/*! 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 54 If on average one in 20 of a cer... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If on average one in 20 of a certain type of column will fail under a given axial load, what are the probabilities that among 16 such columns, (a) at most two, (b) at least four will fail?

Short Answer

Expert verified
(a) P(X ≤ 2) ≈ 0.924, (b) P(X ≥ 4) ≈ 0.032.

Step by step solution

01

Understanding the Problem

We have a probability problem dealing with columns failing under load. Given that the probability of one column failing is \( p = \frac{1}{20} \), we need to calculate the probability for a specified number of columns (out of 16) failing.
02

Defining a Probability Model

This scenario fits a binomial distribution where the number of trials \( n = 16 \) (the total columns), and the probability of a success (a column failing) is \( p = \frac{1}{20} = 0.05 \).
03

Calculating Probability of At Most Two Failures

The probability of "at most two" failures, \( P(X \leq 2) \), is found by summing the probabilities of 0, 1, and 2 failures. Use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] for \( k = 0, 1, 2 \).
04

Formula Details for At Most Two Failures

1. \( P(X = 0) = \binom{16}{0}(0.05)^0(0.95)^{16} \)2. \( P(X = 1) = \binom{16}{1}(0.05)^1(0.95)^{15} \)3. \( P(X = 2) = \binom{16}{2}(0.05)^2(0.95)^{14} \)
05

Computing the Summation for At Most Two Failures

Add the relevant computed probabilities: \[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]
06

Calculating Probability of At Least Four Failures

The probability of "at least four" failures can be calculated as \( P(X \geq 4) = 1 - P(X \leq 3) \). Compute \( P(X \leq 3) \) by adding probabilities for 0 to 3 failures.
07

Formula Details for At Most Three Failures

Add \( P(X = 3) \) to the previous probabilities: \( P(X = 3) = \binom{16}{3}(0.05)^3(0.95)^{13} \)Calculate the total: \[ P(X \leq 3) = P(X =0) + P(X = 1) + P(X = 2) + P(X = 3) \]
08

Final Result for At Least Four Failures

Subtract the result from the total probability: \[ P(X \geq 4) = 1 - P(X \leq 3) \]

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.

Probability Calculation
Calculating probabilities is a fundamental part of understanding statistical behavior in real-world situations. To solve the given problem, we need to determine the probability that a certain number of columns will fail under a load. This probability calculation involves two specific scenarios – calculating the chance of at most two failures and at least four failures.

In our exercise, we have 16 columns and need to understand how likely it is for failures to occur. For at most two failures, we will look at how many ways you can have 0, 1, or 2 failures and then sum these possibilities. Probabilities add up to give you a comprehensive picture of the situation.

Each calculation can be independently straightforward, but they require careful handling and summation of each probability to reach the correct conclusion. Summation offers the total chance of these specific failure scenarios, simplifying the understanding of outcomes.
Binomial Probability Formula
The binomial probability formula is essential for solving problems where the outcome can be classified into two categories, like failure and no failure. Given that each column failing is an independent event, and only two outcomes are possible (fail or not fail), the binomial distribution is suitable here.

The formula is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where:
  • \( \binom{n}{k} \) is the number of ways to choose \( k \) successes from \( n \) trials
  • \( p \) is the probability of a single success
  • \((1-p)\) is the probability of failure
  • \( n \) is the number of trials
  • \( k \) is the number of successes
This formula enables us to calculate the likelihood of exactly \( k \) failures out of \( n = 16 \) columns. Armed with these probabilities, further calculations allow us to describe at most or at least failure scenarios.
Probability of Failure
The probability of failure in our context is essentially the chance that a column fails under the specified load. For this exercise, the probability of a column failing is \( p = \frac{1}{20} = 0.05 \).

Understanding this probability helps us predict behavior under the load. This is known as the probability of success in a binomial distribution context, but here, a 'success' is actually a failure of the column.

The counterpart to this is the probability of not failing, which is \( 1-p = 0.95 \).
  • These probabilities are used directly in the binomial probability formula to calculate the chance of multiple columns failing.
  • They help frame the problem, determining the expected behavior across different numbers of trials.
Probability of failure directly interacts with scenario probability calculations, ensuring the results reflect realistic expectations.
Statistical Analysis
Statistical analysis provides insight into how data behaves under certain conditions. With the binomial distribution, we're equipped to statistically analyze the likelihood of various outcomes for the columns under stress.

The analysis lets us establish overall probabilities for different failure levels, such as at most two or at least four failures. This requires:
  • Summation of individual probabilities, using the probabilities derived from the binomial probability formula.
  • Subtraction methods to calculate complementary probabilities, like the chance of at least four failures.
Statistical analysis simplifies complex data. It turns individual probability results into a coherent narrative about what is likely to happen to these columns. It also helps in decision-making by showing probable ranges of failure rates, thus guiding future actions or designs.

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

Part of an electric circuit consists of three elements \(K, L\) and \(M\) in series. Probabilities of failure for elements \(K\) and \(M\) during operating time \(t\) are \(0.1\) and \(0.2\) respectively. Element \(L\) itself consists of three sub-elements \(L_{1}, L_{2}\) and \(L_{3}\) in parallel, with failure probabilities \(0.4,0.7\) and \(0.5\) respectively, during the same operating time \(t\). Find the probability of failure of the circuit during time \(t\), assuming that all failures of elements are independent.

If \(A\) and \(B\) are mutually exclusive events and \(P(A)=0.2\) and \(P(B)=0.5\), find (a) \(P(A \cup B)\) (b) \(P(\bar{A})\) (c) \(P(\bar{A} \cap B)\)

Suppose that the actual amount of cement that a filling machine puts into 'six-kilogram' bags is a normal random variable with \(\sigma=0.05 \mathrm{~kg}\). If only \(3 \%\) of bags are to contain less than \(6 \mathrm{~kg}\), what must be the mean fill of the bags?

Assume that in the composition of a book there exists a constant probability \(0.0001\) that an arbitrary letter will be set incorrectly. After composition, the proofs are read by a proofreader, who discovers \(90 \%\) of the errors. After the proofreader, the author discovers half of the remaining errors. Find the probability that in a book with 500000 printing symbols there remain after this no more than six unnoticed errors.

Eight babies are born in a hospital on a particular day. Find the probability that exactly half of them are boys. (The probability that a baby is a boy is actually slightly greater than one-half, but you can take it as exactly one- half for this exercise.)

See all solutions

Recommended explanations on Physics 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.