/*! 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 27 Assume that the probability that... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 27

Assume that the probability that there is a significant accident in a nuclear power plant during one year's time is .001. If a country has 100 nuclear plants, estimate the probability that there is at least one such accident during a given year.

Short Answer

Expert verified
The probability of at least one accident in a year is approximately 0.0952.

Step by step solution

01

Understanding the Problem

We need to find the probability that at least one nuclear plant will have a significant accident in a year out of 100 nuclear plants. We know the probability of a nuclear accident for one plant in a year is 0.001.
02

Define Operations

We recognize this as a binomial problem where: \( n = 100 \) (number of trials), \( p = 0.001 \) (probability of success 鈥� an accident), and we need \( P(X \geq 1) \).
03

Utilizing Complementary Probability

Since it鈥檚 easier to calculate the probability of no accidents, we use the complement rule: \( P(X \geq 1) = 1 - P(X = 0) \).
04

Calculate the Probability of No Accidents

The probability of no accidents (\( X = 0 \)) in 100 nuclear plants can be found using the binomial probability formula:\[ P(X = 0) = \binom{100}{0} (0.001)^0 (0.999)^{100-0} \]Since \( \binom{100}{0} = 1 \) and \( (0.001)^0 = 1 \), we have:\[ P(X = 0) = (0.999)^{100} \]
05

Computation

We now compute the probability:\( (0.999)^{100} \approx 0.9048 \)So the probability of no accidents in all 100 plants is approximately 0.9048.
06

Final Calculation

Applying the complement rule, we find:\( P(X \geq 1) = 1 - P(X = 0) = 1 - 0.9048 = 0.0952 \).
07

Conclusion

The probability that there is at least one significant accident in one or more of the 100 nuclear plants during a year is approximately 0.0952.

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 91影视!

Key Concepts

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

Binomial Probability
When dealing with scenarios like predicting accidents in nuclear power plants, we're often looking at binomial probability. This statistical method is useful to determine the likelihood of a certain number of "successes" (in this case, accidents) in a fixed number of trials. Here are the basics you need to know:
  • Trials (n): These are the number of events or instances you're looking at. For example, 100 nuclear plants represent 100 trials.
  • Probability of Success (p): This is the chance a specific outcome (accident) happens in a single trial. Here, it鈥檚 given as 0.001, or 0.1% per plant per year.
Each plant operates independently, and the probability of success or failure remains constant across all trials.
Complement Rule
The complement rule helps find probabilities of an event occurring by first finding the probability of it not occurring. This is especially helpful when calculating "at least one" events, as directly calculating probabilities for multiple successes can be complicated. To clarify:
  • If you want to know the probability of at least one accident occurring, first calculate the probability of zero accidents.
  • The complement rule is expressed as: \( P(A) = 1 - P(A^c) \), where \( P(A^c) \) is the probability of the complement event (no accidents).
By subtracting the probability of no accidents from 1, we find the probability of at least one accident. It's both simpler and intuitive.
Probability of No Accidents
To find the probability of having no accidents in the nuclear plants using the binomial probability, focus on the scenario where all plants run without incident. We apply the formula: \[ P(X = 0) = inom{n}{0} (p)^0 (1-p)^{n} \] This formula helps in calculating the chance of zero successes (accidents) over all trials (plants). Given:
  • \( n = 100 \)
  • \( p = 0.001 \)
  • \( 1-p = 0.999 \)
Use these in our formula: \[ (0.999)^{100} \approx 0.9048 \] So, the probability of no accidents in any of the plants is approximately 90.48%.
Nuclear Power Plant Accidents
Considering the severity of nuclear power plant accidents, countries need to assess risks appropriately. Even though each plant may only have a 0.1% chance of an accident annually, when you consider all plants, the probability of at least one incident increases.
  • Risk Analysis: Evaluating the combined risk across many plants can reveal significant risks, necessitating safety protocols and emergency plans.
  • Communication and Education: Understanding the probabilities involved can help inform the public about safety measures and necessary precautions.
The calculated probability of at least one accident (roughly 9.52%) emphasizes the importance of stringent safety measures in nuclear facilities.

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

Cars coming along Magnolia Street come to a fork in the road and have to choose either Willow Street or Main Street to continue. Assume that the number of cars that arrive at the fork in unit time has a Poisson distribution with parameter \(\lambda=4 .\) A car arriving at the fork chooses Main Street with probability \(3 / 4\) and Willow Street with probability \(1 / 4 .\) Let \(X\) be the random variable which counts the number of cars that, in a given unit of time, pass by Joe's Barber Shop on Main Street. What is the distribution of \(X ?\)

A manufactured lot of brass turnbuckles has \(S\) items of which \(D\) are defective. A sample of \(s\) items is drawn without replacement. Let \(X\) be a random variable that gives the number of defective items in the sample. Let \(p(d)=P(X=d)\) (a) Show that $$ p(d)=\frac{\left(\begin{array}{l} D \\ d \end{array}\right)\left(\begin{array}{l} S-D \\ s-d \end{array}\right)}{\left(\begin{array}{l} S \\ s \end{array}\right)} $$ Thus, \(\mathrm{X}\) is hypergeometric. (b) Prove the following identity, known as Euler's formula: $$ \sum_{d=0}^{\min (D, s)}\left(\begin{array}{l} D \\ d \end{array}\right)\left(\begin{array}{l} S-D \\ s-d \end{array}\right)=\left(\begin{array}{l} S \\ s \end{array}\right) $$

Let \(X\) be a random variable normally distributed with parameters \(\mu=70\), \(\sigma=10 .\) Estimate (a) \(P(X>50)\). (b) \(P(X<60)\). (c) \(P(X>90)\) (d) \(P(60

Feller \(^{5}\) discusses the statistics of flying bomb hits in an area in the south of London during the Second World War. The area in question was divided into \(24 \times 24=576\) small areas. The total number of hits was 537 . There were 229 squares with 0 hits, 211 with 1 hit, 93 with 2 hits, 35 with 3 hits, 7 with 4 hits, and 1 with 5 or more. Assuming the hits were purely random, use the Poisson approximation to find the probability that a particular square would have exactly \(k\) hits. Compute the expected number of squares that would have \(0,1,2,3,4,\) and 5 or more hits and compare this with the observed results.

For which of the following random variables would it be appropriate to assign a uniform distribution? (a) Let \(X\) represent the roll of one die. (b) Let \(X\) represent the number of heads obtained in three tosses of a coin. (c) A roulette wheel has 38 possible outcomes: \(0,00,\) and 1 through \(36 .\) Let \(X\) represent the outcome when a roulette wheel is spun. (d) Let \(X\) represent the birthday of a randomly chosen person. (e) Let \(X\) represent the number of tosses of a coin necessary to achieve a head for the first time.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation
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.