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Chapter 5: Problem 23

A home security system is designed to have a \(99 \%\) reliability rate. Suppose that nine homes equipped with this system experience an attempted burglary. Find the probabilities of these events: a. At least one of the alarms is triggered. b. More than seven of the alarms are triggered. c. Eight or fewer alarms are triggered.

Short Answer

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Question: In a neighborhood with nine homes, there is a 99% chance that a burglar alarm will be triggered in the event of a break-in. What are the probabilities of the following: a. At least one alarm is triggered. b. More than seven alarms are triggered. c. Eight or fewer alarms are triggered. Answer: a. The probability of at least one alarm being triggered is approximately 0.9999999996. b. The probability of more than seven alarms being triggered is approximately 0.7988. c. The probability of eight or fewer alarms being triggered is approximately 0.0865.

Step by step solution

01

Define parameters

Let n = 9 (number of homes), p = 0.99 (probability of success, i.e., alarm triggered), and q = 1 - p = 0.01 (probability of failure, i.e., alarm not triggered).
02

Calculate probabilities#for_at_least_one_success

To find the probability of at least one alarm being triggered, we calculate the complementary probability of no alarms being triggered: 1 - P(X=0) = 1 - (C(9,0) * p^0 * q^9) = 1 - (1 * 0.99^0 * 0.01^9) 鈮� 1 - 0.0000000003874 鈮� 0.9999999996
03

Calculate probabilities for more_than_seven_successes

To find the probability of more than seven alarms being triggered, we find the sum of probabilities of exactly eight and exactly nine alarms being triggered: P(X>7) = P(X=8) + P(X=9) = C(9,8) * p^8 * q^1 + C(9,9) * p^9 * q^0 鈮� 0.387420487 * 0.99^8 * 0.01 + 1 * 0.99^9 * 0.01^0 鈮� 0.7988
04

Calculate probabilities for eight_or_fewer_successes

To find the probability of eight or fewer alarms being triggered, we calculate the complementary probability of exactly nine alarms being triggered: 1 - P(X=9) = 1 - (C(9,9) * p^9 * q^0) = 1 - (1 * 0.99^9 * 0.01^0) 鈮� 1 - 0.9135 鈮� 0.0865
05

Present the results

a. The probability of at least one alarm being triggered is approximately 0.9999999996. b. The probability of more than seven alarms being triggered is approximately 0.7988. c. The probability of eight or fewer alarms being triggered is approximately 0.0865.

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Key Concepts

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

Reliability in Probability
Reliability in probability refers to the likelihood that a system performs its intended function without failure. In this exercise, the security system has a reliability of 99%. This means that each alarm is 99% likely to trigger when needed. High reliability is crucial in systems like security alarms because failures can lead to significant consequences.

Reliability of a system is often expressed as a percentage:
  • This tells us how dependable the system is.
  • For 9 homes, each alarm has the same probability (99%) of going off.
  • Calculating events such as 'at least one alarm triggered' involves understanding this reliability.
When working with probabilities, remember that the reliability percentage is the starting point for all calculations. It defines the success rate and can be adjusted to understand probabilities of multiple events.
Understanding Complementary Probability
Complementary probability is a fundamental concept especially useful for finding the probability of 'at least one' event happening. The complement of an event is the probability that the event does not occur.

In this exercise:
  • The complementary probability is calculated by evaluating the scenario where no alarms are triggered.
  • The probability that at least one alarm is triggered can be found using: \[1 - P(X=0)\]
  • This computes the complement of no alarm triggering, giving us the probability of at least one event occurring.
Using complementary probability simplifies problems, avoids long calculations and is essential for efficiently determining the occurrence of events within a larger sample space.
Using the Combination Formula
The combination formula is used to determine the number of ways to choose items from a larger set without considering the order. In probability, it's crucial for calculating events where order doesn't matter, like how many alarms are triggered.

The formula is:\[C(n, k) = \frac{n!}{k!(n-k)!}\]
  • Here, \(n\) is the total number of items, and \(k\) is the number of items to choose.
  • It helps find probabilities like exactly eight or nine alarms being triggered in the exercise.
  • Calculations, such as \(C(9, 8)\) or \(C(9, 9)\), determine how many combinations exist for these specific numbers of successes.
Understanding this concept is vital because it enables precise calculation of probabilities where multiple outcomes are possible. It ensures you consider all potential scenarios without bias from sequence or arrangement.

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Most popular questions from this chapter

In southern California, a growing number of persons pursuing a teaching credential are choosing paid internships over traditional student teaching programs. A group of eight candidates for three teaching positions consisted of five paid interns and three traditional student teachers. Let us assume that all eight candidates are equally qualified for the positions. Let \(x\) represent the number of paid interns who are hired for these three positions. a. Does \(x\) have a binomial distribution or a hypergeometric distribution? Support your answer. b. Find the probability that three paid interns are hired for these positions. c. What is the probability that none of the three hired was a paid intern? d. Find \(P(x \leq 1)\).

A new study by Square Trade indicates that smartphones are \(50 \%\) more likely to malfunction than simple phones over a 3-year period. \({ }^{10}\) Of smartphone failures, \(30 \%\) are related to internal components not working, and overall, there is a \(31 \%\) chance of having your smartphone fail over 3 years. Suppose that smartphones are shipped in cartons of \(N=50\) phones. Before shipment \(n=10\) phones are selected from each carton and the carton is shipped if none of the selected phones are defective. If one or more are found to be defective, the whole carton is tested. a. What is the probability distribution of \(x\), the number of defective phones related to internal components not working in the sample of \(n=10\) phones? b. What is the probability that the carton will be shipped if two of the \(N=50\) smartphones in the carton have defective internal components? c. What is the probability that the carton will be shipped if it contains four defectives? Six defectives?

A city commissioner claims that \(80 \%\) of all people in the city favor private garbage collection in contrast to collection by city employees. To check the \(80 \%\) claim, you randomly sample 25 people and find that \(x\), the number of people who support the commissioner's claim, is \(22 .\) a. What is the probability of observing at least 22 who support the commissioner's claim if, in fact, \(p=.8 ?\) b. What is the probability that \(x\) is exactly equal to \(22 ?\) c. Based on the results of part a, what would you conclude about the claim that \(80 \%\) of all people in the city favor private collection? Explain.

A Snapshot in USA Today shows that \(60 \%\) of consumers say they have become more conservative spenders. \({ }^{12}\) When asked "What would you do first if you won \(\$ 1\) million tomorrow?" the answers had to do with somewhat conservative measures like "hire a financial advisor," or "pay off my credit card," or "pay off my mortgage.' Suppose a random sample of \(n=15\) consumers is selected and the number \(x\) of those who say they have become conservative spenders recorded. a. What is the probability that more than six consumers say they have become conservative spenders? b. What is the probability that fewer than five of those sampled have become conservative spenders? c. What is the probability that exactly nine of those sampled are now conservative spenders.

Let \(x\) be a binomial random variable with \(n=10\) and \(p=.4 .\) Find these values: a. \(P(x=4)\) b. \(P(x \geq 4)\) c. \(P(x>4)\) d. \(P(x \leq 4)\) e. \(\mu=n p\) f. \(\sigma=\sqrt{n p q}\)
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