travel together in the same vehicle, so either both people will be on time or else both will arrive late). Assume that different couples and individuals are on time or late independently of one another. Let \(X=\) the number of people who arrive late for the seminar. a. Determine the probability mass function of \(X\). b. Obtain the cumulative distribution function of \(X\), and use it to calculate \(P(2 \leq X \leq 6)\).

The pmf for \(X=\) the number of major defects on a randomly selected appliance of a certain type is \begin{tabular}{l|ccccc} \(x\) & 0 & 1 & 2 & 3 & 4 \\ \hline\(p(x)\) & \(.08\) & \(.15\) & \(.45\) & \(.27\) & \(.05\) \end{tabular} Compute the following: a. \(E(X)\) b. \(V(X)\) directly from the definition c. The standard deviation of \(X\) d. \(V(X)\) using the shortcut formula

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.